Đáp án: \(\cos\left(a+b\right)=\cos a\cos b-\sin a\sin b\)

Lời giải:

Công thức đúng là

\(\cos\left(a+b\right)=\cos a\cos b-\sin a\sin b\).

Đáp án: \(\sin(a+b)=\sin a\cos b+\cos a\sin b\)

Lời giải:

Theo công thức cộng cung thì \(\sin(a+b)=\sin a\cos b+\cos a\sin b\).

Đáp án: \(\sin 2a=2\sin a\cos a\)

Lời giải:

Ta có \(\sin 2a=2\sin a\cos a\).

Đáp án: \(\cos 2a=\cos^2 a+\sin^2 a\)

Lời giải:

Ta có \(\cos 2a=\cos^2 a-\sin^2 a = 2\cos^2 a-1 = 1-2\sin^2 a\).

Đáp án: \(P=2\sin 3x\cos 2x\)

Lời giải:

Ta có

\(P=\sin x+\sin 5x\) \(=2\sin\left( \displaystyle\frac{x+5x}{2}\right)\cos\left( \displaystyle\frac{x-5x}{2}\right)\) \(=2\sin 3x \cos (-2x)=2\sin 3x\cos 2x \).

Đáp án: \(\sin a+\sin b=2 \sin \displaystyle\frac{a+b}{2} \cos \displaystyle\frac{a-b}{2}\)

Lời giải:

Công thức đúng là

\(\sin a+\sin b=2 \sin \displaystyle\frac{a+b}{2} \cos \displaystyle\frac{a-b}{2}\).

Đáp án: \(\displaystyle\frac{\sqrt{3}}{2} \)

Lời giải:

Ta có

\(\cos 18^\circ\cdot \cos 12^\circ-\sin 18^\circ\cdot \sin 12^\circ\) \(=\cos(18^\circ+12^\circ)=\cos 30^\circ\) \(=\displaystyle\frac{\sqrt{3}}{2}.\)

Đáp án: \(\displaystyle\frac{\sqrt{3}}{2}\sin a+\displaystyle\frac{1}{2}\cos a\)

Lời giải:

Ta có

\(\sin\left(a+\displaystyle\frac{\pi}{6}\right)=\sin a\cdot \cos\displaystyle\frac{\pi}{6}+\cos a\cdot \sin\displaystyle\frac{\pi}{6}\) \(=\displaystyle\frac{\sqrt{3}}{2}\sin a+\displaystyle\frac{1}{2}\cos a.\)

Đáp án: \(\displaystyle\frac{7}{9}\)

Lời giải:

Ta có

\(\cos 2\alpha=1-2\sin^2 \alpha=1-2\left(\displaystyle\frac{1}{3}\right)^2=\displaystyle\frac{7}{9}\).

Đáp án: \(-\displaystyle\frac{7}{9}\)

Lời giải:

Ta có

\(\cos 2a=2\cos^2a-1=2\left(\displaystyle\frac{1}{3}\right)^2-1\) \(=-\displaystyle\frac{7}{9}\).

Đáp án: \(P=\cot 4 \alpha\)

Lời giải:

\(P=\displaystyle\frac{\cos 2 \alpha+\cos 4 \alpha+\cos 6 \alpha}{\sin 2 \alpha+\sin 4 \alpha+\sin 6 \alpha}\)

\(=\displaystyle\frac{\cos 2 \alpha+\cos 6 \alpha+\cos 4 \alpha}{\sin 2 \alpha+\sin 6 \alpha+\sin 4 \alpha}\)

\(=\displaystyle\frac{2 \cdot \cos 4 \alpha \cdot \cos 2 \alpha+\cos 4 \alpha}{2 \cdot \sin 4 \alpha \cdot \cos 2 \alpha + \sin 4 \alpha}\)

\(=\displaystyle\frac{\cos 4 \alpha \cdot \left( 2 \cdot \cos 2 \alpha +1\right)}{\sin 4 \alpha \cdot \left( 2 \cdot \cos 2 \alpha +1\right)}\)

\(=\cot 4 \alpha.\)

Đáp án: \(A=\displaystyle\frac{1}{2}\cos 2x\)

Lời giải:

\(A=\cos\left(x+45^\circ\right)\cos\left(x-45^\circ\right)\)

\(=\displaystyle\frac{1}{2}\left[ \cos(x+45^\circ+x-45^\circ)+\cos(x+45^\circ-x+45^\circ)\right]\)

\(=\displaystyle\frac{1}{2}\left( \cos 2x+\cos 90^\circ\right)\)

\(=\displaystyle\frac{1}{2}\cos 2x.\)

Đáp án: \(P=\tan \displaystyle\frac{x}{2}\)

Lời giải:

Ta có

\(P=\displaystyle\frac{2\sin \displaystyle\frac{x}{2}\cos \displaystyle\frac{x}{2}+\sin \displaystyle\frac{x}{2}}{2\cos^2 \displaystyle\frac{x}{2}+\cos \displaystyle\frac{x}{2}}\) \(=\displaystyle\frac{\sin \displaystyle\frac{x}{2}\left(2\cos \displaystyle\frac{x}{2}+1 \right)}{\cos \displaystyle\frac{x}{2}\left(2\cos \displaystyle\frac{x}{2}+1 \right)}\) \(=\tan \displaystyle\frac{x}{2}.\)

Đáp án: \(1\)

Lời giải:

Ta có

\(\sin^4 x+\cos^4 x=\left(\sin^2 x+\cos^2x\right)^2-2\sin^2 x\cos^2 x\) \(=1-\displaystyle\frac{1}{2}\sin^2 2x.\)

Suy ra \(a=1\), \(b=2\). Vậy \(3a-b=1\).

Đáp án:

• \(\cos2\alpha=\displaystyle\frac{7}{25}\)
• \(\tan2\alpha=-\displaystyle\frac{24}{7}\)
• \(\cot2\alpha=-\displaystyle\frac{7}{24}\)

Lời giải:

a) \(\sin2\alpha=2\sin\alpha\cos\alpha=2\cdot\displaystyle\frac{3}{5}\cdot\left(-\displaystyle\frac{4}{5}\right)=-\displaystyle\frac{24}{25}\).

b) \(\cos2\alpha=1-2\sin^2\alpha=1-2\left(\displaystyle\frac{3}{5}\right)^2=\displaystyle\frac{7}{25}\).

c) \(\tan2\alpha=\displaystyle\frac{\sin2\alpha}{\cos2\alpha}=-\displaystyle\frac{24}{25}:\displaystyle\frac{7}{25}=-\displaystyle\frac{24}{7}.\)

d) \(\cot2\alpha=\displaystyle\frac{1}{\tan2\alpha}=-\displaystyle\frac{7}{24}.\)

Đáp án:

• \(\cos \displaystyle\frac{\alpha}{2}\approx 0{,}986\) (làm tròn tới chữ số thập phân thứ ba)}

Lời giải:

Ta có \(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow \cos^2\alpha=\displaystyle\frac{8}{9}\Rightarrow \cos\alpha=\displaystyle\frac{2\sqrt{2}}{3}\) (vì \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(\cos\alpha>0\)).

a) Ta có \(\tan\alpha=\displaystyle\frac{\sin \alpha}{\cos\alpha}=\displaystyle\frac{\sqrt{2}}{4}\).

b) Ta có \(\sin\alpha=2\sin\alpha\cos\alpha=2\cdot \displaystyle\frac{1}{3}\cdot \displaystyle\frac{2\sqrt{2}}{3}=\displaystyle\frac{4\sqrt{2}}{9}\).

c) Ta có \(\cos 2 \alpha=1-2\sin^2\alpha=\displaystyle\frac{7}{9}\neq \displaystyle\frac{1}{3}=\sin\alpha\).

d) Ta có \(\cos^2\displaystyle\frac{\alpha}{2}=\displaystyle\frac{1+\cos\alpha}{2}=\displaystyle\frac{1+\displaystyle\frac{2\sqrt{2}}{3}}{2}=\displaystyle\frac{3+2\sqrt{2}}{6}\). \\

Vậy \(\cos \displaystyle\frac{\alpha}{2}=\sqrt{\displaystyle\frac{3+2\sqrt{2}}{6}}\approx 0{,}986\) (vì \(0< \displaystyle\frac{\alpha}{2}< \displaystyle\frac{\pi}{4}\) nên \(\cos\displaystyle\frac{\alpha}{2}>0\)).

Đáp án:

• \(\sin \alpha=-\displaystyle\frac{3}{5}\)
• \(\cos 2\alpha=\displaystyle\frac{7}{25}\)
• \(\tan 4\alpha=\displaystyle\frac{a}{b}\) với \(\displaystyle\frac{a}{b}\) là phân số tối giản, khi đó \(b-a=7\)

Lời giải:

Ta có \(1+\tan^2\alpha=\displaystyle\frac{1}{\cos^2\alpha}\Leftrightarrow \cos^2\alpha=\displaystyle\frac{16}{25}\Rightarrow\cos\alpha=-\displaystyle\frac{4}{5}\) (vì \(\pi< \alpha< \displaystyle\frac{3\pi}{2}\) nên \(\cos\alpha< 0\)).

a) \(\sin\alpha=\tan\alpha\cdot \cos\alpha=-\displaystyle\frac{3}{5}\).

b) \(\cos 2\alpha=2\cos^2\alpha-1=\displaystyle\frac{16}{25}-1=-\displaystyle\frac{9}{25}\).

c) \(\sin 2\alpha=2\sin\alpha\cos \alpha=2\cdot \left(-\displaystyle\frac{3}{5}\right)\cdot \left(-\displaystyle\frac{4}{5}\right)=\displaystyle\frac{24}{25}\).

d) Ta có \(\tan 2\alpha=\displaystyle\frac{\sin 2\alpha}{\cos 2 \alpha}=-\displaystyle\frac{8}{3}\). \\

Vậy \(\tan 4\alpha=\displaystyle\frac{2\tan 2\alpha}{1-\tan^2 2\alpha}=\displaystyle\frac{-\displaystyle\frac{16}{3}}{1-\left(-\displaystyle\frac{8}{3}\right)^2}=\displaystyle\frac{48}{55}\).

Đáp án:

• \(\cos x=\displaystyle\frac{1}{4}\)

Lời giải:

a) Ta có \(-\displaystyle\frac{\pi}{2}< x< 0\Leftrightarrow -\pi< 2x< 0\). Vậy \(\sin 2x< 0\). \\

Mặt khác \(\cos^2 2x+\sin^2 2x=1\Leftrightarrow \sin^2 2x=\displaystyle\frac{15}{64}\Rightarrow \sin 2x=-\displaystyle\frac{\sqrt{15}}{8}\).

b) Ta có \(-\displaystyle\frac{\pi}{2}< x< 0\) nên \(\cos x>0\) và \(\sin x< 0\). \\

Sử dụng công thức hạ bậc ta có

\(\)\cos^2 x=\displaystyle\frac{1+\cos 2x}{2}\Leftrightarrow \cos^2 x=\displaystyle\frac{1}{16}\Rightarrow \cos x=\displaystyle\frac{1}{4}.\(\)

c) Ta có \(\cos 4x=2\cos^2 2x-1=\displaystyle\frac{17}{32}\).

d) Ta có \(\cos^2 x+\sin^2 x=1\Leftrightarrow \sin x=\displaystyle\frac{15}{16}\Rightarrow \sin x=-\displaystyle\frac{\sqrt{15}}{4}\). \\

Vậy \(\tan x=\displaystyle\frac{\sin x}{\cos x}=-\sqrt{15}\).

