Đáp án: \(\cos \alpha< 0\)

Lời giải:

Vì \(\displaystyle\frac{\pi}{2}< \alpha< \pi\) (điểm biểu diễn thuộc góc lượng giác phần tư thứ II) nên \(\cos \alpha< 0\), \(\sin \alpha>0\), \(\tan \alpha< 0\) và \(\cot \alpha< 0\).

Đáp án: \( \cot(\pi-\alpha)< 0 \)

Lời giải:

\( \pi< \alpha < \displaystyle\frac{3\pi}{2}\Leftrightarrow 0>\pi -\alpha >-\displaystyle\frac{\pi}{2} \Rightarrow \cot (\pi -\alpha)< 0\).

Đáp án: \(P=0\)

Lời giải:

Ta có

\(P=\cos\left(\displaystyle\frac{\pi}{2}-x \right)-\sin x =\sin x-\sin x=0.\)

Đáp án: \(-\cos \alpha\)

Lời giải:

Ta có \( \cos (\pi-\alpha) =-\cos \alpha\).

Đáp án: \(1+\cot^2\alpha=\displaystyle\frac{1}{\sin^2\alpha}\)

Lời giải:

Ta có \(1+\cot^2\alpha=\displaystyle\frac{1}{\sin^2\alpha}\).

Đáp án: \(\sin^2\alpha+\cos^2\alpha=1\)

Lời giải:

Công thức đúng là \(\sin^2\alpha+\cos^2\alpha=1\).

Đáp án: \(0\)

Lời giải:

Ta có \(\sin 720^{\circ}=\sin2.360^{\circ}=0\).

Đáp án: \(-\displaystyle\frac{\sqrt{15}}{15}\)

Lời giải:

Do \(M\left(-\displaystyle\frac{1}{4} ; \displaystyle\frac{\sqrt{15}}{4}\right)\) nằm trên đường tròn lượng giác nên

\(\begin{cases}\sin \alpha = -\displaystyle\frac{1}{4}\\ \cos \alpha=\displaystyle\frac{\sqrt{15}}{4}\end{cases}\Rightarrow\tan\alpha=-\displaystyle\frac{\sqrt{15}}{15}\).

Đáp án: \( \cos \alpha =-\displaystyle\frac{5}{13} \)

Lời giải:

Ta có

\(\cos^2 \alpha=1-\sin^2 \alpha=1-\left(\displaystyle\frac{12}{13}\right)^2=\displaystyle\frac{25}{169}.\)

Vì \( \displaystyle\frac{\pi}{2} < \alpha < \pi \) nên \( \cos \alpha =-\displaystyle\frac{5}{13} \).

Đáp án: \(\displaystyle\frac{-2}{\sqrt{5}}\)

Lời giải:

Ta có \(\cos a=\displaystyle\frac{\sqrt{5}}{3}\)

\(\Rightarrow \sin a=-\sqrt{1-\left(\displaystyle\frac{\sqrt{5}}{3}\right)^2}\) \(=-\displaystyle\frac{2}{3}\quad\left(\text{vì }\displaystyle\frac{3\pi}{2} < \alpha < 2\pi\right)\).

\(\Rightarrow \tan \alpha=\displaystyle\frac{\sin \alpha}{\cos \alpha}=\displaystyle\frac{-\displaystyle\frac{2}{3}}{\displaystyle\frac{\sqrt{5}}{3}}=-\displaystyle\frac{2}{\sqrt{5}}\).

Đáp án: \(\displaystyle\frac{9}{13} \)

Lời giải:

Ta có \(\cos^2 x=\displaystyle\frac{1}{1+\tan^2 x}=\displaystyle\frac{1}{5}\). Khi đó

\(M=\displaystyle\frac{2\sin ^2x+3\sin x \cos x+4\cos ^2x}{5\sin ^2x+6\cos ^2x}\)

\(=\displaystyle\frac{2\sin^2x+2\cos^2x+2\cos^2x+3\sin x\cos x}{5\sin^2 x+5\cos^2x+\cos^2x}\)

\(=\displaystyle\frac{2+2\cos^2x+3\sin x\cos x}{5+\cos^2x}\)

\(=\displaystyle\frac{2+\displaystyle\frac{2}{5}+\displaystyle\frac{6}{5}}{5+\displaystyle\frac{1}{5}}\) \(=\displaystyle\frac{9}{13}.\)

Đáp án: \(Q=1\)

Lời giải:

Chia cả tử và mẫu của \(Q\) cho \(\cos x\) ta được

\(Q=\displaystyle\frac{3\cdot\displaystyle\frac{\sin x}{\cos x}-4\cdot\displaystyle\frac{\cos x}{\cos x}}{\displaystyle\frac{\cos x}{\cos x}+2\cdot\displaystyle\frac{\sin x}{\cos x}}\)

\(=\displaystyle\frac{3\cdot\tan x-4\cdot 1}{1+2\cdot\tan x}\)

\(=\displaystyle\frac{3\cdot 5-4}{1+2\cdot 5}=1.\)

Đáp án: \(\cos (A+B)=-\cos C\)

Lời giải:

Ta có

+) \(A+B=180-C\Rightarrow \sin(A+B)=\sin(180-C)=\sin C\).

+) \(A+B=180-C\Rightarrow \cos(A+B)=\cos(180-C)=-\cos C\).

+) \(\displaystyle\frac{A+B}{2}=90-\displaystyle\frac{C}{2}\)

\(\Rightarrow \sin \left(\displaystyle\frac{A+B}{2}\right)=\sin \left(90-\displaystyle\frac{C}{2}\right)\) \(=\sin \displaystyle\frac{C}{2}\).

Đáp án: \(\sin(A+C)=-\sin B\)

Lời giải:

Vì \(A\), \(B\), \(C\) là ba góc của một tam giác nên \(\widehat{A}+\widehat{B}+\widehat{C}=\pi\).

Suy ra

\(\displaystyle\frac{\widehat{A}+\widehat{B}+\widehat{C}}{2}=\displaystyle\frac{\pi}{2}\). Khi đó

+) \(\sin(A+C)= \sin(\pi-B)=\sin B\).

+) \(\cos(A+B)= \cos(\pi-C)=-\cos C\).

+) \(\cos(A+B+2C)=\cos(\pi+C)=-\cos C\).

+) \(\cos\displaystyle\frac{A+B}{2}=\cos\left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{C}{2}\right)=\sin\displaystyle\frac{C}{2}\).

Đáp án: \(\displaystyle\frac{5}{3}\)

Lời giải:

Do \(\tan x=\displaystyle\frac{1}{3}\) nên \(\cos x\neq 0\).

Ta có

\(A=\displaystyle\frac{\displaystyle\frac{1}{\cos^{2} x}}{1-\displaystyle\frac{\sin x}{\cos x}}\) \(=\displaystyle\frac{1+\tan^{2}x}{1-\tan x}\) \(=\displaystyle\frac{1+\left(\displaystyle\frac{1}{3}\right)^{2}}{1-\displaystyle\frac{1}{3}}\) \(=\displaystyle\frac{5}{3}\).

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

Lời giải:

#

Ta có

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

Đáp án:

• \(A=\cos(\alpha+\pi)< 0\)
• \(B=\tan(\alpha-\pi)>0\)

Lời giải:

a) Vì \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(\pi< \alpha+\pi< \displaystyle\frac{3\pi}{2}\Rightarrow \cos(\alpha+\pi)< 0\).

b) Vì \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(-\pi< \alpha-\pi< -\displaystyle\frac{\pi}{2}\Rightarrow \tan(\alpha-\pi)>0\).

c) Vì \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(\displaystyle\frac{2\pi}{5}< \alpha+\displaystyle\frac{2\pi}{5}< \displaystyle\frac{9\pi}{10}\Rightarrow \sin\left(\alpha+\displaystyle\frac{2\pi}{5}\right)>0\).

d) Vì \(0< \alpha< \displaystyle\frac{\pi}{2}\) nên \(-\displaystyle\frac{3\pi}{8}< \alpha-\displaystyle\frac{3\pi}{8}< \displaystyle\frac{\pi}{8}\Rightarrow \cos\left(\alpha-\displaystyle\frac{3\pi}{8}\right)>0\).

