10 câu Trắc nghiệm Công thức lượng giác có đáp án (Vận dụng)

Đáp án A

$$\begin{gathered} \cos \alpha + \cos \beta = m; \sin \alpha + \sin \beta = n \\ \Rightarrow m^2+n^2={(cos\alpha +cos\beta )}^2\text{+(}sin\alpha +sin\beta {)}^2 \\ =co{s}^2\alpha +2cos\alpha cos\beta +co{s}^2\beta \text{+}si{n}^2\alpha +2sin\alpha sin\beta +si{n}^2\beta \\ =\left(co{s}^2\alpha \text{+}si{n}^2\alpha \right)+(co{s}^2\beta +si{n}^2\beta )++2(cos\alpha cos\beta +sin\alpha sin\beta ) \\ =1+1+2cos\left(\alpha -\beta \right)=2+2cos\left(\alpha -\beta \right) \end{gathered}$$

Do đó $cos\left(\alpha -\beta \right)=\frac{m^2+n^2-2}{2}$

Đáp án D

$$\begin{gathered} A\sin \frac{2\pi }{9}=\sin \frac{2\pi }{9}cos\frac{2\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{2}.2\sin \frac{2\pi }{9}cos\frac{2\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{2}\sin \frac{4\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{2}.\frac{1}{2}.2\sin \frac{4\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{4}\sin \frac{8\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{4}.\frac{1}{2}.2\sin \frac{8\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{8}\sin \frac{16\pi }{9}=\frac{1}{8}\sin \left(2\pi -\frac{2\pi }{9}\right) \\ =-\frac{1}{8}\sin \frac{2\pi }{9} \\ \Rightarrow A=-\frac{1}{8} \end{gathered}$$

Đáp án A

$$\begin{gathered} \sin \frac{\pi }{7}\left(\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{6\pi }{7}\right) \\ =\sin \frac{\pi }{7}\sin \frac{2\pi }{7}+\sin \frac{\pi }{7}\sin \frac{4\pi }{7}+\sin \frac{\pi }{7}\sin \frac{6\pi }{7} \\ =\frac{1}{2}\left(cos\frac{\pi }{7}-cos\frac{3\pi }{7}\right)+\frac{1}{2}\left(cos\frac{3\pi }{7}-cos\frac{5\pi }{7}\right)+\frac{1}{2}\left(cos\frac{5\pi }{7}-cos\frac{7\pi }{7}\right) \\ =\frac{1}{2}cos\frac{\pi }{7}+\frac{1}{2}=co{s}^2\frac{\pi }{14} \\ \sin \frac{\pi }{7}=2\sin \frac{\pi }{14}cos\frac{\pi }{14} \\ \Rightarrow \sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{6\pi }{7}=\frac{1}{2}\cot \frac{\pi }{14} \end{gathered}$$

Đáp án C

$$\begin{gathered} A=cos\alpha +cos(\alpha +\frac{\pi }{5})+\mathrm{...}+cos(\alpha +\frac{9\pi }{5}) \\ A=\left[cos\alpha +cos(\alpha +\frac{9\pi }{5})\right]+\mathrm{...}+\left[cos(\alpha +\frac{4\pi }{5})+cos(\alpha +\frac{5\pi }{5})\right] \\ A=2cos(\alpha +\frac{9\pi }{10})cos\frac{9\pi }{10}+2cos(\alpha +\frac{9\pi }{10})cos\frac{7\pi }{10}+\mathrm{...}+2cos(\alpha +\frac{9\pi }{10})cos\frac{\pi }{10} \\ A=2cos(\alpha +\frac{9\pi }{10})\left(cos\frac{9\pi }{10}+cos\frac{7\pi }{10}+cos\frac{5\pi }{10}+cos\frac{3\pi }{10}+cos\frac{\pi }{10}\right) \\ A=2cos(\alpha +\frac{9\pi }{10})\left(2cos\frac{\pi }{2}cos\frac{2\pi }{5}+2cos\frac{\pi }{2}cos\frac{\pi }{5}+cos\frac{\pi }{2}\right) \\ A=2cos(\alpha +\frac{9\pi }{10}).0=0 \end{gathered}$$