Đáp án:

• \(\sin (\pi -\alpha)=\displaystyle\frac{\sqrt{7}}{4}\)
• \(\sin 2\alpha=\displaystyle\frac{3\sqrt{7}}{8}\)

Lời giải:

Ta có \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(\sin \alpha>0\). \\

Mặt khác, \(\cos^2\alpha+\sin^2\alpha=1\Leftrightarrow \sin^2\alpha=\displaystyle\frac{7}{16}\Rightarrow\sin\alpha=\displaystyle\frac{\sqrt{7}}{4}\).

a) Ta có \(\sin (\pi-\alpha)=\sin \alpha=\displaystyle\frac{\sqrt{7}}{4}\).

b) Ta có \(\sin 2\alpha=2\sin\alpha\cos\alpha=2\cdot \displaystyle\frac{\sqrt{7}}{4}\cdot\displaystyle\frac{3}{4}=\displaystyle\frac{3\sqrt{7}}{8}\).

c) Ta có \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(0< \displaystyle\frac{\alpha}{2}< \displaystyle\frac{\pi}{4}\). Vậy \(\sin \displaystyle\frac{\alpha}{2}>0\). \\

Áp dụng công thức hạ bậc ta có

\(\sin^2\displaystyle\frac{\alpha}{2}=\displaystyle\frac{1-\cos \alpha}{2}=\displaystyle\frac{1}{8}\Rightarrow \sin \displaystyle\frac{\alpha}{2}=\displaystyle\frac{\sqrt{2}}{4}.\)

d) Ta có \(\tan \alpha=\displaystyle\frac{\sin \alpha}{\cos \alpha}=\displaystyle\frac{\sqrt{7}}{3}\).

Vậy \(\tan 2\alpha=\displaystyle\frac{2\tan \alpha}{1-\tan^2\alpha}=3\sqrt{7}\).

Đáp án:

• \(\cos2\alpha=\displaystyle\frac{7}{25}\)
• \(\tan2\alpha=-\displaystyle\frac{24}{7}\)
• \(\cot2\alpha=-\displaystyle\frac{7}{24}\)

Lời giải:

a) \(\sin2\alpha=2\sin\alpha\cos\alpha=2\cdot\displaystyle\frac{3}{5}\cdot\left(-\displaystyle\frac{4}{5}\right)=-\displaystyle\frac{24}{25}\).

b) \(\cos2\alpha=1-2\sin^2\alpha=1-2\left(\displaystyle\frac{3}{5}\right)^2=\displaystyle\frac{7}{25}\).

c) \(\tan2\alpha=\displaystyle\frac{\sin2\alpha}{\cos2\alpha}=-\displaystyle\frac{24}{25}:\displaystyle\frac{7}{25}=-\displaystyle\frac{24}{7}.\)

d) \(\cot2\alpha=\displaystyle\frac{1}{\tan2\alpha}=-\displaystyle\frac{7}{24}.\)

Đáp án:

• \(\cos \displaystyle\frac{\alpha}{2}\approx 0{,}986\) (làm tròn tới chữ số thập phân thứ ba)}

Lời giải:

Ta có \(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow \cos^2\alpha=\displaystyle\frac{8}{9}\Rightarrow \cos\alpha=\displaystyle\frac{2\sqrt{2}}{3}\) (vì \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(\cos\alpha>0\)).

a) Ta có \(\tan\alpha=\displaystyle\frac{\sin \alpha}{\cos\alpha}=\displaystyle\frac{\sqrt{2}}{4}\).

b) Ta có \(\sin\alpha=2\sin\alpha\cos\alpha=2\cdot \displaystyle\frac{1}{3}\cdot \displaystyle\frac{2\sqrt{2}}{3}=\displaystyle\frac{4\sqrt{2}}{9}\).

c) Ta có \(\cos 2 \alpha=1-2\sin^2\alpha=\displaystyle\frac{7}{9}\neq \displaystyle\frac{1}{3}=\sin\alpha\).

d) Ta có \(\cos^2\displaystyle\frac{\alpha}{2}=\displaystyle\frac{1+\cos\alpha}{2}=\displaystyle\frac{1+\displaystyle\frac{2\sqrt{2}}{3}}{2}=\displaystyle\frac{3+2\sqrt{2}}{6}\).

Vậy \(\cos \displaystyle\frac{\alpha}{2}=\sqrt{\displaystyle\frac{3+2\sqrt{2}}{6}}\approx 0{,}986\) (vì \(0< \displaystyle\frac{\alpha}{2}< \displaystyle\frac{\pi}{4}\) nên \(\cos\displaystyle\frac{\alpha}{2}>0\)).

Đáp án:

• \(\sin \alpha=-\displaystyle\frac{3}{5}\)
• \(\cos 2\alpha=\displaystyle\frac{7}{25}\)
• \(\tan 4\alpha=\displaystyle\frac{a}{b}\) với \(\displaystyle\frac{a}{b}\) là phân số tối giản, khi đó \(b-a=7\)

Lời giải:

Ta có \(1+\tan^2\alpha=\displaystyle\frac{1}{\cos^2\alpha}\Leftrightarrow \cos^2\alpha=\displaystyle\frac{16}{25}\Rightarrow\cos\alpha=-\displaystyle\frac{4}{5}\) (vì \(\pi< \alpha< \displaystyle\frac{3\pi}{2}\) nên \(\cos\alpha< 0\)).

a) \(\sin\alpha=\tan\alpha\cdot \cos\alpha=-\displaystyle\frac{3}{5}\).

b) \(\cos 2\alpha=2\cos^2\alpha-1=\displaystyle\frac{16}{25}-1=-\displaystyle\frac{9}{25}\).

c) \(\sin 2\alpha=2\sin\alpha\cos \alpha=2\cdot \left(-\displaystyle\frac{3}{5}\right)\cdot \left(-\displaystyle\frac{4}{5}\right)=\displaystyle\frac{24}{25}\).

d) Ta có \(\tan 2\alpha=\displaystyle\frac{\sin 2\alpha}{\cos 2 \alpha}=-\displaystyle\frac{8}{3}\).

Vậy \(\tan 4\alpha=\displaystyle\frac{2\tan 2\alpha}{1-\tan^2 2\alpha}=\displaystyle\frac{-\displaystyle\frac{16}{3}}{1-\left(-\displaystyle\frac{8}{3}\right)^2}=\displaystyle\frac{48}{55}\).

Đáp án:

• \(\cos x=\displaystyle\frac{1}{4}\)

Lời giải:

a) Ta có \(-\displaystyle\frac{\pi}{2}< x< 0\Leftrightarrow -\pi< 2x< 0\). Vậy \(\sin 2x< 0\).

Mặt khác \(\cos^2 2x+\sin^2 2x=1\Leftrightarrow \sin^2 2x=\displaystyle\frac{15}{64}\Rightarrow \sin 2x=-\displaystyle\frac{\sqrt{15}}{8}\).

b) Ta có \(-\displaystyle\frac{\pi}{2}< x< 0\) nên \(\cos x>0\) và \(\sin x< 0\).

Sử dụng công thức hạ bậc ta có

\(\cos^2 x=\displaystyle\frac{1+\cos 2x}{2}\Leftrightarrow \cos^2 x=\displaystyle\frac{1}{16}\Rightarrow \cos x=\displaystyle\frac{1}{4}.\)

c) Ta có \(\cos 4x=2\cos^2 2x-1=\displaystyle\frac{17}{32}\).

d) Ta có \(\cos^2 x+\sin^2 x=1\Leftrightarrow \sin x=\displaystyle\frac{15}{16}\Rightarrow \sin x=-\displaystyle\frac{\sqrt{15}}{4}\).

Vậy \(\tan x=\displaystyle\frac{\sin x}{\cos x}=-\sqrt{15}\).

Đáp án:

• \(\sin (\pi -\alpha)=\displaystyle\frac{\sqrt{7}}{4}\)
• \(\sin 2\alpha=\displaystyle\frac{3\sqrt{7}}{8}\)

Lời giải:

Ta có \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(\sin \alpha>0\).

Mặt khác, \(\cos^2\alpha+\sin^2\alpha=1\Leftrightarrow \sin^2\alpha=\displaystyle\frac{7}{16}\Rightarrow\sin\alpha=\displaystyle\frac{\sqrt{7}}{4}\).

a) Ta có \(\sin (\pi-\alpha)=\sin \alpha=\displaystyle\frac{\sqrt{7}}{4}\).

b) Ta có \(\sin 2\alpha=2\sin\alpha\cos\alpha=2\cdot \displaystyle\frac{\sqrt{7}}{4}\cdot\displaystyle\frac{3}{4}=\displaystyle\frac{3\sqrt{7}}{8}\).

c) Ta có \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(0< \displaystyle\frac{\alpha}{2}< \displaystyle\frac{\pi}{4}\). Vậy \(\sin \displaystyle\frac{\alpha}{2}>0\).

Áp dụng công thức hạ bậc ta có

\(\sin^2\displaystyle\frac{\alpha}{2}=\displaystyle\frac{1-\cos \alpha}{2}=\displaystyle\frac{1}{8}\Rightarrow \sin \displaystyle\frac{\alpha}{2}=\displaystyle\frac{\sqrt{2}}{4}.\)

d) Ta có \(\tan \alpha=\displaystyle\frac{\sin \alpha}{\cos \alpha}=\displaystyle\frac{\sqrt{7}}{3}\).

Vậy \(\tan 2\alpha=\displaystyle\frac{2\tan \alpha}{1-\tan^2\alpha}=3\sqrt{7}\).

Đáp án:

• \(2\cos 3a \cos a=\displaystyle\frac{27}{25}\)
• \(2\sin 3a \cos a=\displaystyle\frac{39}{25}\)
• \(2\sin 3a \sin a+ 2 \sin \displaystyle\frac{3a}{2} \cos \displaystyle\frac{3a}{2}=\displaystyle\frac{3\sqrt{10}}{10}-\displaystyle\frac{7}{25}\)

Lời giải:

a) Đúng.

Ta có \(2\cos 3a \cos a=\cos 2a+ \cos 4a =\cos 2a+ 2\cos^2 a-1=\displaystyle\frac{4}{5}+2\left(\displaystyle\frac{4}{5}\right)^2-1=\displaystyle\frac{27}{25}\).

Vậy \(2\cos 3a \cos a=\displaystyle\frac{27}{25}\).

b) Đúng.

Ta có \(2\sin 3a \cos a=\sin 2a+ \sin 4a =\sin 2a(1+2\cos2a) \).

Mặt khác \(\sin2a=\pm\sqrt{1-\cos^22a}=\pm\sqrt{1-\left(\displaystyle\frac{4}{5}\right)^2}=\pm\displaystyle\frac{3}{5}\).\\ Kết hợp với \(0< \alpha < \displaystyle\frac{\pi}{2}\), tức là \(0< 2a < \pi\), ta nhận \(\sin 2a =\displaystyle\frac{3}{5}\).

Vậy \(2\sin 3a \cos a=\displaystyle\frac{39}{25}\).

c) Đúng.

Ta có \(2\sin 3a \sin a+ 2 \sin \displaystyle\frac{3a}{2} \cos \displaystyle\frac{3a}{2}=\cos 2a- \cos 4a +\cos a -\cos2 a =\cos a-\cos4a \).

Mặt khác \(\cos^2 a=\displaystyle\frac{1+\cos^2(2a)}{2}=\displaystyle\frac{1+\displaystyle\frac{4}{5}}{2}=\displaystyle\frac{9}{10}\).\\ Kết hợp với \(0< a < \displaystyle\frac{\pi}{2}\), ta nhận \(\cos a = \displaystyle\frac{3}{\sqrt{10}}\).

Vậy \(2\sin 3a \sin a+ 2 \sin \displaystyle\frac{3a}{2} \cos \displaystyle\frac{3a}{2}=\displaystyle\frac{3}{\sqrt{10}}+ 2\left(\displaystyle\frac{4}{5}\right)^2-1=\displaystyle\frac{3\sqrt{10}}{10}+\displaystyle\frac{7}{25}\).

d) Sai.

\begin{eqnarray*}\cos7a\cos a-\cos5a\cos a&=&\cos6a+\cos8a-(\cos4a+\cos6a)\\&=&\cos8a-\cos4a=2\cos^24a-1+\cos4a.\end{eqnarray*}

Ta lại có \(\cos4a=2\cos^22a-1=2\left(\displaystyle\frac{4}{5}\right)^2-1=\displaystyle\frac{7}{25}\).

Vậy \(\cos7a\cos a-\cos5a\cos a =\displaystyle\frac{7}{25}-\displaystyle\frac{4}{5}=-\displaystyle\frac{13}{25}\).