Đáp án:

• \(A=\sin(\alpha+90^{\circ})>0\)
• \(B=\cos(\alpha-45^{\circ})>0\)
• \(D=\cos(2\alpha+90^{\circ})< 0\)

Lời giải:

a) Ta có \(0^{\circ}< \alpha< 90^{\circ}\Rightarrow 90^{\circ}< \alpha+90^{\circ}< 180^{\circ}\)

\(\Rightarrow \sin\left(\alpha+90^{\circ}\right)>0\).

b) Ta có \(0^{\circ}< \alpha< 90^{\circ}\Rightarrow -45^{\circ}< \alpha-45^{\circ}< 45^{\circ}\)

\(\Rightarrow \cos\left(\alpha-45^{\circ}\right)>0\).

c) Ta có \(0^{\circ}< \alpha< 90^{\circ}\Rightarrow -90^{\circ}< -\alpha< 0^{\circ}\)

\(\Rightarrow 270^{\circ}+(-90^{\circ})< 270^{\circ}+(-\alpha)< 270^{\circ}+0^{\circ}\)

\(\Rightarrow 180^{\circ}< 270^{\circ}-\alpha< 270^{\circ}\)

\(\Rightarrow \tan(270^{\circ}-\alpha)>0\).

d) Ta có \(0^{\circ}< \alpha< 90^{\circ}\Rightarrow 90^{\circ}< 2\alpha+90^{\circ}< 270^{\circ}\)

\(\Rightarrow \cos(2\alpha+270^{\circ})< 0\).

Đáp án:

• \(\tan\alpha=\sqrt{3}\)

Lời giải:

a) Sai. Ta có \(\sin\left(\displaystyle\frac{\pi}{3}+k2\pi\right)=\sin\displaystyle\frac{\pi}{3}=\displaystyle\frac{\sqrt{3}}{2}\);

b) Sai. \(\cos\left(\displaystyle\frac{\pi}{3}+k2\pi\right)=\cos\displaystyle\frac{\pi}{3}=\displaystyle\frac{1}{2}\);

c) Đúng. \(\tan \left(\displaystyle\frac{\pi}{3}+k2\pi\right)=\tan\displaystyle\frac{\pi}{3}=\sqrt{3}\);

d) Sai. \(\cot\left(\displaystyle\frac{\pi}{3}+k2\pi\right)=\cot\displaystyle\frac{\pi}{3}=\displaystyle\frac{\sqrt{3}}{3}\).

Đáp án:

• \(\cos x=-\displaystyle\frac{4}{5}\)
• \(\tan x=\displaystyle\frac{3}{4}\)
• \(\cot x=\displaystyle\frac{4}{3}\)

Lời giải:

a) Sai. Do \(\pi< x< \displaystyle\frac{3\pi}{2}\) nên \(\cos x< 0\).

b) Đúng. Ta có \(\sin^{2}x+\cos^{2}x=1\Rightarrow \cos^{2}x=1-\sin^{2}x=1-\displaystyle\frac{9}{25}=\displaystyle\frac{16}{25}\).

\(\Rightarrow \cos x=-\displaystyle\frac{4}{5}\) (vì \(\cos x< 0\)).

c) Đúng. \(\tan x=\displaystyle\frac{\sin x}{\cos x}=\displaystyle\frac{3}{4}\).

d) Đúng. \(\cot x=\displaystyle\frac{\cos x}{\sin x}=\displaystyle\frac{4}{3}\).

Đáp án:

• \(\sin\alpha=\displaystyle\frac{\sqrt{2}}{2}\)
• \(\tan \alpha=-1\)
• \(\cot\alpha=-1\)

Lời giải:

a) Đúng. Ta có

\begin{eqnarray*}\sin\left[-\displaystyle\frac{\pi}{4}+(2k+1)\pi\right]&=&\sin\left(-\displaystyle\frac{\pi}{4}+2k\pi+\pi\right)\\&=&\sin\left(-\displaystyle\frac{\pi}{4}+\pi\right)=-\sin\left(-\displaystyle\frac{\pi}{4}\right)=\sin\displaystyle\frac{\pi}{4}=\displaystyle\frac{\sqrt{2}}{2}.\end{eqnarray*}

b) Sai. Ta có

\begin{eqnarray*}\cos\left[-\displaystyle\frac{\pi}{4}+(2k+1)\pi\right]&=&\cos\left(-\displaystyle\frac{\pi}{4}+2k\pi+\pi\right)\\&=&\cos\left(-\displaystyle\frac{\pi}{4}+\pi\right)=-\cos\left(-\displaystyle\frac{\pi}{4}\right)=-\cos\displaystyle\frac{\pi}{4}=-\displaystyle\frac{\sqrt{2}}{2}\end{eqnarray*}

c) Đúng. Ta có

\begin{eqnarray*}\tan\left[-\displaystyle\frac{\pi}{4}+(2k+1)\pi\right]=\tan\left(-\displaystyle\frac{\pi}{4}\right)=-\tan\displaystyle\frac{\pi}{4}=-1\end{eqnarray*}

d) Đúng. Ta có

\begin{eqnarray*}\cot\left[-\displaystyle\frac{\pi}{4}+(2k+1)\pi\right]=\cot\left(-\displaystyle\frac{\pi}{4}\right)=-\cot\displaystyle\frac{\pi}{4}=-1.\end{eqnarray*}

Đáp án:

• \(\tan x=\sqrt{15}\)

Lời giải:

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

b) Sai. Ta có \(\sin^{2}x+\cos{2}x=1\Rightarrow \sin^{2}x=1-\cos^{2}x=1-\displaystyle\frac{1}{16}=\displaystyle\frac{15}{16}\).

\(\Rightarrow \sin x=\displaystyle\frac{\sqrt{15}}{4}\);

c) Đúng. \(\tan x=\displaystyle\frac{\sin x}{\cos x}=\sqrt{15}\);

d) Sai. \(\cot x=\displaystyle\frac{\cos x}{\sin x}=\displaystyle\frac{1}{\sqrt{15}}\).

Đáp án:

• \(\sin \left(\displaystyle\frac{A+B}{2}\right)=\cos \displaystyle\frac{C}{2}\)

Lời giải:

Trong \(\triangle ABC\) có \(\widehat{A}+\widehat{B}+\widehat{C}=180^\circ \Leftrightarrow \begin{cases}\widehat{A}+\widehat{B}=180^\circ-\widehat{C}\\ \displaystyle\frac{\widehat{A}+\widehat{B}}{2}=90^\circ-\displaystyle\frac{\widehat{C}}{2}.\end{cases}\)

Khi đó ta có

a) \(\sin (A+B)=\sin (180^\circ-C)=\sin C\).

b) \(\sin \left(\displaystyle\frac{A+B}{2}\right)=\sin \left(90^\circ-\displaystyle\frac{C}{2}\right)=\cos \displaystyle\frac{C}{2}\).

c) \(\cos (A+B)=\cos (180^\circ-C)=-\cos C\).

d) \(\tan (A+B)=\tan (180^\circ-C)=-\tan C\).

