Ôn tập chương VI (phần 1)

$$\begin{gathered} 2\cos \frac{\mathrm{\pi }}{7}.\cos \frac{2\mathrm{\pi }}{7}+\cos \frac{4\mathrm{\pi }}{7}+\cos \frac{6\mathrm{\pi }}{7} \\ =2\cos \frac{\mathrm{\pi }}{7}.\cos \frac{2\mathrm{\pi }}{7}+2\cos \frac{5\mathrm{\pi }}{7}\cos \frac{\mathrm{\pi }}{7} \\ =2\cos \frac{\mathrm{\pi }}{7}\left(\cos \frac{2\mathrm{\pi }}{7}+\cos \frac{5\mathrm{\pi }}{7}\right) \\ =4\cos \frac{\mathrm{\pi }}{7}\cos \frac{\mathrm{\pi }}{2}\cos \frac{3\mathrm{\pi }}{14} \\ =4\cos \frac{\mathrm{\pi }}{7}.0.\cos \frac{3\mathrm{\pi }}{14}=0 \end{gathered}$$

Chọn B

Ta có

$$\begin{gathered} \sin \frac{6\mathrm{\pi }}{7}+\sin \frac{8\mathrm{\pi }}{7}=2.\sin \left(\frac{{\displaystyle \frac{6\mathrm{\pi }}{7}}+{\displaystyle \frac{8\mathrm{\pi }}{7}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{8\mathrm{\pi }}{7}-\frac{6\mathrm{\pi }}{7}}}{2}\right) \\ =2\mathrm{sin\pi }.\cos \frac{\mathrm{\pi }}{7}=2.0.\cos \frac{\mathrm{\pi }}{7}=0 \end{gathered}$$

$$\begin{gathered} \cos \alpha =2.{\cos }^2\frac{\alpha }{2}-1=2.0-1=-1 \\ \cos 2\alpha =2{\cos }^2\alpha -1=2.{(-1)}^2-1=1 \\ \cos 4\alpha =2{\cos }^{2}2\alpha -1=2.1^2-1=1 \\ \cos 7\alpha =\cos \left(a+6a\right)=\cos \alpha .\cos 6\alpha -\sin \alpha \sin 6\alpha =-\cos (3.2\alpha )=-4{\cos }^{3}2\alpha +3\cos 2\alpha =-4+3=-1 \\ \cos \alpha +\cos 2\alpha +\cos 4\alpha +\cos 7\alpha =-1+1+1+(-1)=0 \end{gathered}$$

$$\begin{gathered} \cos \left(\frac{-\mathrm{\pi }}{4}\right).\cos \left(\frac{3\mathrm{\pi }}{4}\right)+\sin \left(\frac{-\mathrm{\pi }}{4}\right).\sin \left(\frac{3\mathrm{\pi }}{4}\right) \\ =\cos \left(\frac{-\mathrm{\pi }}{4}-\frac{3\mathrm{\pi }}{4}\right)=\cos \left(-\mathrm{\pi }\right)=\mathrm{cos\pi }=-1 \end{gathered}$$

$$\begin{gathered} \sin \left(\frac{\mathrm{\pi }}{12}\right).\sin \left(\frac{7\mathrm{\pi }}{12}\right)=\frac{1}{2}\left[\cos \left(\frac{\mathrm{\pi }}{12}-\frac{7\mathrm{\pi }}{12}\right)-\cos \left(\frac{\mathrm{\pi }}{12}+\frac{7\mathrm{\pi }}{12}\right)\right] \\ =\frac{1}{2}.\left[\cos \frac{-\mathrm{\pi }}{2}-\cos \frac{2\mathrm{\pi }}{3}\right]=\frac{1}{2}.\left(0-\frac{-1}{2}\right)=\frac{1}{4} \\ \Rightarrow \sin \left(\frac{\mathrm{\pi }}{4}\right).\sin \left(\frac{\mathrm{\pi }}{12}\right).\sin \left(\frac{7\mathrm{\pi }}{12}\right)=\frac{\sqrt{2}}{2}.\frac{1}{4}=\frac{\sqrt{2}}{8} \end{gathered}$$

$\left(\alpha +\beta \right)$ và $\gamma$ là hai góc (cung) bù nhau nên A đúng.