Đáp án:

• \(\sin a \cos b=\displaystyle\frac{32}{65}\)
• \(\cos a \cos b=\displaystyle\frac{56}{65}\)
• \(\cos 2a \cos 2b=\displaystyle\frac{2079}{4225}\)

Lời giải:

a)

\(\sin a \cos b=\displaystyle\frac{1}{2}\left[\sin (a-b)+\sin (a+b)\right]=\displaystyle\frac{1}{2}\left(\displaystyle\frac{5}{13}+\displaystyle\frac{3}{5}\right)=\displaystyle\frac{32}{65}\).

b)

\begin{eqnarray*}\cos a \cos b&=&\displaystyle\frac{1}{2}\left[\cos (a-b)+\cos (a+b)\right]=\displaystyle\frac{1}{2}\left[\sqrt{1-\sin^2(a-b)}+\sqrt{1-\sin^2(a+b)}\right]\\&=& \displaystyle\frac{1}{2}\left[\sqrt{1-\left(\displaystyle\frac{5}{13}\right)^2}+\sqrt{1-\left(\displaystyle\frac{3}{5}\right)^2}\right]=\displaystyle\frac{56}{65}.\end{eqnarray*}

c)

\begin{eqnarray*}\cos 2a \cos 2b &=& \displaystyle\frac{1}{2}\left[\cos 2(a-b)+\cos 2(a+b)\right]= \displaystyle\frac{1}{2}\left[1-2\sin^2(a-b)+1-2\sin^2(a+b)\right]\\&=&1-\sin^2(a-b)-\sin^2(a+b)\\&=&1-\left(\displaystyle\frac{5}{13}\right)^2-\left(\displaystyle\frac{3}{5}\right)^2=-\displaystyle\frac{2079}{4225}.\end{eqnarray*}

d)

\begin{eqnarray*}&&\sin a \sin b+\cos2(a+b)+\cos2(a-b)\\&=&\displaystyle\frac{1}{2}\left[\cos (a-b)-\cos (a+b)\right]+1-2\sin^2(a+b)+1-2\sin^2(a-b)\\&=&\displaystyle\frac{1}{2}\left[\sqrt{1-\sin^2(a-b)}-\sqrt{1-\sin^2(a+b)}\right]+2-2\left(\displaystyle\frac{5}{13}\right)^2-2\left(\displaystyle\frac{3}{5}\right)^2\\&=& \displaystyle\frac{1}{2}\left[\sqrt{1-\left(\displaystyle\frac{5}{13}\right)^2}-\sqrt{1-\left(\displaystyle\frac{3}{5}\right)^2}\right]+2-2\left(\displaystyle\frac{5}{13}\right)^2-2\left(\displaystyle\frac{3}{5}\right)^2=\displaystyle\frac{4418}{4225}.\end{eqnarray*}

Đáp án:

• \(\sin\left(\displaystyle\frac{a}{2}+\displaystyle\frac{b}{2}\right)\cos\left(\displaystyle\frac{a}{2}-\displaystyle\frac{b}{2}\right)=\displaystyle\frac{56}{65}\)
• \(\sin (a+b) \sin (a-b)=-\displaystyle\frac{1296}{4225}\)
• \(\sin (a+b) \cos (a-b)=-\displaystyle\frac{1128}{4225}\)

Lời giải:

a)

\(\sin\left(\displaystyle\frac{a}{2}+\displaystyle\frac{b}{2}\right)\cos\left(\displaystyle\frac{a}{2}-\displaystyle\frac{b}{2}\right)=\displaystyle\frac{1}{2}(\sin a + \sin b)=\displaystyle\frac{1}{2}\left(\sqrt{1-\cos^2a}+\sqrt{1-\cos^2b}\right)=\displaystyle\frac{1}{2}\left(\displaystyle\frac{4}{5}+\displaystyle\frac{12}{13}\right)=\displaystyle\frac{56}{65}.\)

b)

\(\cos (a+b) \cos (a-b)=\displaystyle\frac{1}{2}(\cos 2a+\cos 2b)=\cos ^2a+\cos ^2b-1=\left(\displaystyle\frac{1}{3}\right)^2+\left(\displaystyle\frac{1}{4}\right)^2-1=-\displaystyle\frac{2479}{4225}\).

c)

\(\sin (a+b) \sin (a-b)=\displaystyle\frac{1}{2}(\cos 2b-\cos 2a)=\cos ^2b-\cos ^2a=\left(\displaystyle\frac{3}{13}\right)^2-\left(\displaystyle\frac{3}{5}\right)^2=-\displaystyle\frac{1296}{4225}\).

d)

\begin{eqnarray*}\sin (a+b) \cos (a-b)&=&\displaystyle\frac{1}{2}(\sin 2b-\sin 2a)=\sin b\cos b-\sin a\cos a\\&=&\sqrt{1-\cos^2b}\cos b-\sqrt{1-\cos^2a}\cos a\\&=&\displaystyle\frac{3}{13}\cdot\displaystyle\frac{3}{13}-\displaystyle\frac{4}{5}\cdot \displaystyle\frac{3}{5}=-\displaystyle\frac{1128}{4225}.\end{eqnarray*}

Đáp án:

• \(2\cos\alpha\cos 3\alpha+2\sin^2\alpha=2m^2\)
• \(2\sin3\alpha \cos \alpha=\sqrt{1-m^2}\cdot(1+2m)\)
• \(2\sin3\alpha \sin \alpha-\sqrt{2}\cos\left(2\alpha+\displaystyle\frac{\pi}{4}\right)=\sqrt{1-m^2}+1-2m^2\)

Lời giải:

a) \(2 \cos3\alpha \cos \alpha=\cos 2\alpha+\cos 4\alpha=2\cos^22\alpha+\cos 2\alpha-1=2m^2+m-1\).

b) \(2\cos\alpha\cos 3\alpha+2\sin^2\alpha=2\cos^22\alpha+\cos 2\alpha-1+1-2\cos 2\alpha=2\cos^22\alpha=2m^2.\)

c) \(2\sin3\alpha \cos \alpha=\sin 2\alpha +\sin 4\alpha=\sin 2\alpha\left(1+2\cos 2\alpha\right)=\sqrt{1-m^2}(1+2m)\).

d) \begin{eqnarray*}2\sin3\alpha \sin \alpha-\sqrt{2}\cos\left(2\alpha+\displaystyle\frac{\pi}{4}\right)&=&\cos 2 \alpha -\cos 4\alpha- \cos 2\alpha +\sin 2\alpha=-\cos 4\alpha+\sin 2\alpha\\ &=&\sqrt{1-m^2}+1-2m^2.\end{eqnarray*}

Đáp án:

• \(\sin a \sin 5a+\cos 2a \cos4 a=-\displaystyle\frac{351}{1250}\)
• \(\cos 5a \cos a+\sin 4a \sin 2a =-\displaystyle\frac{351}{1250}\)
• \(\cos5a\cos a-\cos3a\cos a=\displaystyle\frac{8064}{15625}\)

Lời giải:

a) \(2\sin a \sin 5a+2\cos 2a \cos4 a=\cos 4a -\cos 6a+\cos6a+\cos 2a=2\cos^22a-1+\cos2a=8m^4-6m^2.\)

Vậy \(\sin a \sin 5a+\cos 2a \cos4 a=-\displaystyle\frac{351}{625}\).

b) \(2\cos 5a \cos a+2\sin 4a \sin 2a =\cos4a+\cos6a+\cos2a-\cos6a=\cos4a+\cos2a=\displaystyle\frac{351}{1250}\).

c) \begin{eqnarray*}2\sin 5a \cos a+2\sin 4x \cos 2x &=&\sin 4a +\sin 6a-(\sin6a+\sin 2a)=\sin 4a-\sin 2a\\&=&2\sin a \cos a \left[2\left(2\cos^2a-1\right)-1\right]=-\displaystyle\frac{936}{625}.\end{eqnarray*}

d) \begin{eqnarray*}2\cos5a\cos a-2\cos3a\cos a&=&\cos6a+\cos4a-(\cos2a+\cos4a)=\cos6a-\cos2a\\&=&4\cos^32a-4\cos2a=4\cos2a(\cos^22a-1)=\displaystyle\frac{8064}{15625}.\end{eqnarray*}

Đáp án:

• \(\cos3x+\cos x=2\cos2x\cos x\)

Lời giải:

a) \(\sin3x+\sin x=2\sin\displaystyle\frac{3x+x}{2}\cdot\cos\displaystyle\frac{3x-x}{2}=2\sin2x\cos x\).

b) \(\sin3x-\sin x=2\cos\displaystyle\frac{3x+x}{2}\cdot\sin\displaystyle\frac{3x-x}{2}=2\cos2x\sin x\).

c) \(\cos3x+\cos x=2\cos\displaystyle\frac{3x+x}{2}\cdot\cos\displaystyle\frac{3x-x}{2}=2\cos2x\cos x\).

d) \(\cos3x-\cos x=-2\sin\displaystyle\frac{3x+x}{2}\cdot\sin\displaystyle\frac{3x-x}{2}=-2\sin2x\sin x\).

}

Đáp án:

• Nếu tam giác \(ABC\) đều thì \(M=\displaystyle\frac{\sqrt{3}}{2}\)
• \(\sin C=2\sin\displaystyle\frac{C}{2}\cos\displaystyle\frac{C}{2}\)
• \(M=4\sin\displaystyle\frac{A}{2}\sin\displaystyle\frac{B}{2}\cos\displaystyle\frac{C}{2}\)

Lời giải:

a) Ta có \(\triangle ABC\) đều suy ra \(\widehat{A}=\widehat{B}=\widehat{C}=60^\circ\).

Nên ta có \(M=\sin60^\circ+\sin60^\circ-\sin60^\circ=\displaystyle\frac{\sqrt{3}}{2}\).

b) Áp dụng công thức tổng thành tích ta được \(\sin A+\sin B=2\sin\displaystyle\frac{A+B}{2}\cos\displaystyle\frac{A-B}{2}\).

c) Áp dụng công thức nhân đôi ta được \(\sin C=2\sin\displaystyle\frac{C}{2}\cos\displaystyle\frac{C}{2}\).

d) Ta có

\begin{eqnarray*}M&=&\sin A+\sin B-\sin C\\ &=& 2\sin\displaystyle\frac{A+B}{2}\cos\displaystyle\frac{A-B}{2}-2\sin\displaystyle\frac{C}{2}\cos\displaystyle\frac{C}{2}\\&= & 2\cos\displaystyle\frac{C}{2}\cos\displaystyle\frac{A-B}{2}-2\sin\displaystyle\frac{C}{2}\cos\displaystyle\frac{C}{2}\\&=&2\cos\displaystyle\frac{C}{2}\left(\cos\displaystyle\frac{A-B}{2}-\cos\displaystyle\frac{A+B}{2}\right)\\&= & 4\sin\displaystyle\frac{A}{2}\sin\displaystyle\frac{B}{2}\cos\displaystyle\frac{C}{2}.\end{eqnarray*}

}

Đáp án:

• Nếu \(a=\displaystyle\frac{\pi}{12}\) thì \(A=2\)
• \(\cos 2a-\cos 4a=2\sin 3a \cdot \sin a\)

Lời giải:

a) Thay \(a=\displaystyle\frac{\pi}{12}\) ta được \(A=2\).

b) Áp dụng công thức biến đổi tổng thành tích ta được \(\cos 2a-\cos 4a=2\sin 3a \cdot \sin a\).

c) Ta có \(\displaystyle\frac{\cos a-\cos 5a}{\sin 5a-\sin a}=\displaystyle\frac{2\sin 3a \cdot \sin 2a}{2\cos 3a \cdot \sin 2a} =\tan 3a\)

d) Ta có

\begin{eqnarray*}& A& =\displaystyle\frac{\cos 2a-\cos 4a}{\sin 4a-\sin 2a}+\displaystyle\frac{\cos a-\cos 5a}{\sin 5a-\sin a} \\& & =\displaystyle\frac{2\sin 3a \cdot \sin a}{2\cos 3a \cdot \sin a}+\displaystyle\frac{2\sin 3a \cdot \sin 2a}{2\cos 3a \cdot \sin 2a}\\& & =\tan 3a+\tan 3a=2\tan 3a.\end{eqnarray*}

\(\sin15^\circ=\sin\left(45^\circ-30^\circ\right)=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ\) \(=\displaystyle\frac{\sqrt{2}}{2}\cdot\displaystyle\frac{\sqrt{3}}{2}-\displaystyle\frac{\sqrt{2}}{2}\cdot\displaystyle\frac{1}{2}=\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\).