Đáp án:

• \(\tan (-\alpha)=-2\)
• \(\cot\left(\displaystyle\frac{\pi}{2}-\alpha\right)=2\)
• \(\cot(\pi+\alpha)=\displaystyle\frac12\)

Lời giải:

a) Sai. Ta có \(\tan (\pi-\alpha)=-\tan\alpha=-2\).

b) Đúng. Ta có \(\tan (-\alpha)=-\tan\alpha=-2\).

c) Đúng. Ta có \(\cot\left(\displaystyle\frac{\pi}{2}-\alpha\right)=\tan \alpha=2\).

d) Đúng. Ta có \(\cot(\pi+\alpha)=\cot \alpha=\displaystyle\frac{1}{\tan\alpha}=\displaystyle\frac12\).

Đáp án:

• \(\sin \alpha>0\)
• \(\cos \alpha<0\)
• \(\tan\left(\displaystyle\frac{\pi}{2}-\alpha\right)<0\)

Lời giải:

a) Nếu \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) thì \(\sin\alpha>0\).

b) Nếu \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) thì \(\cos\alpha <0\).

c) Ta có \(\cot\left(\displaystyle\frac{\pi}{2}+\alpha\right)=\tan(-\alpha)=-\tan\alpha\).

Nếu \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) thì \(\tan\alpha <0\), suy ra \(-\tan\alpha>0\). Do đó \(\cot\left(\displaystyle\frac{\pi}{2}+\alpha\right)>0\).

d) Ta có \(\tan\left(\displaystyle\frac{\pi}{2}-\alpha\right)=\cot\alpha\).

Nếu \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) thì \(\cot\alpha <0\). Do đó \(\tan\left(\displaystyle\frac{\pi}{2}-\alpha\right)<0\).

Đáp án:

• \(A=\cos\alpha-\sin\alpha\)
• \(A-B=-\displaystyle\frac{29}{20}\)

Lời giải:

a) Ta có \(A=\sin\left(\displaystyle\frac{\pi}{2}-\alpha\right)+\sin(\pi+\alpha)=\cos\alpha-\sin\alpha=-\displaystyle\frac{4}{5}-\displaystyle\frac{3}{5}=-\displaystyle\frac{7}{5}\).

b) Ta có \(B=\cos(\pi-\alpha)+\cot\left(\displaystyle\frac{\pi}{2}-\alpha\right)=-\cos\alpha+\tan\alpha=-\cos\alpha+\displaystyle\frac{\sin\alpha}{\cos\alpha}=\displaystyle\frac{4}{5}+\displaystyle\frac{\displaystyle\frac{3}{5}}{-\displaystyle\frac{4}{5}}=\displaystyle\frac{1}{20}\).

c) \(A+B=-\displaystyle\frac{7}{5}+\displaystyle\frac{1}{20}=-\displaystyle\frac{27}{20}\).

d) \(A-B=-\displaystyle\frac{7}{5}-\displaystyle\frac{1}{20}=-\displaystyle\frac{29}{20}\).

Đáp án:

• \(\cos\left(-\displaystyle\frac{2024\pi}{9}-x\right)=\cos\left(\displaystyle\frac{8\pi}{9}+x\right)\)
• \(\tan\left(\displaystyle\frac{2023\pi}{2}+2\alpha\right)=-\cot2\alpha\)

Lời giải:

Ta có

a) \(\sin\left(\displaystyle\frac{2024\pi}{3}-x\right)=\sin\left(288\pi+\pi+\displaystyle\frac{\pi}{7}-x\right)=\sin\left(\pi+\displaystyle\frac{\pi}{7}-x\right)=-\sin\left(\displaystyle\frac{\pi}{7}-x\right)\).

b) \(\cos\left(-\displaystyle\frac{2024\pi}{9}-x\right)=\cos\left(-224\pi-\displaystyle\frac{8\pi}{9}-x\right)=\cos\left(-\displaystyle\frac{8\pi}{9}-x\right)=\cos\left(\displaystyle\frac{8\pi}{9}+x\right)\).

c) \(\tan\left(\displaystyle\frac{2023\pi}{2}+2\alpha\right)=\tan\left(1011\pi+\displaystyle\frac{\pi}{2}+2\alpha\right)=\tan\left(\displaystyle\frac{\pi}{2}+2\alpha\right)=-\cot2\alpha\).

d) \(\cot\left(\alpha+\displaystyle\frac{2024\pi}{13}\right)=\cot\left(\displaystyle\frac{9\pi}{13}+\alpha\right)=\cot\left[\pi-\left(\displaystyle\frac{4\pi}{13}-\alpha\right)\right]=-\cot\left(\displaystyle\frac{4\pi}{13}-\alpha\right)\).

Đáp án:

• \(\cos\displaystyle\frac{A+B}{2}=\sin\displaystyle\frac{C}{2}\)
• \(\sin(A+B)=\sin C\)

Lời giải:

a) Sai. Ta có \(\cos A=\sin B\) là khẳng định sai.

b) Sai. Ta có \(\tan A=\cot\left(B+\displaystyle\frac{C}{2}\right)\) là khẳng định sai vì hai góc \(\widehat{A}\) và \(\widehat{B}+\displaystyle\frac{\widehat{C}}{2}\) không liên quan đến nhau.

c) Đúng. Vì \(\widehat{A}+\widehat{B}+\widehat{C}=180^\circ\) nên \(\cos\left(\displaystyle\frac{A+B}{2}\right)=\sin \displaystyle\frac{C}{2}\).

d) Đúng. Vì \(\widehat{A}+\widehat{B}+\widehat{C}=180^\circ\) nên \(\sin(A+B)=\sin C\).

Đáp án:

• \(\tan\alpha<0\)
• \(\sin\alpha=-\displaystyle\frac{4}{5}\)

Lời giải:

a) Vì \(-\displaystyle\frac{\pi}{2}<\alpha<0\) nên \(\sin\alpha<0\). Do đó \(\sin\alpha>0\) nên là sai.

b) Vì \(-\displaystyle\frac{\pi}{2}<\alpha<0\) nên \(\tan\alpha<0\). Do đó \(\tan\alpha<0\) là đúng.

c) Vì \(-\displaystyle\frac{\pi}{2}<\alpha<0\) nên \(\sin\alpha=-\sqrt{1-\cos^2\alpha}=-\displaystyle\frac{4}{5}\). Do đó \(\sin\alpha=-\displaystyle\frac{4}{5}\) là đúng.

d) Vì \(\sin\left(\displaystyle\frac{\pi}{2}-\alpha\right)-\sin (-\alpha)=\cos\alpha+\sin\alpha=-\displaystyle\frac{1}{5}\). Do đó \(\sin\left(\displaystyle\frac{\pi}{2}-\alpha\right)-\sin (-\alpha)=\displaystyle\frac{7}{5}\) là sai.

Đáp án:

• \(\cos\alpha<0\)
• \(\cos\alpha=\displaystyle\frac{-\sqrt{5}}{3}\)

Lời giải:

a) Đúng.

Vì \(\alpha\) nằm ở góc phần tư thứ hai của đường tròn lượng giác nên \(\cos\alpha<0\).

b) Sai.

Ta có \(\sin(\pi-\alpha)=\sin\alpha=\displaystyle\frac23\).

c) Đúng.

Vì \(\sin^2\alpha+\cos^2\alpha=1\) nên \(\cos\alpha=\pm\sqrt{1-\sin^2\alpha}=\sqrt{1-\displaystyle\frac{4}{9}}=\displaystyle\frac{\pm\sqrt{5}}{3}\) mà \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) nên \(\cos\alpha<0\), suy ra \(\cos\alpha=\displaystyle\frac{-\sqrt{5}}{3}\).

d) Sai.

