Ôn tập chương VI (phần 2)

$\frac{23\mathrm{\pi }}{6}=-\frac{\mathrm{\pi }}{6}+4\mathrm{\pi } \mathrm{nên} \sin \frac{23\mathrm{\pi }}{6}=\sin \left(-\frac{\mathrm{\pi }}{6}\right)=-\sin \frac{\mathrm{\pi }}{6}=-\frac{1}{2}$

$\cos \frac{23\mathrm{\pi }}{6}=\cos \left(\frac{-\mathrm{\pi }}{6}+4\mathrm{\pi }\right)=\cos \left(-\frac{\mathrm{\pi }}{6}\right)=\cos \frac{\mathrm{\pi }}{6}=\frac{\sqrt{3}}{2}$

$\frac{17\mathrm{\pi }}{6}=-\frac{\mathrm{\pi }}{6}+\mathrm{\pi }+2\mathrm{\pi } \mathrm{nên} \sin \frac{17\mathrm{\pi }}{6}=\sin \left(-\frac{\mathrm{\pi }}{6}+\mathrm{\pi }\right)=-\sin \frac{-\mathrm{\pi }}{6}=\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}$

$\cos \frac{17\mathrm{\pi }}{6}=\cos \left(\frac{-\mathrm{\pi }}{6}+\mathrm{\pi }+2\mathrm{\pi }\right)=\cos \left(\frac{-\mathrm{\pi }}{6}+\mathrm{\pi }\right)=-\cos \frac{-\mathrm{\pi }}{6}=-\cos \frac{\mathrm{\pi }}{6}=-\frac{\sqrt{3}}{2}$

$\frac{157\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{3}+52\mathrm{\pi } \mathrm{nên} \tan \frac{157\mathrm{\pi }}{3}=\tan \frac{\mathrm{\pi }}{3}=\sqrt{3}$

$-\frac{105\mathrm{\pi }}{6}=-\frac{\mathrm{\pi }}{2}-17\mathrm{\pi } \mathrm{nên} \cot \left(-\frac{105\mathrm{\pi }}{6}\right)=\cot \left(-\frac{\mathrm{\pi }}{2}\right)=-\cot \frac{\mathrm{\pi }}{2}=0$

Ta có:  $\cos 2\alpha =2{\cos }^2\alpha -1=2.\frac{9}{16}-1=\frac{1}{8}$

Vì $\alpha \in \left(\frac{3\mathrm{\pi }}{2};2\mathrm{\pi }\right)\Rightarrow 2\alpha \in \left(3\mathrm{\pi };4\mathrm{\pi }\right)\Rightarrow \sin 2\alpha <0$

$$\begin{gathered} {\sin }^{2}2\alpha +{\cos }^{2}2\alpha =1\Rightarrow {\sin }^{2}2\alpha =1-{\cos }^{2}2\alpha =1-\frac{1}{64}=\frac{63}{64} \\ \Rightarrow \sin 2\alpha =\frac{-3\sqrt{7}}{8} \end{gathered}$$

Do $\alpha \in \left(\mathrm{\pi };\frac{3\mathrm{\pi }}{2}\right)\Rightarrow \cos \alpha <0$

Chọn C.

$$\begin{gathered} {\tan }^3\alpha +co{t}^3\alpha =\left(\tan \alpha +cot\alpha \right).\left({\tan }^2\alpha -\tan \alpha .cot\alpha +co{t}^2\alpha \right) \\ =4.\left[{\left(\tan \alpha +cot\alpha \right)}^2-3\tan \alpha .cot\alpha \right]=4.\left(4^2-3.1\right)=52 \end{gathered}$$

Ta có: $\sin \frac{4\mathrm{\pi }}{5}=\sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{5}\right)=\sin \frac{\mathrm{\pi }}{5}$

Và  $cos\frac{4\mathrm{\pi }}{5}=\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{5}\right)=-\cos \frac{\mathrm{\pi }}{5}$

$\Rightarrow \sin \frac{\mathrm{\pi }}{5}+cos\frac{\mathrm{\pi }}{5}-\sin \frac{4\mathrm{\pi }}{5}+\cos \frac{4\mathrm{\pi }}{5}=\sin \frac{\mathrm{\pi }}{5}+\cos \frac{\pi }{5}-\sin \frac{\mathrm{\pi }}{5}-cos\frac{\mathrm{\pi }}{5}=0$.