\(\cos15^\circ=\cos\left(45^\circ-30^\circ\right)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ\) \(=\displaystyle\frac{\sqrt{2}}{2}\cdot\displaystyle\frac{\sqrt{3}}{2}+\displaystyle\frac{\sqrt{2}}{2}\cdot\displaystyle\frac{1}{2}=\displaystyle\frac{\sqrt{6}+\sqrt{2}}{4}\).

\(\tan15^\circ=\displaystyle\frac{\sin15^\circ}{\cos15^\circ}=2-\sqrt{3}\).

\(\cot15^\circ=2+\sqrt{3}\).

Ta có \(\displaystyle\frac{\sqrt{2}}{2} = \cos \displaystyle\frac{\pi}{4} = \cos \left(2 \cdot \displaystyle\frac{\pi}{8}\right) = 1 - 2 \sin ^2 \displaystyle\frac{\pi}{8}\).

Suy ra \(\sin ^2 \displaystyle\frac{\pi}{8} = \displaystyle\frac{2 - \sqrt{2}}{4}\).

Vì \(0 < \displaystyle\frac{\pi}{8} < \displaystyle\frac{\pi}{2}\) nên \(\sin \displaystyle\frac{\pi}{8} > 0\).

Suy ra \(\sin \displaystyle\frac{\pi}{8} = \displaystyle\frac{\sqrt{2 - \sqrt{2}}}{2}\).

\(\cos \displaystyle\frac{\pi}{12} = \cos \left(\displaystyle\frac{\pi}{3} - \displaystyle\frac{\pi}{4}\right) = \cos \displaystyle\frac{\pi}{3} \cos \displaystyle\frac{\pi}{4} + \sin \displaystyle\frac{\pi}{3} \sin \displaystyle\frac{\pi}{4}\) \(= \displaystyle\frac{1}{2} \cdot \displaystyle\frac{\sqrt{2}}{2} + \displaystyle\frac{\sqrt{3}}{2} \cdot \displaystyle\frac{\sqrt{2}}{2} = \displaystyle\frac{\sqrt{2} + \sqrt{6}}{4}\).

Ta có

\(\sin \displaystyle\frac{\pi}{12} = \sin \left(\displaystyle\frac{\pi}{3} - \displaystyle\frac{\pi}{4}\right) = \sin \displaystyle\frac{\pi}{3} \cos \displaystyle\frac{\pi}{4} - \cos \displaystyle\frac{\pi}{3} \sin \displaystyle\frac{\pi}{4}\) \( = \displaystyle\frac{\sqrt{3}}{2} \cdot \displaystyle\frac{\sqrt{2}}{2} - \displaystyle\frac{1}{2} \cdot \displaystyle\frac{\sqrt{2}}{2} = \displaystyle\frac{\sqrt{6} - \sqrt{2}}{4}\).

\(\tan \displaystyle\frac{\pi}{12} = \tan \left(\displaystyle\frac{\pi}{3} - \displaystyle\frac{\pi}{4}\right)\) \( = \displaystyle\frac{\tan \displaystyle\frac{\pi}{3} - \tan \displaystyle\frac{\pi}{4} }{1 + \tan \displaystyle\frac{\pi}{3} \tan \displaystyle\frac{\pi}{4}} = \displaystyle\frac{\sqrt{3} -1}{1 + \sqrt{3}}\).

}

Ta có \(\displaystyle\frac{\sqrt{2}}{2} = \cos \displaystyle\frac{\pi}{4} = \cos \left(2 \cdot \displaystyle\frac{\pi}{8}\right) = 2 \cos ^2 \displaystyle\frac{\pi}{8} - 1\).

Suy ra \(\cos ^2 \displaystyle\frac{\pi}{8} = \displaystyle\frac{2 + \sqrt{2}}{4}\).

Vì \(0 < \displaystyle\frac{\pi}{8} < \displaystyle\frac{\pi}{2}\) nên \(\cos \displaystyle\frac{\pi}{8} > 0\). Suy ra \(\cos \displaystyle\frac{\pi}{8} = \displaystyle\frac{\sqrt{2 + \sqrt{2}}}{2}\).

Ta có \(1 = \tan \displaystyle\frac{\pi}{4} = \tan \left(2 \cdot \displaystyle\frac{\pi}{8}\right) = \displaystyle\frac{2\tan \displaystyle\frac{\pi}{8}}{1 - \tan^2 \displaystyle\frac{\pi}{8}}\).

Suy ra \(\tan^2 \displaystyle\frac{\pi}{8} + 2\tan \displaystyle\frac{\pi}{8} - 1= 0\) \(\Leftrightarrow \tan \displaystyle\frac{\pi}{8} = -1 + \sqrt{2}\) hoặc \(\tan \displaystyle\frac{\pi}{8} = -1 - \sqrt{2}.\)

Vì \(0 < \displaystyle\frac{\pi}{8} < \displaystyle\frac{\pi}{2}\) nên \(\tan \displaystyle\frac{\pi}{8} > 0\). Suy ra \(\tan \displaystyle\frac{\pi}{8} = -1 + \sqrt{2}\).

}

\(\cos \displaystyle\frac{11 \pi}{12} \cos \displaystyle\frac{7 \pi}{12} = \displaystyle\frac{1}{2}\left[\cos \left(\displaystyle\frac{11 \pi}{12} - \displaystyle\frac{7 \pi}{12}\right) + \cos \left(\displaystyle\frac{11 \pi}{12} + \displaystyle\frac{7 \pi}{12}\right)\right]\) \(= \displaystyle\frac{1}{2}\left(\cos \displaystyle\frac{\pi}{3} + \cos \displaystyle\frac{3 \pi}{2}\right) = \displaystyle\frac{1}{4}.\)

Ta có

\(\sin \displaystyle\frac{\pi}{24} \cos \displaystyle\frac{5 \pi}{24} = \displaystyle\frac{1}{2}\left[\sin \left(\displaystyle\frac{\pi}{24} - \displaystyle\frac{5 \pi}{24}\right) + \sin \left(\displaystyle\frac{\pi}{24} + \displaystyle\frac{5 \pi}{24}\right)\right]\) \( = \displaystyle\frac{1}{2}\left[\sin \left(-\displaystyle\frac{\pi}{6}\right) + \sin \displaystyle\frac{ \pi}{4}\right] = \displaystyle\frac{-1 + \sqrt{2}}{4}.\)

\(\sin \displaystyle\frac{7 \pi}{8} \sin \displaystyle\frac{5 \pi}{8} = \displaystyle\frac{1}{2}\left[\cos \left(\displaystyle\frac{7 \pi}{8} - \displaystyle\frac{5 \pi}{8}\right) - \cos \left(\displaystyle\frac{7 \pi}{8} + \displaystyle\frac{5 \pi}{8}\right)\right]\) \(= \displaystyle\frac{1}{2}\left(\cos \displaystyle\frac{\pi}{4} - \cos \displaystyle\frac{3 \pi}{2}\right) = \displaystyle\frac{\sqrt{2}}{4}.\)

}

Ta có

\(\sin \displaystyle\frac{5 \pi}{12} + \sin \displaystyle\frac{\pi}{12} = 2 \sin \displaystyle\frac{\displaystyle\frac{5 \pi}{12} + \displaystyle\frac{\pi}{12}}{2} \cos \displaystyle\frac{\displaystyle\frac{5 \pi}{12} - \displaystyle\frac{\pi}{12}}{2}\) \(= 2 \sin \displaystyle\frac{\pi}{4} \cos \displaystyle\frac{\pi}{6} = 2 \cdot \displaystyle\frac{\sqrt{2}}{2} \cdot \displaystyle\frac{\sqrt{3}}{2} = \displaystyle\frac{\sqrt{6}}{2}.\)

Ta có

\(\cos \displaystyle\frac{7 \pi}{12} + \cos \displaystyle\frac{\pi}{12} = 2\cos \displaystyle\frac{\displaystyle\frac{7 \pi}{12} + \displaystyle\frac{\pi}{12}}{2} \cos \displaystyle\frac{\displaystyle\frac{7 \pi}{12} - \displaystyle\frac{\pi}{12}}{2}\) \(= 2 \cos \displaystyle\frac{\pi}{3} \cos \displaystyle\frac{\pi}{4} = 2 \cdot \displaystyle\frac{1}{2} \cdot \displaystyle\frac{\sqrt{2}}{2} = \displaystyle\frac{\sqrt{2}}{2}.\)

}

a) Áp dụng công thức cộng đối với tang, ta có:

\(\tan \displaystyle\frac{7 \pi}{12}=\tan \left(\displaystyle\frac{\pi}{4}+\displaystyle\frac{\pi}{3}\right)\) \(=\displaystyle\frac{\tan \frac{\pi}{4}+\tan \frac{\pi}{3}}{1-\tan \frac{\pi}{4} \tan \frac{\pi}{3}}=\displaystyle\frac{1+\sqrt{3}}{1-\sqrt{3}}=-2-\sqrt{3}\).

b) Áp dụng công thức cộng đối với tang, ta có:

\(\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)=\displaystyle\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}\) \(=\displaystyle\frac{1-\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}=2-\sqrt{3}\).

a) Áp dụng công thức cộng, ta có:

\(\sin 75^{\circ}=\sin \left(30^{\circ}+45^{\circ}\right)\) \(=\sin 30^{\circ} \cos 45^{\circ}+\cos 30^{\circ} \sin 45^{\circ}=\displaystyle\frac{\sqrt{6}+\sqrt{2}}{4}\).

b) Áp dụng công thức cộng, ta có:

\(\sin \displaystyle\frac{\pi}{12}=\sin \left(\displaystyle\frac{\pi}{3}-\displaystyle\frac{\pi}{4}\right)\) \(=\sin \displaystyle\frac{\pi}{3} \cos \displaystyle\frac{\pi}{4}-\cos \displaystyle\frac{\pi}{3} \sin \displaystyle\frac{\pi}{4}=\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\).

a) Áp dụng công thức cộng, ta có:

\(\cos \displaystyle\frac{5 \pi}{12}=\cos \left(\displaystyle\frac{\pi}{6}+\displaystyle\frac{\pi}{4}\right)\) \(=\cos \displaystyle\frac{\pi}{6} \cos \displaystyle\frac{\pi}{4}-\sin \displaystyle\frac{\pi}{6} \sin \displaystyle\frac{\pi}{4}=\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\).

b) Áp dụng công thức cộng, ta có:

\(\sin 15^{\circ}=\sin \left(60^{\circ}-45^{\circ}\right)\) \(=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}=\displaystyle\frac{\sqrt{6}+\sqrt{2}}{4}\).

Do \(\pi < \alpha < \displaystyle\frac{3 \pi}{2} \Rightarrow \cos \alpha < 0\).

Ta có

\(\sin^2 \alpha + \cos^2 \alpha = 1 \Rightarrow \cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left(-\displaystyle\frac{5}{13}\right)^2} = -\displaystyle\frac{12}{13}\).

Suy ra

\(\sin \left(\alpha + \displaystyle\frac{\pi}{6}\right) = \sin \alpha \cos \displaystyle\frac{\pi}{6} + \sin \displaystyle\frac{\pi}{6} \cos \alpha\) \(= -\displaystyle\frac{5}{13} \cdot \displaystyle\frac{\sqrt{3}}{2} - \displaystyle\frac{1}{2} \cdot \displaystyle\frac{12}{13} = \displaystyle\frac{-5\sqrt{3} -12}{26}\);

\(\cos \left(\displaystyle\frac{\pi}{4} - \alpha\right) = \cos \displaystyle\frac{\pi}{4} \cos \alpha + \sin \displaystyle\frac{\pi}{4} \sin \alpha\) \(= -\displaystyle\frac{\sqrt{2}}{2} \cdot \displaystyle\frac{12}{13} - \displaystyle\frac{\sqrt{2}}{2} \cdot \displaystyle\frac{5}{13} = -\displaystyle\frac{17\sqrt{2}}{26}\).