Vì \(\sin\left(\displaystyle\frac{5\pi}{2}-\alpha\right)=\sin\left(\displaystyle\frac{\pi}{2}-\alpha+2\pi\right)=\sin\left(\displaystyle\frac{\pi}{2}-\alpha\right)=\cos\alpha\) mà \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) nên \(\cos\alpha<0\) mà \(\displaystyle\frac{\sqrt{5}}{3}>0\) nên mệnh đề sai.

Đáp án:

• \(\sin^2x+\cos^2 x=1\)
• \((\sin x-\cos x)^2=1-2\sin x\cos x\)

Lời giải:

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

b) Sai. Ta có \((\sin x+\cos x)^2=\sin^2x+2\sin x\cos x+\cos^2x\).

c) Đúng. Ta có \((\sin x-\cos x)^2=\sin^2x-2\sin x\cos x+\cos^2 x=(\sin^2 x+\cos^2 x)-2\sin x\cos x=1-2\sin x\cos x\).

d) Sai. Ta có \(\left(\sin x+\cos x\right)^2+\left(\sin x-\cos x\right)^2=\sin^2x+2\sin x\cos x+\cos^2x+\sin^2x-2\sin x\cos x+\cos^2x=2\left(\sin^2x+\cos^2x\right)=2\).

Đáp án:

• \(\sin^2x+\cos^2x=1\)
• \(\tan x\cdot\cot x=1\)

Lời giải:

a) Đúng. Ta có \(\sin^2x+\cos^2x=1\).

b) Sai. Ta có \(\displaystyle\frac{1}{\cos^2x}=1+\tan^2x\).

c) Sai. Ta có \(\displaystyle\frac{1}{\sin^2x}=1+\cot^2x\).

d) Đúng. Ta có \(\tan x\cdot\cot x=1\).

Đáp án:

• \(\tan^2 x=\displaystyle\frac{\sin^2 x}{\cos^2 x}\)
• \(\tan^2x-\sin^2x=\tan^2x\cdot\sin^2x\)

Lời giải:

a) Sai. Ta có \(\tan x=\displaystyle\frac{\sin x}{\cos x}\).

b) Đúng. Ta có \(\tan^2 x=\left(\displaystyle\frac{\sin x}{\cos x}\right)^2=\displaystyle\frac{\sin^2 x}{\cos^2 x}\).

c) Sai. Ta có \(1+\tan^2 x=\displaystyle\frac{1}{\cos^2 x}\). Suy ra \(\displaystyle\frac{1}{\cos^2 x}-1=\tan^2 x\).

d) Đúng. Ta có \(\tan^2x-\sin^2x=\displaystyle\frac{\sin^2 x}{\cos^2 x}-\sin^2x=\sin^2x\left(\displaystyle\frac{1}{\cos^2 x}-1\right)=\sin^2x\cdot\tan^2x\).

Đáp án:

• \(\sin^2 x+\cos ^2 x=1\)
• \(\cos^4 x-\sin^4 x=\cos^2x-\sin^2x\)
• \(A=2\)

Lời giải:

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

b) Đúng. Ta có \(\cos^4 x-\sin^4x=(\cos^2 x-\sin^2x)(\cos^2 x+\sin^2x)=\cos^2 x-\sin^2x\).

c) Sai. Ta có

\begin{eqnarray*}A&=&2\cos^4x-\sin^4x+\sin^2x\cos^2x+3\sin^2x\\&=&\cos^4x-\sin^4x+\cos^2x\left(\cos^2x+\sin^2x\right)+3\sin^2x\\&=&\cos^2x-\sin^2x+\cos^2x+3\sin^2x\\&=&2\cos^2x+2\sin^2x.\end{eqnarray*}

d) Đúng. Ta có \(A=2\cos^2x+2\sin^2x=2(\cos^2 x+\sin^2x)=2\).

Đáp án:

• \(\tan ^2 \alpha+\cos ^2 \alpha-1=\sin^2\alpha \left(\displaystyle\frac{1}{\cos^2\alpha}-1 \right)\)
• \(\displaystyle\frac{\tan ^2 \alpha+\cos ^2 \alpha-1}{\cot ^2 \alpha+\sin ^2 \alpha-1}=\tan ^6 \alpha\)

Lời giải:

a) Sai. Ta có \(\sin^2\alpha+\cos^2\alpha=1\) nên \(\cos^2\alpha-1=-\sin^2\alpha\).

b) Đúng. Ta có \(\tan ^2 \alpha+\cos ^2 \alpha-1=\tan^2\alpha-\sin^2\alpha=\displaystyle\frac{\sin^2\alpha}{\cos^2\alpha}-\sin^2\alpha=\sin^2\alpha \left(\displaystyle\frac{1}{\cos^2\alpha}-1 \right)\).

c) Sai. Ta có \(\displaystyle\frac{1}{\cos^2\alpha}-1=\tan^2\alpha\).

d) Đúng. Ta có

\(\tan ^2 \alpha+\cos ^2 \alpha-1=\sin^2\alpha \left(\displaystyle\frac{1}{\cos^2\alpha}-1 \right)=\sin^2\alpha \cdot\tan^2\alpha\).\\

Tương tự, ta có \(\cot ^2 \alpha+\sin ^2 \alpha-1=\cot^2\alpha -\cos^2\alpha =\cos^2\alpha\left(\displaystyle\frac{1}{\sin^2\alpha}-1\right)=\cos^2\alpha\cdot \cot^2\alpha\).\\

Do đó

\(\displaystyle\frac{\tan ^2 \alpha+\cos ^2 \alpha-1}{\cot ^2 \alpha+\sin ^2 \alpha-1}= \displaystyle\frac{\sin^2\alpha \cdot\tan^2\alpha}{\cos^2\alpha\cdot \cot^2\alpha}=\tan^2\alpha \cdot \displaystyle\frac{\tan^2\alpha}{\cot^2\alpha}=\tan ^6 \alpha\).

Đáp án:

• Nếu \(\sin x+\cos x=m\) thì \(\sin x\cdot \cos x=\displaystyle\frac{m^2-1}{2}\)
• Nếu \(\sin x-\cos x=m\) thì \(\sin x\cdot \cos x=\displaystyle\frac{1-m^2}{2}\)
• Nếu \(\tan x+\cot x=m\) thì \(\tan^2x+\cot^2x=m^2-2\)
• Nếu \(\tan x-\cot x=m\) thì \(\tan^2x+\cot^2x=m^2+2\)

Lời giải:

Ta có

a) \(\sin x+\cos x=m\Rightarrow (\sin x+\cos x)^2=m^2\Leftrightarrow \sin x\cos x=\displaystyle\frac{m^2-1}{2}\).

b) \(\sin x-\cos x=m\Rightarrow (\sin x-\cos x)^2=m^2\Leftrightarrow \sin x\cos x=\displaystyle\frac{1-m^2}{2}\).

c) \(\tan x+\cot x=m\Rightarrow (\tan x+\cot x)^2=m^2\Leftrightarrow \tan^2x+\cot^2x=m^2-2\).

d) \(\tan x-\cot x=m\Rightarrow (\tan x-\cot x)^2=m^2\Leftrightarrow \tan^2x+\cot^2x=m^2+2\).