Ta có: $\sin \frac{3\mathrm{\pi }}{10}=\sin \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{5}\right)=\cos \frac{\mathrm{\pi }}{5}$

Và  $\cos \frac{3\mathrm{\pi }}{10}=\cos \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{5}\right)=\sin \frac{\mathrm{\pi }}{5}$

$\Rightarrow \sin \frac{\mathrm{\pi }}{5}+\cos \frac{\mathrm{\pi }}{5}-\sin \frac{3\mathrm{\pi }}{10}-\cos \frac{3\mathrm{\pi }}{10}=\sin \frac{\mathrm{\pi }}{5}+\cos \frac{\pi }{5}-\cos \frac{\mathrm{\pi }}{5}-\sin \frac{\mathrm{\pi }}{5}=0$.

* Ta có ${\sin }^2\alpha +{\cos }^2\alpha =1\Rightarrow {\sin }^2\alpha +\frac{16}{25}=1\Rightarrow {\sin }^2\alpha =\frac{9}{25}$

Mà $\sin \alpha <0\Rightarrow \sin \alpha =-\frac{3}{5}$

* Vì ${\sin }^2\beta +{\cos }^2\beta =1\Rightarrow \frac{9}{16}+{\cos }^2\beta =1\Rightarrow co{s}^2\beta =\frac{7}{16}$

 $\cos \beta >0\Rightarrow \cos \beta =\frac{\sqrt{7}}{4}$

* $\sin \left(\alpha +\beta \right)=\sin \alpha .\cos \beta +cos\alpha .\sin \beta =\frac{-3}{5}.\frac{\sqrt{7}}{4}+\frac{4}{5}.\frac{3}{4}=\frac{12-3\sqrt{7}}{20}$

Vì $cos\alpha =1$ nên $\alpha =k2\pi$, từ đó

$\sin \left(\alpha -\frac{\mathrm{\pi }}{6}\right)=\sin \left(-\frac{\mathrm{\pi }}{6}\right)=-\frac{1}{2};\sin \left(\alpha +\frac{\mathrm{\pi }}{6}\right)=\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}$.

${\sin }^2\left(\alpha -\frac{\mathrm{\pi }}{6}\right)+{\sin }^2\left(\alpha +\frac{\mathrm{\pi }}{6}\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Do tổng ba góc của tam giác bằng 180$°$ nên:

     M + N + P = 180$°$ =>  M + N = 180$°$- P

Suy ra:  2M + 2N = 360$°$ – 2P

Do đó, cos(2M + 2N ) = cos(360$°$-  2P) =  cos(- 2P) = cos2P

 $$\begin{gathered} {\cos }^{2}2M+{\cos }^{2}2N+{\cos }^{2}2P=\frac{1+cos4M}{2}+\frac{1+\cos 4N}{2}+co{s}^{2}2P \\ =1+\frac{1}{2}.\left(cos4M+\cos 4N\right)+{\cos }^{2}2P \\ =1+\frac{1}{2}.2.cos\left(2M+2N\right).\cos \left(2M-2N\right)+{\cos }^{2}2P \\ =1+\cos 2P.\cos (2M-2N)+{\cos }^{2}2P=1+\cos 2P.\left[\cos \left(2M-2N\right)+\cos 2P\right] \\ =1+\cos 2P.\left[\cos \left(2M-2N\right)+\cos \left(2M+2N\right)\right] \\ =1+2\cos 2P.\cos 2M.\cos 2N \end{gathered}$$