}

a) Ta có

\(\sin \displaystyle\frac{5 \pi}{12}=\sin\left( \displaystyle\frac{3\pi}{12}+\displaystyle\frac{2\pi}{12} \right)=\sin\left(\displaystyle\frac{\pi}{4}+\displaystyle\frac{\pi}{6}\right)\) \(=\sin\displaystyle\frac{\pi}{4}\cos\displaystyle\frac{\pi}{6}+\cos\displaystyle\frac{\pi}{4} \sin\displaystyle\frac{\pi}{6}=\displaystyle\frac{\sqrt{6}+\sqrt{2}}{4}\).

Ta có \(\cos \displaystyle\frac{5 \pi}{12}=\cos\left(\displaystyle\frac{\pi}{4}+\displaystyle\frac{\pi}{6}\right)\) \(=\cos\displaystyle\frac{\pi}{4}\cos\displaystyle\frac{\pi}{6}-\sin\displaystyle\frac{\pi}{4} \sin\displaystyle\frac{\pi}{6}=\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\).

Suy ra \(\tan\displaystyle\frac{5\pi}{12}=\displaystyle\frac{\sin\displaystyle\frac{5\pi}{12}}{\cos\displaystyle\frac{5\pi}{12}}\) \(=\displaystyle\frac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2}\Rightarrow \cot\displaystyle\frac{5\pi}{12}=\displaystyle\frac{\sqrt 6-\sqrt 2}{\sqrt 6+\sqrt 2}\).

b) Ta có \(-555^{\circ}=-2\cdot360^\circ+165^\circ\).

Suy ra \begin{eqnarray*}\sin(-555^\circ)&=&\sin 165^\circ \\ &=&\sin 15^\circ=\sin\left(45^\circ-30^\circ\right)\\ &=&\sin 45^\circ\cos 30^\circ-\sin 30^\circ \cos 45^\circ\\ &=&\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4};\end{eqnarray*}

\begin{eqnarray*}\cos(-555^\circ)&=&\cos 165^\circ \\ &=&-\cos 15^\circ=-\cos\left(45^\circ-30^\circ\right)\\ &=&-\left( \cos 45^\circ\cos 30^\circ+\sin 30^\circ \sin 45^\circ\right) \\ &=&-\displaystyle\frac{\sqrt{6}+\sqrt{2}}{4}.\end{eqnarray*}

Suy ra \(\tan (-555^\circ)=\displaystyle\frac{\sin (-555^\circ)}{\cos(-555^\circ)}\) \(=-\displaystyle\frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\Rightarrow \cot(-555^\circ)=-\displaystyle\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2}\).

}

a) Do \(-\displaystyle\frac{\pi}{2} < \alpha < 0 \Rightarrow \begin{cases}\cos \alpha > 0\\ \sin \alpha < 0\\ \tan \alpha < 0\\ \cot \alpha < 0.\end{cases}\)

Ta có

\(\cos 2 \alpha = \displaystyle\frac{2}{5} \Leftrightarrow 2\cos^2 \alpha - 1 = \displaystyle\frac{2}{5} \Leftrightarrow \cos^2 \alpha = \displaystyle\frac{7}{10}\) \(\Leftrightarrow \cos \alpha = \sqrt{\displaystyle\frac{7}{10}}\) hoặc \(\cos \alpha = -\sqrt{\displaystyle\frac{7}{10}}\) (loại).

Do \(\sin^2 \alpha + \cos^2 \alpha = 1 \Rightarrow \sin \alpha = - \sqrt{1 -\cos^2 \alpha} = - \sqrt{1 - \displaystyle\frac{7}{10}} = -\sqrt{\displaystyle\frac{3}{10}}\).

Suy ra \(\tan \alpha = \displaystyle\frac{\sin \alpha}{\cos \alpha} = - \sqrt{\displaystyle\frac{3}{7}} \Rightarrow \cot \alpha = - \sqrt{\displaystyle\frac{7}{3}}\).

b) Do \(\displaystyle\frac{\pi}{2} < \alpha < \displaystyle\frac{3 \pi}{4} \Rightarrow \begin{cases}\sin \alpha > 0\\ \cos \alpha < 0\\ \tan \alpha < 0\\ \cot \alpha < 0.\end{cases}\)

Mặt khác \(\displaystyle\frac{\pi}{2} < \alpha < \displaystyle\frac{3 \pi}{4} \Rightarrow \pi < 2\alpha < \displaystyle\frac{3 \pi}{2} \Rightarrow \cos 2\alpha < 0\).

Ta có \(\sin^2 2\alpha + \cos^2 2\alpha = 1 \Rightarrow \cos 2 \alpha = -\sqrt{1 - \sin^2 2\alpha} = -\sqrt{1 - \left(-\displaystyle\frac{4}{9}\right)^2} = -\displaystyle\frac{\sqrt{65}}{9}.\)

Ta có

\(\cos 2 \alpha = -\displaystyle\frac{\sqrt{65}}{9} \Leftrightarrow 2\cos^2 \alpha - 1 = -\displaystyle\frac{\sqrt{65}}{9} \Leftrightarrow \cos^2 \alpha = \displaystyle\frac{9 - \sqrt{65}}{18}\) \(\Leftrightarrow \cos \alpha = \sqrt{\displaystyle\frac{9 - \sqrt{65}}{18}}\) (loại) hoặc \(\cos \alpha = -\sqrt{\displaystyle\frac{9 - \sqrt{65}}{18}}.\)

Do \(\sin^2 \alpha + \cos^2 \alpha = 1 \Rightarrow \sin \alpha = \sqrt{1 -\cos^2 \alpha} = \sqrt{1 - \displaystyle\frac{9 - \sqrt{65}}{18}} = \sqrt{\displaystyle\frac{9 + \sqrt{65}}{18}}\).

Suy ra \(\tan \alpha = \displaystyle\frac{\sin \alpha}{\cos \alpha} = - \sqrt{\displaystyle\frac{9 + \sqrt{65}}{9 - \sqrt{65}}} \Rightarrow \cot \alpha = - \sqrt{\displaystyle\frac{9 - \sqrt{65}}{9 + \sqrt{65}}}\).

}

a) \(A=\displaystyle\frac{1}{2}\left(\cos 2 x+\cos \displaystyle\frac{\pi}{3}\right)\) \(=\displaystyle\frac{1}{2}\left(\displaystyle\frac{1}{4}+\displaystyle\frac{1}{2}\right)=\displaystyle\frac{3}{8}\).

b) \(B=-\displaystyle\frac{1}{2}\left(\cos 2 x-\cos \displaystyle\frac{2\pi}{3}\right)\) \(=-\displaystyle\frac{1}{2}\left(\displaystyle\frac{1}{4}+\displaystyle\frac{1}{2}\right)=-\displaystyle\frac{3}{8}\).

+) \(\cos 2a=1-2\sin^2a=1-2\cdot \left(\displaystyle\frac{2}{\sqrt 5}\right)^2=-\displaystyle\frac{3}{5}\).

+) \(\cos 4a=2\cos^2(2a)-1=2\cdot \left(-\displaystyle\frac{3}{5} \right)^2-1=-\displaystyle\frac{7}{25}\).

Ta có

\(B=\left(\cos \displaystyle\frac{\pi}{9}+\cos \displaystyle\frac{5 \pi}{9}\right)+\cos \displaystyle\frac{11 \pi}{9}\) \(=2 \cos \displaystyle\frac{2 \pi}{9} \cos \displaystyle\frac{\pi}{3}+\cos \displaystyle\frac{11 \pi}{9}=\cos \displaystyle\frac{2 \pi}{9}-\cos \displaystyle\frac{2 \pi}{9}=0.\)

Ta có:

\(\begin{aligned}A & =\cos \displaystyle\frac{5 \pi}{12} \cos \displaystyle\frac{7 \pi}{12}=\displaystyle\frac{1}{2}\left[\cos \left(\displaystyle\frac{5 \pi}{12}-\displaystyle\frac{7 \pi}{12}\right)+\cos \left(\displaystyle\frac{5 \pi}{12}+\displaystyle\frac{7 \pi}{12}\right)\right] \\ & =\displaystyle\frac{1}{2}\left[\cos \left(-\displaystyle\frac{\pi}{6}\right)+\cos \pi\right]=\displaystyle\frac{1}{2}\left(\displaystyle\frac{\sqrt{3}}{2}-1\right)=\displaystyle\frac{\sqrt{3}-2}{4}.\\ B & =\cos 75^{\circ} \sin 15^{\circ}=\displaystyle\frac{1}{2}\left[\sin \left(15^{\circ}-75^{\circ}\right)+\sin \left(15^{\circ}+75^{\circ}\right)\right] \\ & =\displaystyle\frac{1}{2}\left[\sin \left(-60^{\circ}\right)+\sin 90^{\circ}\right]=\displaystyle\frac{1}{2}\left(-\displaystyle\frac{\sqrt{3}}{2}+1\right)=\displaystyle\frac{2-\sqrt{3}}{4}.\end{aligned}\)

\(\begin{aligned}A & =\cos 75^{\circ} \cos 15^{\circ}=\displaystyle\frac{1}{2}\left[\cos \left(75^{\circ}-15^{\circ}\right)+\cos \left(75^{\circ}+15^{\circ}\right)\right] \\ & =\displaystyle\frac{1}{2}\left[\cos \left(60^{\circ}\right)+\cos 90^{\circ}\right]=\displaystyle\frac{1}{2}\left(\displaystyle\frac{1}{2}+0\right)=\displaystyle\frac{1}{4}.\\ B & =\sin \displaystyle\frac{5 \pi}{12} \cos \displaystyle\frac{7 \pi}{12}=\displaystyle\frac{1}{2}\left[\sin \left(\displaystyle\frac{5 \pi}{12}-\displaystyle\frac{7 \pi}{12}\right)+\sin \left(\displaystyle\frac{5 \pi}{12}+\displaystyle\frac{7 \pi}{12}\right)\right] \\ & =\displaystyle\frac{1}{2}\left[\sin \left(-\displaystyle\frac{\pi}{6}\right)+\sin \pi\right]=\displaystyle\frac{1}{2}\left(-\displaystyle\frac{1}{2}+0\right)=-\displaystyle\frac{1}{4}.\end{aligned}\)

Ta có

\(A=\left(\sin \displaystyle\frac{\pi}{9}+\sin \displaystyle\frac{7 \pi}{9}\right)-\sin \displaystyle\frac{5 \pi}{9}\) \(=2 \sin \displaystyle\frac{4 \pi}{9} \cos \displaystyle\frac{\pi}{3}-\sin \displaystyle\frac{5 \pi}{9}=\sin \displaystyle\frac{4 \pi}{9}-\sin \displaystyle\frac{5 \pi}{9}=0.\)

Ta có

\(\cos^{2} \displaystyle\frac{\pi}{8}=\displaystyle\frac{1+\cos \left(2\cdot\displaystyle\frac{\pi}{8}\right)}{2}\) \(=\displaystyle\frac{1+\cos \displaystyle\frac{\pi}{4}}{2}=\displaystyle\frac{1+\displaystyle\frac{\sqrt{2}}{2}}{2}=\displaystyle\frac{2+\sqrt{2}}{4}\).