Đáp án:

• \(\cot x=-\displaystyle\frac{1}{2}\)
• \(A_1=-\displaystyle\frac{21}{11}\)

Lời giải:

a) Đúng. Ta có \(\tan x=-2\Rightarrow \cot x=-\displaystyle\frac{1}{2}\).

b) Sai. Vì \(\tan x=-2\) nên \(\cos x\ne 0\).

c) Đúng. \(A_1=\displaystyle\frac{-\displaystyle\frac{5}{2}+4\cdot(-2)}{-\displaystyle\frac{5}{2}-4\cdot (-2)}=-\displaystyle\frac{21}{11}\).

d) Sai. Chia cả tử và mẫu của biểu thức \(A_2\) cho \(\cos x\), ta được

\[A_2=\displaystyle\frac{\displaystyle\frac{2\sin x}{\cos x}+\displaystyle\frac{\cos x}{\cos x}}{\displaystyle\frac{\cos x}{\cos x}-\displaystyle\frac{3\sin x}{\cos x}}=\displaystyle\frac{2\tan x+1}{1-3\tan x}=\displaystyle\frac{2\cdot(-2)+1}{1-3\cdot(-2)}=-\displaystyle\frac{3}{7}.\]

Đáp án:

• \(\sin x\ne 0\)
• \(B_1=-8\)

Lời giải:

a) Đúng. Vì \(\cot x=2\) nên \(\sin x\ne 0\).

b) Đúng. Chia cả tử và mẫu của biểu thức \(B_1\) cho \(\sin x\), ta được

\[B_1=\displaystyle\frac{2\displaystyle\frac{\sin x}{\sin x}+3\displaystyle\frac{\cos x}{\sin x}}{3\displaystyle\frac{\sin x}{\sin x}-2\displaystyle\frac{\cos x}{\sin x}}=\displaystyle\frac{2+3\cot x}{3-2\cot x}=\displaystyle\frac{2+3\cdot 2}{3-2\cdot 2}=-8.\]

c) Sai. Chia cả tử và mẫu của biểu thức \(B_2\) cho \(\sin^{2}x\), ta được

\[B_2=\displaystyle\frac{\displaystyle\frac{2}{\sin^{2}x}}{\displaystyle\frac{\cos^{2}x}{\sin^{2}x}-\displaystyle\frac{\sin x\cdot \cos x}{\sin^{2}x}}=\displaystyle\frac{2\left(1+\cot^{2}x\right)}{\cot^2 x-\cot x}=\displaystyle\frac{2(1+2^2)}{2^2-2}=5.\]

d) Sai. \(B_1+B_2=-8+5=-3\).

a) Điểm \(M\) trên đường tròn lượng giác biểu diễn góc lượng giác có số đo là \(-\displaystyle\frac{\pi}{3}\) được xác định trong hình bên.

b) Ta có

\(\cos\left(-\displaystyle\frac{\pi}{3}\right)=\displaystyle\frac{1}{2}\); \(\sin\left(-\displaystyle\frac{\pi}{3}\right)=-\displaystyle\frac{\sqrt{3}}{2}\);

\(\tan\left(-\displaystyle\frac{\pi}{3}\right)=\displaystyle\frac{\sin\left(-\displaystyle\frac{\pi}{3}\right)}{\cos\left(-\displaystyle\frac{\pi}{3}\right)}=-\sqrt{3}\); \(\cot\left(-\displaystyle\frac{\pi}{3}\right)=\displaystyle\frac{\cos\left(-\displaystyle\frac{\pi}{3}\right)}{\sin\left(-\displaystyle\frac{\pi}{3}\right)}=-\displaystyle\frac{1}{\sqrt{3}}\).

Lấy điểm \(M\) trên đường tròn lượng giác sao cho \((OA,OM)=\alpha=120^\circ\) (hình bên). Gọi \(H\), \(K\) lần lượt là hình chiếu của điểm \(M\) trên các trục \(Ox\), \(Oy\).

Khi đó, ta có \(\widehat{AOM}=120^\circ\), suy ra \(\widehat{BOM}=\widehat{KOM}=30^\circ\).

Theo hệ thức trong tam giác vuông \(KOM\), ta có

\(OK=OM\cdot \cos \widehat{KOM}=\cos 30^\circ=\displaystyle\frac{\sqrt{3}}{2}\)

và

\(MK=OM\cdot \sin \widehat{KOM}=\sin 30^\circ=\displaystyle\frac{1}{2}.\)

Do đó, \(M\left(-\displaystyle\frac{1}{2};\displaystyle\frac{\sqrt{3}}{2}\right)\).

Vậy \(\sin 120^\circ=\displaystyle\frac{\sqrt{3}}{2}\); \(\cos 120^\circ=-\displaystyle\frac{1}{2}\); \(\tan 120^\circ=\displaystyle\frac{\sin 120^\circ}{\cos 120^\circ}=-\sqrt{3}\); \(\cot 120^\circ=\displaystyle\frac{\cos 120^\circ}{\sin 120^\circ}=-\displaystyle\frac{\sqrt{3}}{3}\).

Lấy điểm \(M\) trên đường tròn lượng giác sao cho \((OA,OM)=\alpha=-\displaystyle\frac{\pi}{4}=-45^\circ\) (hình bên). Gọi \(H\), \(K\) lần lượt là hình chiếu của điểm \(M\) trên các trục \(Ox\), \(Oy\).

Khi đó, ta có \(\widehat{AOM}=45^\circ\), suy ra \(\widehat{B'OM}=\widehat{KOM}=45^\circ\).

Theo hệ thức trong tam giác vuông \(KOM\), ta có

\(OK=OM\cdot \cos \widehat{KOM}=\cos 45^\circ=\displaystyle\frac{\sqrt{2}}{2}.\)

và

\(MK=OM\cdot \sin \widehat{KOM}=\sin 45^\circ=\displaystyle\frac{\sqrt{2}}{2}.\)

Do đó, \(M\left(-\displaystyle\frac{\sqrt{2}}{2};\displaystyle\frac{\sqrt{2}}{2}\right)\).

Vậy \(\sin \left(-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{\sqrt{2}}{2}\); \(\cos \left(-\displaystyle\frac{\pi}{4}\right)=-\displaystyle\frac{\sqrt{2}}{2}\); \(\tan \left(-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{\sin \left(-\displaystyle\frac{\pi}{4}\right)}{\cos \left(-\displaystyle\frac{\pi}{4}\right)}=-1\); \(\cot \left(-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{\cos \left(-\displaystyle\frac{\pi}{4}\right)}{\sin \left(-\displaystyle\frac{\pi}{4}\right)}=-1\).

a) Vì \(\displaystyle\frac{13 \pi}{3}=\displaystyle\frac{\pi}{3}+4 \pi\) nên

\(\begin{array}{ll}\sin \displaystyle\frac{13 \pi}{3}=\sin \displaystyle\frac{\pi}{3}=\displaystyle\frac{\sqrt{3}}{2}; & \cos \displaystyle\frac{13 \pi}{3}=\cos \displaystyle\frac{\pi}{3}=\displaystyle\frac{1}{2}; \\ \tan \displaystyle\frac{13 \pi}{3}=\displaystyle\frac{\sin \displaystyle\frac{13 \pi}{3}}{\cos \displaystyle\frac{13 \pi}{3}}=\sqrt{3}; & \cot \displaystyle\frac{13 \pi}{3}=\displaystyle\frac{\cos \displaystyle\frac{13 \pi}{3}}{\sin \displaystyle\frac{13 \pi}{3}}=\displaystyle\frac{\sqrt{3}}{3}.\end{array}\)

b) Vì điểm biểu diễn của góc \(-45^\circ\) và góc \(45^\circ\) trên đường tròn lượng giác đối xứng nhau qua trục hoành nên chúng có cùng hoành độ và tung độ đối nhau. Do đó ta có

\(\begin{array}{ll}\sin \left(-45^\circ\right)=-\sin 45^\circ=\displaystyle\frac{-\sqrt{2}}{2} ; & \cos \left(-45^\circ\right)=\cos 45^\circ=\displaystyle\frac{\sqrt{2}}{2}; \\ \tan \left(-45^\circ\right)=\displaystyle\frac{\sin \left(-45^\circ\right)}{\cos \left(-45^\circ\right)}=-1 ; & \cot \left(-45^\circ\right)=\displaystyle\frac{\cos \left(-45^\circ\right)}{\sin \left(-45^\circ\right)}=-1.\end{array}\)