Ta có: $\alpha +\beta +\gamma =\pi$. Suy ra:

 $\left(2\alpha +2\beta \right)=-2\gamma +2\pi$ nên  $\cos \left(2\alpha +2\beta \right)=\cos \left(-2\gamma \right)=\cos \left(2\gamma \right)$

nhưng $\sin \left(2\alpha +2\beta \right)=\sin \left(-2\gamma \right)=-\sin \left(2\gamma \right)$.

 Từ đó D đúng và A, B, C sai.

Ta có: $\alpha +\beta +\gamma =\pi \Rightarrow \alpha +\beta =\pi -\gamma \Rightarrow \frac{\alpha +\beta }{2}=\frac{\pi }{2}-\frac{\gamma }{2}$

 $\frac{\alpha +\beta }{2}$ và $\frac{\gamma }{2}$ là hai góc (cung) phụ nhau nên B đúng và A, C, D sai.

Ta có: $\alpha +\beta +\gamma =\pi \Rightarrow \alpha +\beta =\pi -\gamma$

$\Rightarrow 3\left(\alpha +\beta \right)=3\pi -3\gamma =2\pi +\pi -3\gamma$.

 Do đó,

$$\begin{gathered} \cos \left(3\alpha +3\beta \right)=\cos \left(2\pi +\pi -3\gamma \right)=\cos \left(\pi -3\gamma \right)=-\cos \left(-3\gamma \right)=-\cos 3\gamma \\ \sin \left(3\alpha +3\beta \right)=\sin \left(2\pi +\pi -3\gamma \right)=\sin \left(\pi -3\gamma \right)=-\sin \left(-3\gamma \right)=\sin 3\gamma \\ \tan \left(3\alpha +3\beta \right)=\frac{\sin \left(3\alpha +3\beta \right)}{\cos \left(3\alpha +3\beta \right)}=\frac{\sin 3\gamma }{-\cos 3\gamma }=-\tan 3\gamma \\ cot\left(3\alpha +3\beta \right)=-cot3\gamma \end{gathered}$$

Tổng ba góc trong 1 tam giác bằng 180$°$  nên:

$$\begin{gathered} \hat{M}+\hat{N}+\hat{P}=180°\iff \hat{M}+\hat{N}=180°-\hat{P} \\ \iff \frac{\hat{M}+\hat{N}}{2}=90°-\frac{\hat{P}}{2}\Rightarrow \cos \frac{M+N}{2}=\sin \frac{P}{2};\sin \frac{M+N}{2}=\cos \frac{P}{2} \\ \sin M+\sin N+\sin P=2\sin \frac{M+N}{2}.\cos \frac{M-N}{2}+2\sin \frac{P}{2}.\cos \frac{P}{2} \\ =2\cos \frac{P}{2}.\cos \frac{M-N}{2}+2\sin \frac{P}{2}.\cos \frac{P}{2}=2\cos \frac{P}{2}.\left(\cos \frac{M-N}{2}+\sin \frac{P}{2}\right) \\ =2\cos \frac{P}{2}.\left(\cos \frac{M-N}{2}+\cos \frac{M+N}{2}\right)=2\cos \frac{P}{2}.2\cos \frac{M}{2}.\cos \frac{N}{2} \\ =4\cos \frac{M}{2}.\cos \frac{N}{2}.\cos \frac{P}{2} \end{gathered}$$

Tổng ba góc trong 1 tam giác bằng 180$°$  nên:

$$\begin{gathered} \hat{M}+\hat{N}+\hat{P}=180°\iff \hat{M}+\hat{N}=180°-\hat{P} \\ \Rightarrow \sin \left(M+N\right)=\sin \left(180°-P\right)=\sin P; \cos \left(M+N\right)=-cosP \end{gathered}$$

Ta có: 

$$\begin{gathered} \sin 2M+\sin 2N+\sin 2P=2.\sin \left(M+N\right).\cos \left(M-N\right)+2\sin P.\cos P \\ =2\sin P.\cos \left(M-N\right)+2\sin P.\cos P=2\sin P.\left(\cos \left(M-N\right)+\cos P\right) \\ =2\sin P.\left(\cos \left(M-N\right)-\cos \left(M+N\right)\right)=2\sin P.\left(-2\sin M.\sin \left(-N\right)\right) \\ =4.\sin M.\sin N.\sin P \end{gathered}$$