Suy ra \(\cos \displaystyle\frac{\pi}{8}=\displaystyle\frac{\sqrt{{2+\sqrt{2}}}}{2}\).

a) Ta có

\(\begin{aligned}\cos 75^{\circ}&=\cos \left(45^{\circ}+30^{\circ}\right)=\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ}\\ &=\displaystyle\frac{\sqrt{2}}{2} \cdot \displaystyle\frac{\sqrt{3}}{2}-\displaystyle\frac{\sqrt{2}}{2} \cdot \displaystyle\frac{1}{2}=\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}.\end{aligned}\)

b) Ta có

\(\begin{aligned}\tan \displaystyle\frac{\pi}{12}&=\tan \left(\displaystyle\frac{\pi}{3}-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{\tan \displaystyle\frac{\pi}{3}-\tan \displaystyle\frac{\pi}{4}}{1+\tan \displaystyle\frac{\pi}{3} \cdot \tan \displaystyle\frac{\pi}{4}}\\ &=\displaystyle\frac{\sqrt{3}-1}{1+\sqrt{3}}=\displaystyle\frac{(\sqrt{3}-1)^2}{3-1}=2-\sqrt{3}.\end{aligned}\)

a) \(A=\displaystyle\frac{\sin\left(\displaystyle\frac{\pi}{15}+\displaystyle\frac{\pi}{10}\right)}{\cos\left(\displaystyle\frac{2\pi}{15}+\displaystyle\frac{\pi}{5}\right)}\) \(=\displaystyle\frac{\sin\displaystyle\frac{\pi}{6}}{\cos\displaystyle\frac{\pi}{3}}=\displaystyle\frac{\sin\displaystyle\frac{\pi}{6}}{\sin\left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{\pi}{3}\right)}=\displaystyle\frac{\sin\displaystyle\frac{\pi}{6}}{\sin\displaystyle\frac{\pi}{6}}=1\).

b) Áp dụng công thức nhân đôi \(\sin2a=2\sin a\cos a\), ta có

\begin{eqnarray*}B&=&\sin \displaystyle\frac{\pi}{32} \cos \displaystyle\frac{\pi}{32} \cos \displaystyle\frac{\pi}{16} \cos \displaystyle\frac{\pi}{8}\\&=&\displaystyle\frac{1}{2}\cdot2\sin \displaystyle\frac{\pi}{32} \cos \displaystyle\frac{\pi}{32} \cos \displaystyle\frac{\pi}{16} \cos \displaystyle\frac{\pi}{8}\\&=&\displaystyle\frac{1}{4}\cdot2\sin \displaystyle\frac{\pi}{16} \cos \displaystyle\frac{\pi}{16} \cos \displaystyle\frac{\pi}{8}\\&=&\displaystyle\frac{1}{8}\cdot2\sin \displaystyle\frac{\pi}{8} \cos \displaystyle\frac{\pi}{8}\\&=&\displaystyle\frac{1}{8}\cdot\sin\displaystyle\frac{\pi}{4}\\&=&\displaystyle\frac{1}{8}\cdot\displaystyle\frac{\sqrt{2}}{2}=\displaystyle\frac{\sqrt{2}}{16}.\end{eqnarray*}

Ta có \(\sin \displaystyle\frac{7\pi}{9}+\sin \displaystyle\frac{\pi}{9}=2 \sin \displaystyle\frac{\frac{7 \pi}{9}+\frac{\pi}{9}}{2} \cos \displaystyle\frac{\frac{7\pi}{9}-\frac{\pi}{9}}{2}\) \(=2 \sin \displaystyle\frac{4 \pi}{9} \cos \displaystyle\frac{\pi}{3}=\sin \displaystyle\frac{4 \pi}{9}\).

\(\cos \displaystyle\frac{7\pi}{9}-\cos \displaystyle\frac{\pi}{9}=-2 \sin \displaystyle\frac{\frac{7 \pi}{9}+\frac{\pi}{9}}{2} \sin \displaystyle\frac{\frac{7\pi}{9}-\frac{\pi}{9}}{2}\) \(=-2 \sin \displaystyle\frac{4 \pi}{9} \sin \displaystyle\frac{\pi}{3}=-\sqrt{3}\sin \displaystyle\frac{4 \pi}{9}\).

Suy ra \(D=-\displaystyle\frac{1}{\sqrt{3}}=-\displaystyle\frac{\sqrt{3}}{3}\).

a) \(\sin \displaystyle\frac{11 \pi}{12}-\sin \displaystyle\frac{5 \pi}{12}=2 \cos \displaystyle\frac{\frac{11 \pi}{12}+\frac{5 \pi}{12}}{2} \sin \displaystyle\frac{\frac{11 \pi}{12}-\frac{5 \pi}{12}}{2}\) \(=2 \cos \displaystyle\frac{2 \pi}{3} \sin \displaystyle\frac{\pi}{4}=2 \cdot\left(-\displaystyle\frac{1}{2}\right) \cdot \displaystyle\frac{\sqrt{2}}{2}=-\displaystyle\frac{\sqrt{2}}{2}\).

b) \(\cos 105^{\circ}+\cos 15^{\circ}=2 \cos \displaystyle\frac{105^{\circ}+15^{\circ}}{2} \cos \displaystyle\frac{105^{\circ}-15^{\circ}}{2}\) \(=2 \cos 60^{\circ} \cos 45^{\circ}=2 \cdot \displaystyle\frac{1}{2} \cdot \displaystyle\frac{\sqrt{2}}{2}=\displaystyle\frac{\sqrt{2}}{2}\).

\(\begin{aligned}A&=\sin \left(x+\frac{\pi}{4}\right) \cos \left(x-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{1}{2}\left[\sin \left(x+\displaystyle\frac{\pi}{4}+x-\displaystyle\frac{\pi}{4}\right)+\sin \left(x+\displaystyle\frac{\pi}{4}-x+\displaystyle\frac{\pi}{4}\right)\right] \\ &=\displaystyle\frac{1}{2}\left(\sin 2 x+\sin \displaystyle\frac{\pi}{2}\right)=\displaystyle\frac{1}{2}\left(-\displaystyle\frac{1}{3}+1\right)=\displaystyle\frac{1}{3}.\end{aligned}\)

Ta có \(\cos2a=2\cos^2 a-1=-\displaystyle\frac{1}{9}\).

Do đó

\(B=\cos \displaystyle\frac{3 a}{2} \cos \displaystyle\frac{a}{2}=\displaystyle\frac{1}{2}\left[\cos \left(\displaystyle\frac{3 a}{2}+\displaystyle\frac{a}{2}\right)+\cos \left(\displaystyle\frac{3 a}{2}-\displaystyle\frac{a}{2}\right)\right]\) \(=\displaystyle\frac{1}{2}\left(\cos 2a+\cos a \right)=\displaystyle\frac{5}{18}.\)

Ta có

\(\sin^2\alpha+\cos^2\alpha=1\Rightarrow \sin^2\alpha=1-\cos^2\alpha=1-\left(\displaystyle\frac{1}{3}\right)^2=\displaystyle\frac{8}{9}\Rightarrow \sin\alpha=\pm \displaystyle\frac{2\sqrt{2}}{3}\).

Mặt khác, \(-\displaystyle\frac{\pi}{2}<\alpha<0\) nên \(\sin\alpha<0\). Do đó \(\sin\alpha=-\displaystyle\frac{2\sqrt{2}}{3}\).

\(\sin 2 \alpha=2\sin\alpha\cos\alpha=2\cdot \displaystyle\frac{-2\sqrt{2}}{3}\cdot \displaystyle\frac{1}{3}=-\displaystyle\frac{4\sqrt{2}}{9}\).

\(\cos \left(\alpha+\displaystyle\frac{\pi}{3}\right)=\cos\alpha\cos\displaystyle\frac{\pi}{3}-\sin\alpha\sin\displaystyle\frac{\pi}{3}\) \(=\displaystyle\frac{1}{3}\cdot \displaystyle\frac{1}{2}-\displaystyle\frac{-2\sqrt{2}}{3}\cdot \displaystyle\frac{\sqrt{3}}{2}=\displaystyle\frac{1+2\sqrt{6}}{6}\).

a) Do \(0 < \alpha < \displaystyle\frac{\pi}{2} \Rightarrow \cos \alpha > 0\).

Ta có \(\sin^2 \alpha + \cos^2 \alpha = 1 \Rightarrow \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\displaystyle\frac{\sqrt{3}}{3}\right)^2} = \displaystyle\frac{\sqrt{6}}{3}\).

Suy ra \(\cos 2\alpha = 2\cos^2 \alpha - 1 = 2 \left(\displaystyle\frac{\sqrt{6}}{3}\right)^2 - 1 = \displaystyle\frac{1}{3}\);

\(\sin 2\alpha = 2\sin \alpha \cos \alpha = 2 \cdot \displaystyle\frac{\sqrt{3}}{3} \cdot \displaystyle\frac{\sqrt{6}}{3} = \displaystyle\frac{2\sqrt{2}}{3}\);

\(\tan 2\alpha = \displaystyle\frac{\sin 2\alpha}{\cos 2\alpha} =2\sqrt{2}\Rightarrow \cot 2\alpha = \displaystyle\frac{1}{2\sqrt{2}} \).

b) Ta có \(\cos \alpha=\cos \left( 2\cdot\displaystyle\frac{\alpha}{2}\right) =1-2\sin^2 \displaystyle\frac{\alpha}{2}=1-2\cdot \left(\displaystyle\frac{3}{4}\right)^2=-\displaystyle\frac{1}{8}\).

Ta có \(\sin^2\alpha=1-\cos^2\alpha=1-\displaystyle\frac{1}{64}=\displaystyle\frac{63}{64}\).

Vì \(\pi<\alpha<2\pi\) nên \(\sin \alpha<0\). Suy ra \(\sin \alpha=-\sqrt{\displaystyle\frac{63}{64}}=-\displaystyle\frac{\sqrt{63}}{8}\).

Do đó \(\sin 2\alpha=2\sin\alpha \cos \alpha=2\left(-\displaystyle\frac{\sqrt{63}}{8}\right)\left(-\displaystyle\frac{1}{8}\right)=\displaystyle\frac{\sqrt{63}}{32}\);

\(\cos 2\alpha=2\cos^2\alpha-1=2\left(-\displaystyle\frac{1}{8}\right)^2-1=-\displaystyle\frac{31}{32}\).

Suy ra \(\tan 2\alpha=\displaystyle\frac{\sin2\alpha}{\cos 2\alpha}=-\displaystyle\frac{\sqrt{63}}{31}\) và \(\cot 2\alpha=-\displaystyle\frac{31}{\sqrt{63}} \).

}

Vì \(\displaystyle\frac{\pi}{2}< a<\pi\) nên \(\sin a>0\).

Do đó \(\sin a=\sqrt{1-\cos ^2 a}=\sqrt{1-\left(-\displaystyle\frac{1}{3}\right)^2}=\sqrt{1-\displaystyle\frac{1}{9}}=\sqrt{\displaystyle\frac{8}{9}}=\displaystyle\frac{2 \sqrt{2}}{3}\) .

Vậy \(\sin 2 a=2 \sin a \cos a=2 \cdot \displaystyle\frac{2 \sqrt{2}}{3} \cdot\left(-\displaystyle\frac{1}{3}\right)=-\displaystyle\frac{4 \sqrt{2}}{9}\) .

a) Ta có \(\cos^2a+\sin^2a=1\) nên \(\cos a=-\sqrt{1-\sin^2a}=-\sqrt{1-\displaystyle\frac{1}{9}}=-\displaystyle\frac{2\sqrt{2}}{3}\) (vì \(\displaystyle\frac{\pi}{2}< a<\pi\)).

Suy ra \(\sin2a=2\sin a\cos a=2\cdot\displaystyle\frac{1}{3}\cdot\displaystyle\frac{-2\sqrt{2}}{3}=-\displaystyle\frac{4\sqrt{2}}{9}\).

\(\cos2a=1-2\sin^2a=1-2\left(\displaystyle\frac{1}{3}\right)^2=\displaystyle\frac{7}{9}\).

\(\tan2a=\displaystyle\frac{\sin2a}{\cos2a}=-\displaystyle\frac{4\sqrt{2}}{7}\).

b) Ta có

\(\sin a+\cos a=\displaystyle\frac{1}{2}\) nên

\begin{eqnarray*}&&\left(\sin a+\cos a\right)^2=\sin^2a+\cos^2a+2\sin a\cos a\\&\Leftrightarrow& \left(\displaystyle\frac{1}{2}\right)^2=1+\sin2a\\&\Leftrightarrow&\sin2a=-\displaystyle\frac{3}{4}.\end{eqnarray*}

Vì \(\displaystyle\frac{\pi}{2}< a<\displaystyle\frac{3 \pi}{4}\) nên \(\pi<2a<\displaystyle\frac{3\pi}{2}\). Suy ra \(\cos2a<0\).

Do đó

\(\cos2a=-\sqrt{1-\sin^22a}=-\sqrt{1-\left(-\displaystyle\frac{3}{4}\right)^2}=-\displaystyle\frac{\sqrt{7}}{4}\).

\(\tan2a=\displaystyle\frac{\sin2a}{\cos2a}=\displaystyle\frac{3\sqrt{7}}{7}\).

a) Ta có \(\cos^2a+\sin^2a=1\) nên \(\cos a=-\sqrt{1-\sin^2a}=-\sqrt{1-\displaystyle\frac{1}{3}}=-\displaystyle\frac{\sqrt{6}}{3}\) (vì \(\displaystyle\frac{\pi}{2}< a<\pi\)).