Ta có

\begin{eqnarray*}P&=&\tan^2 \displaystyle\frac{\pi}{3}+\sin^2\displaystyle\frac{\pi}{4}+\cot \displaystyle\frac{\pi}{4}+\cos \displaystyle\frac{\pi}{2}\\ &=&(\sqrt{3})^2+\left(\displaystyle\frac{\sqrt{2}}{2}\right)^2+1+0\\ &=&\displaystyle\frac{9}{2}.\end{eqnarray*}

Ta có

\begin{eqnarray*}P&=& \cos^2 \displaystyle\frac{\pi}{3}+\tan \displaystyle\frac{\pi}{4}+\cot^2\displaystyle\frac{\pi}{6}+\sin \displaystyle\frac{\pi}{2}\\ &=&\left(\displaystyle\frac{1}{2}\right)^2+1+\left(\sqrt{3}\right)^2+1=\displaystyle\frac{21}{4}.\end{eqnarray*}

a) Xét \(\alpha=\displaystyle\frac{\pi}{3}+k2\pi\) (\(k\in \mathbb{Z}\)), ta có

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

+) \(\cos \alpha=\cos \displaystyle\frac{\pi}{3}=\displaystyle\frac{1}{2}\);

+) \(\tan \alpha=\tan \displaystyle\frac{\pi}{3}=\sqrt{3}\);

+) \(\cot \alpha=\tan \displaystyle\frac{\pi}{3}=\displaystyle\frac{\sqrt{3}}{3}\).

b) Xét \(\alpha=k\pi\) (\(k\in \mathbb{Z}\)), ta có

+) \(\sin \alpha=0\);

+) \(\cos \alpha=(-1)^k\);

+) \(\tan \alpha=0\);

+) \(\cot \alpha\) không xác định.

c) Xét \(\alpha=\displaystyle\frac{\pi}{2}+k\pi\) (\(k\in \mathbb{Z}\)), ta có

+) \(\sin \alpha=(-1)^k\);

+) \(\cos \alpha=0\);

+) \(\cot \alpha=0\);

+) \(\tan \alpha\) không xác định.

d) Xét \(\alpha=\displaystyle\frac{\pi}{4}+k\pi\) (\(k\in \mathbb{Z}\)), ta có

+) \(\sin \alpha=(-1)^k\cdot \displaystyle\frac{\sqrt{2}}{2}\);

+) \(\cos \alpha=(-1)^k\cdot \displaystyle\frac{\sqrt{2}}{2}\);

+) \(\tan \alpha=1\);

+) \(\cot \alpha=1\).

+) \(\sin 225^\circ=\sin (45^\circ+180^\circ)=-\sin 45^\circ=-\displaystyle\frac{\sqrt{2}}{2}\);

\(\cos 225^\circ=\cos (45^\circ+180^\circ)=-\cos 45^\circ=-\displaystyle\frac{\sqrt{2}}{2}\);

\(\tan 225^\circ=\displaystyle\frac{\sin 225^\circ}{\cos 225^\circ}=1\);

\(\cot 225^\circ=\displaystyle\frac{\cos 225^\circ}{\sin 225^\circ}=1\).

+) \(\sin (-225^\circ)=\sin (-45^\circ+180^\circ)=-\sin (-45^\circ)=\sin 45^\circ=\displaystyle\frac{\sqrt{2}}{2}\);

\(\cos (-225^\circ)=\cos (-45^\circ+180^\circ)=-\cos (-45^\circ)=-\cos 45^\circ=-\displaystyle\frac{\sqrt{2}}{2}\);

\(\tan (-225^\circ)=\displaystyle\frac{\sin (-225^\circ)}{\cos (-225^\circ)}=-1\);

\(\cot (-225^\circ)=\displaystyle\frac{\cos (-225^\circ)}{\sin (-225^\circ)}=-1\).

+) \(\sin (-1~035^\circ)=\sin (45^\circ -3\cdot 360^\circ)=\sin 45^\circ=\displaystyle\frac{\sqrt{2}}{2}\);

\(\cos (-1~035^\circ)=\cos (45^\circ -3\cdot 360^\circ)=\cos 45^\circ=\displaystyle\frac{\sqrt{2}}{2}\);

\(\tan (-1~035^\circ)=\displaystyle\frac{\sin (-1~035^\circ)}{\cos (-1~035^\circ)}=1\);

\(\cot (-1~035^\circ)=\displaystyle\frac{\cos (-1~035^\circ)}{\sin (-1~035^\circ)}=1\);

+) \(\sin \displaystyle\frac{5\pi}{3}=\sin \left(2\pi-\displaystyle\frac{\pi}{3}\right)=\sin \left(-\displaystyle\frac{\pi}{3}\right)=-\sin \displaystyle\frac{\pi}{3}=-\displaystyle\frac{\sqrt{3}}{2}\);

\(\cos \displaystyle\frac{5\pi}{3}=\cos \left(2\pi-\displaystyle\frac{\pi}{3}\right)=\cos \left(-\displaystyle\frac{\pi}{3}\right)=\cos \displaystyle\frac{\pi}{3}=\displaystyle\frac{1}{2}\);

\(\tan \displaystyle\frac{5\pi}{3}=\displaystyle\frac{\sin \displaystyle\frac{5\pi}{3}}{\cos \displaystyle\frac{5\pi}{3}}=-\sqrt{3}\);

\(\cot \displaystyle\frac{5\pi}{3}=\displaystyle\frac{\cos \displaystyle\frac{5\pi}{3}}{\sin \displaystyle\frac{5\pi}{3}}=-\displaystyle\frac{\sqrt{3}}{3}\).

+) \(\sin \displaystyle\frac{19\pi}{2}=\sin \left(10\pi-\displaystyle\frac{\pi}{2}\right)=\sin \left(-\displaystyle\frac{\pi}{2}\right)=-\sin \displaystyle\frac{\pi}{2}=-1\);

\(\cos \displaystyle\frac{19\pi}{2}=\cos \left(10\pi-\displaystyle\frac{\pi}{2}\right)=\cos \left(-\displaystyle\frac{\pi}{2}\right)=\cos \displaystyle\frac{\pi}{2}=0\);

\(\cot \displaystyle\frac{19\pi}{2}=\displaystyle\frac{\cos \displaystyle\frac{19\pi}{2}}{\sin \displaystyle\frac{19\pi}{2}}=0\).

+) \(\sin \left(-\displaystyle\frac{159\pi}{4}\right)=\sin \left(\displaystyle\frac{\pi}{4}-40\pi\right)=\sin \displaystyle\frac{\pi}{4}=\displaystyle\frac{\sqrt{2}}{2}\);

\(\cos \left(-\displaystyle\frac{159\pi}{4}\right)=\cos \left(\displaystyle\frac{\pi}{4}-40\pi\right)=\cos \displaystyle\frac{\pi}{4}=\displaystyle\frac{\sqrt{2}}{2}\);

\(\tan \left(-\displaystyle\frac{159\pi}{4}\right)=\displaystyle\frac{\sin \left(-\displaystyle\frac{159\pi}{4}\right)}{\cos \left(-\displaystyle\frac{159\pi}{4}\right)}=1\);

\(\cot \left(-\displaystyle\frac{159\pi}{4}\right)=\displaystyle\frac{\cos \left(-\displaystyle\frac{159\pi}{4}\right)}{\sin \left(-\displaystyle\frac{159\pi}{4}\right)}=1\).

