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.