\(\cos \left(a+\displaystyle\frac{\pi}{6}\right)=\cos a\cos\displaystyle\frac{\pi}{6}-\sin a\sin\displaystyle\frac{\pi}{6}\) \(=-\displaystyle\frac{\sqrt{6}}{3}\cdot\displaystyle\frac{\sqrt{3}}{2}-\displaystyle\frac{1}{\sqrt{3}}\cdot\displaystyle\frac{1}{2}=-\displaystyle\frac{\sqrt{3}+3\sqrt{2}}{6}\).

b) Ta có \(\cos^2a+\sin^2a=1\) nên \(\sin a=-\sqrt{1-\cos^2a}=-\sqrt{1-\displaystyle\frac{1}{9}}=-\displaystyle\frac{2\sqrt{2}}{3}\) (vì \(\pi< a<\displaystyle\frac{3 \pi}{2}\)).

Suy ra \(\tan a=\displaystyle\frac{\sin a}{\cos a}=2\sqrt{2}\).

Do đó \(\tan\left(a-\displaystyle\frac{\pi}{6}\right)=\displaystyle\frac{\tan a-\tan\displaystyle\frac{\pi}{6}}{1+\tan a\tan\displaystyle\frac{\pi}{6}}\) \(=\displaystyle\frac{2\sqrt{2}-\displaystyle\frac{\sqrt{3}}{3}}{1+2\sqrt{2}\cdot\displaystyle\frac{\sqrt{3}}{3}}=\displaystyle\frac{9\sqrt{3}-8\sqrt{2}}{5}\).

Áp dụng công thức nhân đôi, ta có

\(\tan a=\displaystyle\frac{2 \tan\displaystyle\frac{a}{2}}{1-\tan ^2 \displaystyle\frac{a}{2}}=\displaystyle\frac{2 \cdot (-2)}{1-(-2)^2}=-\displaystyle\frac{4}{3}.\)

Do \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) nên \(\begin{cases}\cos\alpha <0\\ \sin\alpha >0.\end{cases}\)

Khi đó \(\sin \alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\left(-\displaystyle\frac{1}{\sqrt{3}}\right)^2}=\displaystyle\frac{\sqrt{6}}{3}\).

a) Ta có \(\sin\left(\alpha+\displaystyle\frac{\pi}{6}\right)=\sin\alpha\cos\displaystyle\frac{\pi}{6}+\cos\alpha\sin\displaystyle\frac{\pi}{6}\) \(=\displaystyle\frac{\sqrt{6}}{3}\cdot \displaystyle\frac{\sqrt{3}}{2}-\displaystyle\frac{1}{\sqrt{3}}\cdot \displaystyle\frac{1}{2}=\displaystyle\frac{3\sqrt{2}-\sqrt{3}}{6}\).

b) Ta có \(\cos\left(\alpha+\displaystyle\frac{\pi}{6}\right)=\cos\alpha\cos\displaystyle\frac{\pi}{6}-\sin\alpha\sin\displaystyle\frac{\pi}{6}\) \(=-\displaystyle\frac{1}{\sqrt{3}}\cdot \displaystyle\frac{\sqrt{3}}{2}-\displaystyle\frac{\sqrt{6}}{3}\cdot \displaystyle\frac{1}{2}=\displaystyle\frac{-3-\sqrt{6}}{6}\).

c) Ta có \(\sin\left(\alpha-\displaystyle\frac{\pi}{6}\right)=\sin\alpha\cos\displaystyle\frac{\pi}{6}-\cos\alpha\sin\displaystyle\frac{\pi}{6}\) \(=\displaystyle\frac{\sqrt{6}}{3}\cdot \displaystyle\frac{\sqrt{3}}{2}+\displaystyle\frac{1}{\sqrt{3}}\cdot \displaystyle\frac{1}{2}=\displaystyle\frac{3\sqrt{2}+\sqrt{3}}{6}\).

d) Ta có \(\cos\left(\alpha-\displaystyle\frac{\pi}{6}\right)=\cos\alpha\cos\displaystyle\frac{\pi}{6}+\sin\alpha\sin\displaystyle\frac{\pi}{6}\) \(=-\displaystyle\frac{1}{\sqrt{3}}\cdot \displaystyle\frac{\sqrt{3}}{2}+\displaystyle\frac{\sqrt{6}}{3}\cdot \displaystyle\frac{1}{2}=\displaystyle\frac{-3+\sqrt{6}}{6}\).

Ta có: \(\sin ^2 a+\cos ^2 a=1 \Rightarrow \sin ^2 a=1-\cos ^2 a=1-\left(\displaystyle\frac{3}{5}\right)^2=\displaystyle\frac{16}{25}\).

Do \(0< a<\displaystyle\frac{\pi}{2}\) nên \(\sin a>0\). Suy ra \(\sin a=\displaystyle\frac{4}{5}\),\, \(\tan a=\displaystyle\frac{\sin a}{\cos a}=\displaystyle\frac{4}{3}\).

\(\sin \left(a+\displaystyle\frac{\pi}{6}\right)=\sin a \cos \displaystyle\frac{\pi}{6}+\cos a \sin \displaystyle\frac{\pi}{6}\) \(=\displaystyle\frac{\sqrt{3}}{2} \sin a+\displaystyle\frac{1}{2} \cos a=\displaystyle\frac{\sqrt{3}}{2} \cdot \displaystyle\frac{4}{5}+\displaystyle\frac{1}{2} \cdot \displaystyle\frac{3}{5}=\displaystyle\frac{3+4 \sqrt{3}}{10}\).

\(\cos \left(a-\displaystyle\frac{\pi}{3}\right)=\cos a \cos \displaystyle\frac{\pi}{3}+\sin a \sin \displaystyle\frac{\pi}{3}\) \(=\displaystyle\frac{1}{2} \cos a+\displaystyle\frac{\sqrt{3}}{2} \sin a=\displaystyle\frac{1}{2} \cdot \displaystyle\frac{3}{5}+\displaystyle\frac{\sqrt{3}}{2} \cdot \displaystyle\frac{4}{5}=\displaystyle\frac{3+4 \sqrt{3}}{10}\).

\(\tan \left(a+\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{\tan a+\tan \frac{\pi}{4}}{1-\tan a \cdot \tan \frac{\pi}{4}}\) \(=\displaystyle\frac{\frac{4}{3}+1}{1-\frac{4}{3} \cdot 1}=-7\).

a) \(A=\sin \left[\left(a-17^{\circ}\right)-\left(a+13^{\circ}\right)\right]=\sin \left(-30^{\circ}\right)\) \(=-\sin 30^{\circ}=-\displaystyle\frac{1}{2}\).

b) \(B=\cos \left[\left(b+\displaystyle\frac{\pi}{3}\right]+\left(\displaystyle\frac{\pi}{6}-b\right)\right)=\sin \displaystyle\frac{\pi}{2}=1\).

\(\tan 2a=\tan\left[(a+b)+(a-b)\right]=\displaystyle\frac{\tan(a+b)+\tan(a-b)}{1-\tan(a+b)\tan(a-b)}=\displaystyle\frac{3+2}{1-3\cdot 2}=-1\).

\(\tan 2b=\tan\left[(a+b)-(a-b)\right]=\displaystyle\frac{\tan(a+b)-\tan(a-b)}{1+\tan(a+b)\tan(a-b)}=\displaystyle\frac{3-2}{1+3\cdot 2}=\displaystyle\frac{1}{7}\).

Ta có: \(1=(\sin a+\cos a)^2=\sin^2a+2\sin a\cos a+\cos^2a=1+\sin 2a\).

Suy ra: \(\sin 2a=0\).

Do \(\displaystyle\frac{\pi}{2}< a<\pi\) nên \(\sin a>0,\, \cos a>0\) và \(\tan a>0\).

Ta có

+) \(\sin^2a=\displaystyle\frac{1-\cos 2a}{2}=\displaystyle\frac{1-\frac{1}{3}}{2}=\displaystyle\frac{1}{3}\Rightarrow \sin a=\displaystyle\frac{\sqrt{3}}{3}\).

+) \(\cos^2a=\displaystyle\frac{1+\cos 2a}{2}=\displaystyle\frac{1+\frac{1}{3}}{2}=\displaystyle\frac{2}{3}\Rightarrow \cos a=\displaystyle\frac{\sqrt{6}}{3}\).

+) \(\tan^2a=\displaystyle\frac{1-\cos 2a}{1+\cos 2a}=\displaystyle\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\displaystyle\frac{1}{2}\Rightarrow \tan a=\displaystyle\frac{\sqrt{2}}{2}\).

+) Do \(\sin a+\cos a=\displaystyle\frac{1}{2}\) nên \((\sin a+\cos a)^{2}=\displaystyle\frac{1}{4} \Leftrightarrow \sin ^{2} a+\cos ^{2} a+2 \sin a \cos a=\displaystyle\frac{1}{4}\) hay \(1+2 \sin a \cos a=\displaystyle\frac{1}{4}\).

Suy ra \(\sin 2 a=\displaystyle\frac{1}{4}-1=-\displaystyle\frac{3}{4}\).

+) Áp dụng công thức nhân đôi, ta có: \(\cos 4 a=\cos (2 \cdot 2 a)=1-2 \sin ^{2} 2 a=-\displaystyle\frac{1}{8}\).

a) \(\left(\sin \alpha+\cos\alpha\right)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=1+\sin 2\alpha\);

b) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha-\sin^2\alpha\right)\left(\cos^2\alpha+\sin^2\alpha\right)=\cos 2\alpha\).

Chứng minh \(\sin (a+b) \sin (a-b)=\cos ^2 b-\cos ^2 a\).

Ta có

\begin{eqnarray*}\sin(a+b)\sin(a-b)&=&\displaystyle\frac{1}{2}\left[\cos(2b)-\cos(2a)\right]\\&=&\displaystyle\frac{1}{2}\left[\left(2\cos^2b-1\right)-\left(2\cos^2a-1\right)\right]\\&=&\cos^2b-\cos^2a.\end{eqnarray*}

Chứng minh \(\sin ^2 a-\sin ^2 b=\cos ^2 b-\cos ^2 a\).

Ta có \(\sin ^2 a-\sin ^2 b=1-\cos^2a-\left(1-\cos^2b\right)=\cos^2b-\cos^2a\).

Vậy \(\sin (a+b) \sin (a-b)=\sin ^2 a-\sin ^2 b=\cos ^2 b-\cos ^2 a\).

Ta có

\(\begin{aligned}\sqrt{2} \sin \left(x+\displaystyle\frac{\pi}{4}\right) &=\sqrt{2}\left(\sin x \cos \displaystyle\frac{\pi}{4}+\cos x \sin \displaystyle\frac{\pi}{4}\right)\\ &=\sqrt{2}\left(\sin x \cdot \displaystyle\frac{\sqrt{2}}{2}+\cos x \cdot \displaystyle\frac{\sqrt{2}}{2}\right)\\ &=\sin x+\cos x.\end{aligned}\)

Đẳng thức được chứng minh.

a) Ta có

\[\begin{aligned}\sqrt{2} \sin \left(x-\displaystyle\frac{\pi}{4}\right)&=\sqrt{2}\left(\sin x \cos \displaystyle\frac{\pi}{4}-\cos x \sin \displaystyle\frac{\pi}{4}\right)\\ &=\sqrt{2}\left(\sin x \cdot \displaystyle\frac{\sqrt{2}}{2}-\cos x \cdot \displaystyle\frac{\sqrt{2}}{2}\right)\\ &=\sin x-\cos x.\end{aligned}\]

Đẳng thức được chứng minh.

b) Ta có

\(\begin{aligned}\tan \left(\displaystyle\frac{\pi}{4}-x\right)&=\displaystyle\frac{\tan\displaystyle\frac{\pi}{4}-\tan x}{1+\tan\displaystyle\frac{\pi}{4}\tan x}\\ &=\displaystyle\frac{1-\tan x}{1+\tan x}~\left(x \neq \displaystyle\frac{\pi}{2}+k \pi, x \neq \displaystyle\frac{3 \pi}{4}+k \pi, k \in \mathbb{Z}\right).\end{aligned}\)

Đẳng thức được chứng minh.