Do \(-\pi<-\displaystyle\frac{3\pi}{4}<-\displaystyle\frac{\pi}{2}\) nên \(\sin \left(-\displaystyle\frac{3\pi}{4}\right)<0\); \(\cos \left(-\displaystyle\frac{3\pi}{4}\right)<0\); \(\tan \left(-\displaystyle\frac{3\pi}{4}\right)>0\); \(\cot \left(-\displaystyle\frac{3\pi}{4}\right)>0\).

Do \(\displaystyle\frac{\pi}{2}<\displaystyle\frac{5\pi}{6}<\pi\) nên \(\sin \left(\displaystyle\frac{5\pi}{6}\right)>0\); \(\cos \left(\displaystyle\frac{5\pi}{6}\right)<0\); \(\tan \left(\displaystyle\frac{5\pi}{6}\right)<0\); \(\cot \left(\displaystyle\frac{5\pi}{6}\right)<0\).

Ta có \(\sin ^2 \alpha=1-\cos ^2 \alpha=\displaystyle\frac{7}{16}\).

Do đó \(\sin \alpha=-\displaystyle\frac{\sqrt{7}}{4}\) hoặc \(\sin \alpha=\displaystyle\frac{\sqrt{7}}{4}\).

Vì \(-\displaystyle\frac{\pi}{2}<\alpha<0\) nên điểm biểu diễn của góc \(\alpha\) trên đường tròn lượng giác thuộc góc phần tư thứ \(IV\) (Hình 6), do đó \(\sin \alpha < 0\).

Suy ra \(\sin \alpha=-\displaystyle\frac{\sqrt{7}}{4}\).

Do đó \(\tan \alpha=\displaystyle\frac{\sin \alpha}{\cos \alpha}=-\displaystyle\frac{\sqrt{7}}{3}\) và \(\cot \alpha=\displaystyle\frac{1}{\tan \alpha}=-\displaystyle\frac{3 \sqrt{7}}{7}\).

Do \(\tan \alpha=-2\) nên \(\displaystyle\frac{\sin \alpha}{\cos \alpha}=-2\), suy ra \(\sin \alpha=-2\cos \alpha\).

Vì \(\cos^2\alpha+\sin^2\alpha=1\) nên \(\cos^2\alpha+(-2\cos\alpha)^2=1\), suy ra \(\cos^2 \alpha=\displaystyle\frac{1}{5}\).

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

Từ đó ta có \(\cos \alpha=\sqrt{\displaystyle\frac{1}{5}}=\displaystyle\frac{1}{\sqrt{5}}\), suy ra \(\sin \alpha=-2\cdot \displaystyle\frac{1}{\sqrt{5}}=-\displaystyle\frac{2}{\sqrt{5}}\).

Vì \(\cos^2\alpha+\sin^2\alpha=1\) nên \(\cos^2\alpha=1-\sin^2\alpha=1-\left(-\displaystyle\frac{4}{5}\right)^2=\displaystyle\frac{9}{25}\).

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

Từ đó ta có \(\cos \alpha=-\sqrt{\displaystyle\frac{9}{25}}=-\displaystyle\frac{3}{4}\).

a) Ta có \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow \sin^2\alpha=1-\cos^2\alpha\) \(\Leftrightarrow \sin\alpha=\sqrt{1-\cos^2\alpha}\) hoặc \(\sin\alpha=-\sqrt{1-\cos^2\alpha}.\)

Vì \(0<\alpha<\displaystyle\frac{\pi}{2}\) nên \(\sin\alpha>0\), suy ra \(\sin\alpha=\sqrt{1-\cos^2\alpha}\)

\(\Rightarrow\sin\alpha=\sqrt{1-\left(\displaystyle\frac{1}{5}\right)^2}=\sqrt{\displaystyle\frac{24}{25}}=\displaystyle\frac{2\sqrt{6}}{5}\).

Mà \(\tan\alpha=\displaystyle\frac{\sin\alpha}{\cos\alpha}\) nên \(\tan\alpha=\displaystyle\frac{2\sqrt{6}}{5}:\displaystyle\frac{1}{5}=2\sqrt{6}\) và \(\cot\alpha=\displaystyle\frac{\sqrt{6}}{12}\).

b) Ta có \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow \cos^2\alpha=1-\sin^2\alpha\) \(\Leftrightarrow \cos\alpha=\sqrt{1-\sin^2\alpha}\) hoặc \(\cos\alpha=-\sqrt{1-\sin^2\alpha}.\)

Vì \(\displaystyle\frac{\pi}{2}<\alpha<\pi\) nên \(\cos\alpha<0\), suy ra \(\cos\alpha=-\sqrt{1-\sin^2\alpha}\)

\(\Rightarrow\cos\alpha=-\sqrt{1-\left(\displaystyle\frac{2}{5}\right)^2}=-\sqrt{\displaystyle\frac{21}{25}}=-\displaystyle\frac{\sqrt{21}}{5}\).

Mà \(\tan\alpha=\displaystyle\frac{\sin\alpha}{\cos\alpha}\) nên \(\tan\alpha=\displaystyle\frac{2}{5}:\left(-\displaystyle\frac{\sqrt{21}}{5}\right)=-\displaystyle\frac{2\sqrt{21}}{21}\) và \(\cot\alpha=-\displaystyle\frac{\sqrt{21}}{2}\).

c) Vì \(\pi<\alpha<\displaystyle\frac{3\pi}{2}\) nên \(\cos\alpha<0\), do đó từ công thức:

\(1+\tan^2\alpha=\displaystyle\frac{1}{\cos^2\alpha}\), suy ra \(\cos^2\alpha=\displaystyle\frac{1}{1+\tan^2\alpha}\).

\(\Rightarrow\cos\alpha=-\displaystyle\frac{1}{\sqrt{1+\tan^2\alpha}}=-\displaystyle\frac{1}{\sqrt{1+(\sqrt{5})^2}}=-\displaystyle\frac{\sqrt{6}}{6}\);

\(\cot\alpha=\displaystyle\frac{1}{\tan\alpha}=\displaystyle\frac{\sqrt{5}}{5}\) và \(\sin\alpha=\tan\alpha\cdot\cos\alpha=-\displaystyle\frac{\sqrt{30}}{6}\).

d) Vì \(\displaystyle\frac{3\pi}{2}<\alpha<2\pi\) nên \(\sin\alpha<0\), do đó từ công thức:

\(1+\cot^2\alpha=\displaystyle\frac{1}{\sin^2\alpha}\), suy ra \(\sin^2\alpha=\displaystyle\frac{1}{1+\cot^2\alpha}\).

\(\Rightarrow\sin\alpha=-\displaystyle\frac{1}{\sqrt{1+\cot^2\alpha}}=-\displaystyle\frac{1}{\sqrt{1+\left(-\displaystyle\frac{1}{\sqrt{2}}\right)^2}}=-\displaystyle\frac{\sqrt{6}}{3}\);

\(\tan\alpha=\displaystyle\frac{1}{\cot\alpha}=-\sqrt{2}\) và \(\cos\alpha=\cot\alpha\cdot\sin\alpha=\displaystyle\frac{\sqrt{3}}{3}\).

a) Xét \(\sin \alpha=\displaystyle\frac{\sqrt{15}}{4}\) với \(\displaystyle\frac{\pi}{2}<\alpha<\pi\);

Vì \(\sin^2\alpha+\cos^2\alpha=1\) nên \(\cos^2\alpha=1-\sin^2\alpha=1-\left(\displaystyle\frac{\sqrt{15}}{4}\right)^2=\displaystyle\frac{1}{16}\).