Đáp án D

Với k = 1, 2, 3, 4, 5 ta có:

$$\begin{gathered} \cos \frac{\left(2k\right)\pi }{11}\sin \frac{\pi }{11}=\frac{1}{2}\left[\sin \frac{(2k+1)\pi }{11}-\sin \frac{\left(2k-1\right)\pi }{11}\right] \\ \Rightarrow C.\sin \frac{\pi }{11}=\frac{1}{2}\left[\left(\sin \frac{3\pi }{11}-\sin \frac{\pi }{11}\right)+\left(\sin \frac{5\pi }{11}-\sin \frac{3\pi }{11}\right)+\mathrm{...}+\left(\sin \frac{11\pi }{11}-\sin \frac{9\pi }{11}\right)\right] \\ =-\frac{1}{2}\sin \frac{\pi }{11} \\ \Rightarrow C=-\frac{1}{2} \end{gathered}$$

Đáp án A

Ta có:

$$\begin{gathered} {\sin }^{4}x+{\cos }^{4}x={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x \\ =1-2{\left(\sin \text{x}\cos x\right)}^2=1-2{\left(\frac{1}{2}\sin 2x\right)}^2 \\ =1-2.\frac{1}{4}{\sin }^{2}2=x1-\frac{1}{2}{\sin }^{2}2x \\ =1-\frac{1}{2}.\frac{1-\cos 4x}{2}=1-\frac{1}{4}\left(1-\cos 4x\right) \\ =1-\frac{1}{4}+\frac{1}{4}\cos 4x=\frac{1}{4}\cos 4x+\frac{3}{4} \\ \Rightarrow S=m+n=1 \end{gathered}$$

Đáp án C

Ta có:

$$\begin{gathered} \frac{\cos B+\cos C}{\sin B+\sin C}=\frac{2\cos \frac{B+C}{2}.\cos \frac{B-C}{2}}{2\sin \frac{B+C}{2}\cos \frac{B-C}{2}} \\ =\frac{\cos \frac{B+C}{2}}{\sin \frac{B-C}{2}}=\frac{\cos \left(\frac{\pi }{2}-\frac{A}{2}\right)}{\sin \left(\frac{\pi }{2}-\frac{A}{2}\right)} \\ =\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}\Rightarrow \sin A=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} \\ \Rightarrow 2\sin \frac{A}{2}\cos \frac{A}{2}=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} \\ \Rightarrow 2{\cos }^2\frac{A}{2}=1\Rightarrow \cos A=0\Rightarrow A={90}^0 \end{gathered}$$

Đáp án B

$$\begin{gathered} A={\sin }^6\alpha +{\cos }^6\alpha \\ ={\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^3-3{\sin }^2\alpha {\cos }^2\alpha \left({\sin }^2\alpha +{\cos }^2\alpha \right) \\ =1-3{\sin }^2\alpha {\cos }^2\alpha =1-\frac{3}{4}{\sin }^{2}2\alpha \end{gathered}$$

Vì $0\le {\sin }^{2}2\alpha \le 1\Rightarrow A\ge \frac{1}{4}$ nên $\min A=\frac{1}{4}$ khi ${\sin }^{2}2\alpha =1$

Đáp án D

$$\begin{gathered} {\left(\tan \alpha +\cot \alpha \right)}^2={\tan }^2\alpha +{\cot }^2\alpha +2\tan \alpha .\cot \alpha \\ \Rightarrow {\tan }^2\alpha +{\cot }^2\alpha ={\left(\tan \alpha +\cot \alpha \right)}^2-2\tan \alpha .\cot \alpha \\ {m}^2-2(do\tan \alpha .\cot \alpha =1) \end{gathered}$$

Do đó:

$$\begin{gathered} {\left(\tan \alpha -\cot \alpha \right)}^2={\tan }^2\alpha +{\cot }^2\alpha -2\tan \alpha .\cot \alpha \\ {m}^2-2-2=m^2-4 \end{gathered}$$

Vậy $\left|\tan \alpha -\cot \alpha \right|=\sqrt{m^2-4}$

Đáp án C

Ta có:

$$\begin{gathered} \tan a+\tan b=\frac{\sin a}{\cos a}+\frac{\sin b}{\cos b} \\ =\frac{\sin a\cos b+\sin b\cos a}{\cos a\cos b}=\frac{\sin \left(a+b\right)}{\cos a\cos b} \end{gathered}$$

Suy ra A đúng

Tương tự ta có B đúng.

$$\begin{gathered} \tan a+\cot a=\frac{\sin a}{\cos a}+\frac{\cos a}{\sin a} \\ =\frac{{\sin }^{2}a+{\cos }^{2}a}{\sin a\cos a}=\frac{2}{\sin 2a} \end{gathered}$$

nên D đúng.

$\cot a+\cot b=\frac{\cos a}{\sin a}+\frac{\cos b}{\sin b}=\frac{\sin \left(a+b\right)}{\sin a\sin b}$

nên C sai