a) Ta có

\begin{eqnarray*}\sin (\alpha+\beta)&=&\sin\alpha\cos\beta+\cos\alpha\sin\beta\\ \sin (\alpha-\beta)&=&\sin\alpha\cos\beta-\cos\alpha\sin\beta\\ \Rightarrow \sin (\alpha+\beta) \sin (\alpha-\beta)&=&(\sin\alpha\cos\beta+\cos\alpha\sin\beta)(\sin\alpha\cos\beta-\cos\alpha\sin\beta)\\ &=& \sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta\\ &=&\sin^2\alpha\left(1-\sin^2\beta\right)+\left(1-\sin^2\alpha\right)\sin^2\beta\\ &=&\sin^2\alpha-\sin^2\beta.\end{eqnarray*}

b) Ta có

\begin{eqnarray*}&&\cos ^4 \alpha-\cos ^4\left(\alpha-\displaystyle\frac{\pi}{2}\right)\\ &=&\cos^4\alpha-\cos^4\left(\displaystyle\frac{\pi}{2}-\alpha\right)\\ &=&\cos^4\alpha-\sin^4\alpha \\ &=&\left(\cos^2\alpha-\sin^2\alpha\right)\left(\cos^2\alpha+\sin^2\alpha\right)\\ &=&\cos^2\alpha-\sin^2\alpha=\cos 2\alpha.\end{eqnarray*}

\(A=\displaystyle\frac{(\sin x+\sin 3x)+\sin 2x}{(\cos x+\cos 3x)+\cos 2x}\) \(=\displaystyle\frac{2\sin2x\cos x+\sin 2x}{2\cos 2x\cos x+\cos 2x}=\displaystyle\frac{\sin 2x(2\cos x+1)}{\cos 2x(2\cos x+1)}=\tan 2x\).

a) Ta có

\begin{eqnarray*}\sqrt{2} \sin \left(\alpha + \displaystyle\frac{\pi}{4}\right) - \cos \alpha&=&\sqrt{2}\sin \alpha \cos \displaystyle\frac{\pi}{4} + \sqrt{2}\cos \alpha \sin \displaystyle\frac{\pi}{4} - \cos \alpha\\ &=&\sqrt{2}\sin \alpha \cdot \displaystyle\frac{\sqrt{2}}{2} + \sqrt{2}\cos \alpha \cdot \displaystyle\frac{\sqrt{2}}{2} - \cos \alpha\\ &=&\sin \alpha + \cos \alpha - \cos \alpha\\ &=& \sin \alpha.\end{eqnarray*}

b) Ta có

\begin{eqnarray*}(\cos \alpha+\sin \alpha)^2 - \sin 2 \alpha&=&\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha - \sin 2 \alpha\\ &=&(\sin^2 \alpha + \cos^2 \alpha) + \sin 2 \alpha - \sin 2 \alpha\\ &=&1.\end{eqnarray*}

}

Trong \(\triangle ABC\) ta có \(A + B + C = \pi \Leftrightarrow A = \pi - (B + C)\).

Suy ra \(\sin A = \sin \left(\pi - (B + C)\right) = \sin (B + C) = \sin B \cos C + \sin C \cos B\).

}

a) Ta có \(\displaystyle\frac{a}{\sin A}=\displaystyle\frac{b}{\sin B}\) nên \(b=\displaystyle\frac{a\sin B}{\sin A}\).

Khi đó \(S=\displaystyle\frac{1}{2} a b \sin C=\displaystyle\frac{1}{2}a\cdot\displaystyle\frac{a\sin B}{\sin A}\sin C=\displaystyle\frac{a^2 \sin B \sin C}{2 \sin A} \).

b) Ta có \(\sin B\sin C=\displaystyle\frac{1}{2}\left[\cos\left(B-C\right)-\cos\left(B+C\right)\right]\) \(=\displaystyle\frac{1}{2}\left(\cos30^\circ-\cos120^\circ\right)=\displaystyle\frac{1+\sqrt{3}}{4}\).

Lại có \(\sin A=\sin\left(180^\circ-B-C\right)=\sin60^\circ=\displaystyle\frac{\sqrt{3}}{2}\).

Diện tích \(S\) của tam giác \(A B C\) là

\(S=\displaystyle\frac{a^2 \sin B \sin C}{2 \sin A}\) \(=\displaystyle\frac{12^2\cdot\displaystyle\frac{1+\sqrt{3}}{4}}{2\cdot\displaystyle\frac{\sqrt{3}}{2}}=36+12\sqrt{3}\,\mathrm{cm}^2\).

a) Ta có: \(\tan\widehat{AOH}=\displaystyle\frac{AH}{OH}=\displaystyle\frac{14}{15}\), \(\tan\widehat{BOH}=\displaystyle\frac{BH}{OH}=\displaystyle\frac{12}{15}\). Suy ra:

\(\tan\alpha=\tan\widehat{AOB}=\tan\left(\widehat{AOH}-\widehat{BOH}\right)=\displaystyle\frac{\tan\widehat{AOH}-\tan\widehat{BOH}}{1+\tan\widehat{AOH}\cdot\tan\widehat{BOH}}=\displaystyle\frac{\frac{14}{15}-\frac{12}{15}}{1+\frac{14}{15}\cdot\frac{12}{15}}=\displaystyle\frac{10}{131}\).

b) \(\tan\alpha=\displaystyle\frac{10}{131}\Rightarrow\alpha\approx 4^\circ\).

Ta có: \(CD=CK-BH=8\), \(CI=CK-AH=26\), \(BD=AI=HK=20\).

\(\tan\widehat{BCD}=\displaystyle\frac{BD}{CD}=\displaystyle\frac{20}{8}=\displaystyle\frac{5}{2}\), \(\tan\widehat{ACI}=\displaystyle\frac{AI}{CI}=\displaystyle\frac{20}{26}=\displaystyle\frac{10}{13}\).

\(\tan\widehat{BCA}=\tan\left(\widehat{BCD}-\widehat{ACI}\right)=\displaystyle\frac{\tan\widehat{BCD}-\tan\widehat{ACI}}{1+\tan\widehat{BCD}.\tan\widehat{ACI}}=\displaystyle\frac{\frac{5}{2}-\frac{10}{13}}{1+\frac{5}{2}\cdot\frac{10}{13}}\)

\(\tan\widehat{BCA}=\displaystyle\frac{45}{76}\Rightarrow \widehat{BCA}\approx 30,6^\circ \).

}

Ta có:

\(\begin{aligned}P & =u \cdot i=[40 \sin (120 \pi t)+10 \sin (360 \pi t)] \cdot[4 \sin (120 \pi t)+\sin (360 \pi t)] \\ & =160 \sin ^{2}(120 \pi t)+10 \sin ^{2}(360 \pi t)+80 \sin (120 \pi t) \sin (360 \pi t) \\ & =80[1-\cos (240 \pi t)]+5[1-\cos (720 \pi t)]+40[\cos (360 \pi t-120 \pi t)-\cos (360 \pi t+120 \pi t)] \\& =85-80 \cos (240 \pi t)-5 \cos (720 \pi t)+40 \cos (240 \pi t)-40 \cos (480 \pi t) \\ & =85-40 \cos (240 \pi t)-5 \cos (720 \pi t)-40 \cos (480 \pi t)(\mathrm{W}).\end{aligned}\)

Dao động tổng hợp \(x(t)\) có phương trình \(x(t)=2 \cos \left(\displaystyle\frac{\pi}{3} t+\displaystyle\frac{\pi}{6}\right)+2 \cos \left(\displaystyle\frac{\pi}{3} t-\displaystyle\frac{\pi}{3}\right)\).

Khi đó \(x(t)=2 \left[\cos \left(\displaystyle\frac{\pi}{3} t+\displaystyle\frac{\pi}{6}\right)+ \cos \left(\displaystyle\frac{\pi}{3} t-\displaystyle\frac{\pi}{3}\right)\right]\) \(=2\cdot2\cos\left(\displaystyle\frac{\pi}{3}t-\displaystyle\frac{\pi}{12}\right)\cos\left(\displaystyle\frac{\pi}{4}\right)=2\sqrt{2}\cos\left(\displaystyle\frac{\pi}{3}t-\displaystyle\frac{\pi}{12}\right)\).

Biên độ và pha ban đầu của dao động tổng hợp này lần lượt là \(A=2\sqrt{2}\) và \(\varphi=-\displaystyle\frac{\pi}{12}\).

a) Nhấn phím \(4\) sẽ tạo ra sóng âm \(y=\sin \left(2 \pi f_1 t\right)+\sin \left(2 \pi f_2 t\right)\) với \(f_{1}=770 \mathrm{\mathrm{Hz}}\) và \(f_{2}=1209 \mathrm{Hz}\).

Khi đó \(y=\sin \left(2 \pi 770 t\right)+\sin \left(2 \pi 1209 t\right)\).

Hay \(y=\sin \left(1540 \pi t\right)+\sin \left(2418 \pi t\right)\).

b) Ta có

\(\begin{aligned}y&=\sin \left(1540 \pi t\right)+\sin \left(2418 \pi t\right)\\ &= 2\sin \displaystyle\frac{1540 \pi t+2418 \pi t}{2}\cos \displaystyle\frac{1540 \pi t-2418 \pi t}{2}\\ &=2\sin(1979 \pi t)\cos (439 \pi t).\end{aligned}\)

a) Ta có \(OM=IH\) nên \(x_M = IA \cdot \cos \alpha = 8\cos \alpha\).

b) Sau 1 phút chuyển động, góc quay \(\alpha\) thì ta có

\(x_M = 8\cos \alpha = - 3 \Leftrightarrow \cos \alpha = -\displaystyle\frac{3}{8}.\)

Sau 2 phút chuyển động, ta có góc quay là \(2\alpha\) thì

\(x_M = 8\cos 2\alpha = 8(2\cos^2 \alpha - 1)\) \(= -\displaystyle\frac{23}{4} \approx -5{,}8\ (\mathrm{cm}).\)

a) Ta có tọa độ điểm \(M\) trong hệ trục tọa độ \(Oxy\) là \(M(x_M; y_M)\).

Với \(\sin \alpha = \displaystyle\frac{y_M}{OM} = -\displaystyle\frac{30}{31}\) và \(\cos \alpha =\sqrt{1 - \sin^2 \alpha} = \displaystyle\frac{\sqrt{61}}{31}\).

b) Ta có \((OA, OP)=(OA,OM)+(OM,OP)\) \(=\alpha+\displaystyle\frac{2\pi}{3}\).

Suy ra

\(\sin (OA,OP)=\sin\left(\alpha+\displaystyle\frac{2\pi}{3}\right)\)

\(=\sin\alpha\cos\displaystyle\frac{2\pi}{3}+\sin\displaystyle\frac{2\pi}{3}\cos\alpha\)

\(=-\displaystyle\frac{30}{31}\cdot \left(-\displaystyle\frac{1}{2}\right)+\displaystyle\frac{\sqrt{3}}{2}\cdot\displaystyle\frac{\sqrt{61}}{31}=\displaystyle\frac{30+\sqrt{183}}{62}.\)

Do đó chiều cao của điểm \(P\) so với mặt đất là

\(60+31\sin(OA,OP)\) \(=60+31\cdot \displaystyle\frac{30+\sqrt{183}}{62}\approx 81{,}76\ \mathrm{m}\).

Tương tự, ta có \((OA, ON)=(OA,OM)+(OM,ON)\) \(=\alpha+\displaystyle\frac{4\pi}{3}\).

Suy ra

\(\sin (OA,ON)=\sin\left(\alpha+\displaystyle\frac{4\pi}{3}\right)\)

\(=\sin\alpha\cos\displaystyle\frac{4\pi}{3}+\sin\displaystyle\frac{4\pi}{3}\cos\alpha\)

\(=-\displaystyle\frac{30}{31}\cdot \left(-\displaystyle\frac{1}{2}\right)+\left(-\displaystyle\frac{\sqrt{3}}{2}\right) \cdot\displaystyle\frac{\sqrt{61}}{31}\) \(=\displaystyle\frac{30-\sqrt{183}}{62}.\)

Do đó chiều cao của điểm \(N\) so với mặt đất là

\(60+31\sin(OA,ON)=60+31\cdot \displaystyle\frac{30-\sqrt{183}}{62}\) \(\approx 68{,}24 \ \mathrm{m}\).