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

Từ đó ta có \(\cos \alpha=-\sqrt{\displaystyle\frac{1}{16}}=-\displaystyle\frac{1}{4}\).

Ta có \(\tan \alpha=\displaystyle\frac{\sin \alpha}{\cos \alpha}=-\sqrt{15}\); \(\cot \alpha=\displaystyle\frac{\cos \alpha}{\sin \alpha}=-\displaystyle\frac{\sqrt{15}}{15}\).

b) Xét \(\cos \alpha=-\displaystyle\frac{2}{3}\) với \(-\pi<\alpha<0\);

Vì \(\sin^2\alpha+\cos^2\alpha=1\) nên \(\sin^2\alpha=1-\cos^2\alpha=1-\left(-\displaystyle\frac{2}{3}\right)^2=\displaystyle\frac{5}{9}\).

Do \(-\pi<\alpha<0\) nên \(\sin \alpha<0\).

Từ đó ta có \(\sin \alpha=-\sqrt{\displaystyle\frac{5}{9}}=-\displaystyle\frac{\sqrt{5}}{3}\).

Ta có \(\tan \alpha=\displaystyle\frac{\sin \alpha}{\cos \alpha}=-\displaystyle\frac{\sqrt{5}}{2}\); \(\cot \alpha=\displaystyle\frac{\cos \alpha}{\sin \alpha}=-\displaystyle\frac{2\sqrt{5}}{5}\).

c) Xét \(\tan \alpha=3\) với \(-\pi<\alpha<0\);

Ta có \(\cot \alpha=\displaystyle\frac{1}{3}\).

Do \(\tan \alpha=3\) nên \(\displaystyle\frac{\sin \alpha}{\cos \alpha}=3\), suy ra \(\sin \alpha=3\cos \alpha\).

Vì \(\cos^2\alpha+\sin^2\alpha=1\) nên \(\cos^2\alpha+(3\cos\alpha)^2=1\), suy ra \(\cos^2 \alpha=\displaystyle\frac{1}{10}\).

+) Do \(-\displaystyle\frac{\pi}{2}<\alpha<0\) nên \(\cos \alpha>0\).

Từ đó ta có \(\cos \alpha=\displaystyle\frac{\sqrt{10}}{10}\).

+) Do \(-\pi<\alpha<-\displaystyle\frac{\pi}{2}\) nên \(\cos \alpha<0\).

Từ đó ta có \(\cos \alpha=-\displaystyle\frac{\sqrt{10}}{10}\).

d) Xét \(\cot \alpha=-2\) với \(0<\alpha<\pi\);

Ta có \(\tan \alpha=-\displaystyle\frac{1}{2}\).

Do \(\cot \alpha=-2\) nên \(\displaystyle\frac{\cos \alpha}{\sin \alpha}=-2\), suy ra \(\cos \alpha=-2\sin \alpha\).

Vì \(\cos^2\alpha+\sin^2\alpha=1\) nên \((-2\sin \alpha)^2+\sin^2\alpha=1\), suy ra \(\sin^2 \alpha=\displaystyle\frac{1}{5}\).

Do \(0<\alpha<\pi\) nên \(\sin \alpha>0\).

Từ đó ta có \(\sin \alpha=\displaystyle\frac{\sqrt{5}}{5}\).

a) Ta có

\(\sin \displaystyle\frac{13\pi}{4}=\sin \left(3\pi+\displaystyle\frac{\pi}{4}\right)\) \(=\sin \left(\pi+\displaystyle\frac{\pi}{4}\right)=-\sin \displaystyle\frac{\pi}{4}=-\displaystyle\frac{\sqrt{2}}{2}\);

b) Ta có

\(\sin \displaystyle\frac{\pi}{10}-\cos \displaystyle\frac{2\pi}{5}=\sin \displaystyle\frac{\pi}{10}-\sin \left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{2\pi}{5}\right)\) \(=\sin \displaystyle\frac{\pi}{10}-\sin \displaystyle\frac{\pi}{10}=0\).

a) Ta có

\(\cos^2 \displaystyle\frac{\pi}{8}+\cos^2 \displaystyle\frac{3\pi}{8} =\cos^2 \displaystyle\frac{\pi}{8}+\cos^2 \left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{\pi}{8}\right)\) \(=\cos^2 \displaystyle\frac{\pi}{8}+\sin^2 \displaystyle\frac{\pi}{8}=1.\)

b) Ta có

\begin{eqnarray*}& &\tan 1^\circ\cdot \tan 2^\circ\cdot \tan 45^\circ\cdot \tan 88^\circ\cdot \tan 89^\circ\\ &=&\tan 1^\circ\cdot \tan 2^\circ\cdot \tan 45^\circ\cdot \tan (90^\circ-2^\circ)\cdot \tan (90^\circ-1^\circ)\\ &=&\tan 1^\circ\cdot \tan 2^\circ\cdot \tan 45^\circ\cdot \cot 2^\circ\cdot \cot 1^\circ\\ &=&\tan 1^\circ\cdot \cot 1^\circ\cdot \tan 2^\circ\cdot \cot 2^\circ\cdot \tan 45^\circ\\ &=&\tan 45^\circ=1.\end{eqnarray*}

a) \(\sin \displaystyle\frac{61 \pi}{8}=\sin \left(8 \pi-\displaystyle\frac{3 \pi}{8}\right)=\sin \left(-\displaystyle\frac{3 \pi}{8}\right)\) \(=-\sin \displaystyle\frac{3 \pi}{8}=-\cos \left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{3 \pi}{8}\right)=-\cos \displaystyle\frac{\pi}{8}\);

b) \(\tan 258^\circ =\tan \left(180^\circ +78^\circ \right)=\tan 78^\circ\) \(=\cot \left(90^\circ -12^\circ \right)=\cot 12^\circ\).

a) Ta có

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

b) Ta có

\begin{eqnarray*}\displaystyle\frac{\cos^2\alpha+\tan^2\alpha-1}{\sin^2\alpha}&=& \displaystyle\frac{\cos^2\alpha-1}{\sin^2\alpha}+\displaystyle\frac{\tan^2\alpha}{\sin^2\alpha}\\ &= & \displaystyle\frac{-\sin^2\alpha}{\sin^2\alpha}+\displaystyle\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\displaystyle\frac{1}{\sin^2\alpha}\\ &= & \displaystyle\frac{1}{\cos^2\alpha}-1\\ &=&\displaystyle\frac{1-\cos^2\alpha}{\cos^2\alpha}\\ &=&\displaystyle\frac{\sin^2\alpha}{\cos^2\alpha}=\tan^2\alpha.\end{eqnarray*}

Ta có \(\displaystyle\frac{\sin i}{\sin r}=\displaystyle\frac{n_2}{n_1}\Leftrightarrow \displaystyle\frac{\sin 50^\circ}{\sin r}\) \(=\displaystyle\frac{1{,}33}{1}\Leftrightarrow \sin r=\displaystyle\frac{\sin 50^\circ}{1{,}33}\Rightarrow r\approx 35{,}17^\circ\).

Sau một phút quay được \(660\) rad.

Khoảng cách từ \(V\) đến mặt đất là

\(OA-VV'=OA-OA\sin 660=58-58\sin 660 \approx 42{,}8\) cm.

Gắn hệ trục toạ độ \(Oxy\) như hình vẽ.

Khi thanh quay \(3\) vòng thì trở về vị trí ban đầu \(OA\).

Quay thêm \(\displaystyle\frac{1}{10}\) vòng thì ta được \(\widehat{AOM}=\displaystyle\frac{360^\circ}{10}=36^\circ\).

Khi đó, \(O'M'=OP=15\cdot \cos 36^\circ \approx 12{,}1\) cm.