12 câu Trắc nghiệm Công thức lượng giác có đáp án (Tổng hợp)

Đáp án A

$$\begin{gathered} 4cos{15}^{0}cos{24}^{0}cos{21}^0-cos{12}^0-cos{18}^0 \\ =2\cos {15}^0\left(\cos \left({24}^0+{21}^0\right)+\cos \left({24}^0-{21}^0\right)\right)-2\cos \left(\frac{{12}^0+{18}^0}{2}\right)\cos \left(\frac{{12}^0-{18}^0}{2}\right) \\ =2\cos {15}^0\left(\cos {45}^0+\cos 3^0\right)-2\cos {15}^0\cos 3^0 \\ =2\cos {15}^0.\cos {45}^0=\cos {60}^0+\cos {30}^0 \\ =\frac{1}{2}+\frac{\sqrt{3}}{2} \end{gathered}$$

Đáp án A

$$\begin{gathered} \frac{\sin \alpha +\sin \beta \cos \left(\alpha +\beta \right)}{\cos \alpha -\sin \beta \sin \left(\alpha +\beta \right)} \\ =\frac{\sin \alpha +\frac{1}{2}\left[\sin \left(\alpha +2\beta \right)+\sin (-\alpha )\right]}{\cos \alpha +\frac{1}{2}\left[\cos \left(\alpha +2\beta \right)-\cos (-\alpha )\right]} \\ =\frac{\sin \alpha +\frac{1}{2}\left[\sin \left(\alpha +2\beta \right)-\sin \left(\alpha \right)\right]}{\cos \alpha +\frac{1}{2}\left[\cos \left(\alpha +2\beta \right)-\cos \left(\alpha \right)\right]} \\ =\frac{\sin \left(\alpha +2\beta \right)+\sin \left(\alpha \right)}{\cos \left(\alpha +2\beta \right)+\cos \left(\alpha \right)} \\ =\frac{2\sin \left(\alpha +\beta \right)\cos \beta }{2\cos \left(\alpha +\beta \right)\cos \beta }=\tan \left(\alpha +\beta \right) \end{gathered}$$

Đáp án A

$$\begin{gathered} A=\cos (x-a)\left[\cos (x-a)-2\cos a\cos x\right]+{\cos }^{2}x \\ =\cos (x-a)\left(-\cos x\cos a+\sin x\sin a\right)+{\cos }^{2}x \\ =-\cos (x-a).\cos (x+a)+{\cos }^{2}x \\ =-\frac{1}{2}\left(\cos 2x+\cos 2a\right)+\frac{1+\cos 2x}{2} \\ =\frac{1-\cos 2a}{2}={\sin }^{2}a \end{gathered}$$

Đáp án C

Thực nghiệm

$\cos \frac{5\pi }{2}\cos \frac{3\pi }{2}+\sin \frac{7\pi }{2}\sin \frac{\pi }{2}-\cos \pi \cos 2\pi =0$

Đáp án C

Thực nghiệm ${\cos }^3\pi \cos 3\pi -{\sin }^3\pi \sin 3\pi -\frac{3}{4}\cos 4\pi =\frac{1}{4}$

Đáp án A

Với k = 1, 2, 3, 4, 5 ta có:

$$\begin{gathered} \cos \frac{\left(2k-1\right)\pi }{11}\sin \frac{\pi }{11}=\frac{1}{2}\left[\sin \frac{2k\pi }{11}-\sin \frac{\left(2k-2\right)\pi }{11}\right] \\ \Rightarrow B.\sin \frac{\pi }{11}=\frac{1}{2}\left[\left(\sin \frac{2\pi }{11}-\sin 0\right)+\left(\sin \frac{4\pi }{11}-\sin \frac{2\pi }{11}\right)+\mathrm{...}+\left(\sin \frac{10\pi }{11}-\sin \frac{8\pi }{11}\right)\right] \\ =\frac{1}{2}\sin \frac{10\pi }{11}=\frac{1}{2}\sin \frac{\pi }{11} \\ \Rightarrow B=\frac{1}{2} \end{gathered}$$

Đáp án B

Ta có:

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

Lại có:

$$\begin{gathered} \sin \left(2\alpha +\beta \right)=3\sin \beta \Rightarrow \sin 2\alpha \cos \beta +\cos 2\alpha \sin \beta =3\sin \beta \\ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta +(1-2{\sin }^2\alpha )\sin \beta =3\sin \beta \\ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta -2{\sin }^2\alpha \sin \beta =2\sin \beta \\ \Rightarrow 2\sin \alpha (\cos \alpha \cos \beta -\sin \beta \sin \alpha )=2\sin \beta \\ \Rightarrow \sin \alpha \cos (\alpha +\beta )=\sin \beta \end{gathered}$$

Từ đó suy ra $\frac{\cos \alpha \sin \left(\alpha +\beta \right)}{\sin \alpha \cos \left(\alpha +\beta \right)}=\frac{2\sin \beta }{\sin \beta }$ hay $\cot \alpha \tan \left(\alpha +\beta \right)=2$

$\Rightarrow \tan \left(\alpha +\beta \right)=2\tan \alpha$

Đáp án A

Ta có:

$$\begin{gathered} A=\frac{1+2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}=\frac{{\sin }^{2}x+2\sin x\cos x+{\cos }^{2}x}{{\cos }^{2}x-{\sin }^{2}x} \\ =\frac{{\left(\mathrm{s}\text{inx}+\cos x\right)}^2}{\left(\cos x-s\text{inx}\right)\left(\cos x+s\text{inx}\right)} \\ =\frac{\mathrm{s}\text{inx}+\cos x}{\cos x-s\text{inx}}=\frac{\sqrt{2}\sin \left(x+\frac{\pi }{4}\right)}{\sqrt{2}\cos \left(x+\frac{\pi }{4}\right)} \\ =\tan \left(x+\frac{\pi }{4}\right) \end{gathered}$$

Đáp án D

Ta có:

$$\begin{gathered} {\sin }^{6}x+{\cos }^{6}x={\left({\sin }^{2}x+{\cos }^{2}x\right)}^3-3{\sin }^{2}x{\cos }^{2}x\left({\sin }^{2}x+{\cos }^{2}x\right) \\ =1-3{\left(\frac{1}{2}\sin 2x\right)}^2=1-\frac{3}{4}\frac{1-\cos 4x}{2}=\frac{3}{8}\cos 4x+\frac{5}{8} \\ \Rightarrow S=m+n=1 \end{gathered}$$

Đáp án D

Ta có:

$$\begin{gathered} 3{\sin }^{4}x+2{\cos }^{4}x=\frac{98}{81} \\ 2{\sin }^{4}x+3{\cos }^{4}x=A \end{gathered}$$

Trừ vế cho vế hai đẳng thức trên ta được:

$$\begin{gathered} {\sin }^{4}x-{\cos }^{4}x=\frac{98}{81}-A \\ \iff \left({\sin }^{2}x-{\cos }^{2}x\right)\left({\sin }^{2}x+{\cos }^{2}x\right)=\frac{98}{81}-A \\ \iff \left({\sin }^{2}x-{\cos }^{2}x\right)=\frac{98}{81}-A \\ \iff \left({\cos }^{2}x-{\sin }^{2}x\right)=A-\frac{98}{81} \\ \iff \cos 2x=A-\frac{98}{81} \end{gathered}$$

Cộng vế với vế hai đẳng thức đầu bài ta được:

$$\begin{gathered} 5{\sin }^{4}x+5{\cos }^{4}x=A+\frac{98}{81} \\ \iff 5\left({\sin }^{4}x+{\cos }^{4}x\right)=A+\frac{98}{81} \\ \iff 5\left[{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\right]=A+\frac{98}{81} \\ \iff 5\left(1-\frac{1}{2}.4{\sin }^{2}x{\cos }^{2}x\right)=A+\frac{98}{81} \\ \iff 5\left(1-\frac{1}{2}{\sin }^{2}2x\right)=A+\frac{98}{81} \\ \iff 1-\frac{1}{2}{\sin }^{2}2x=\frac{1}{5}\left(A+\frac{98}{81}\right) \\ \iff 1+{\cos }^{2}2x=\frac{2}{5}\left(A+\frac{98}{81}\right) \end{gathered}$$

Thay ta được:

$1+{\left(A-\frac{98}{81}\right)}^2=\frac{2}{5}\left(A+\frac{98}{81}\right)=\frac{2}{5}\left(A-\frac{98}{81}\right)+\frac{392}{405}$

Đặt $A-\frac{98}{81}=t\Rightarrow t^2-\frac{2}{5}t+\frac{13}{405}=0\iff \left[\begin{array}{c}t=\frac{13}{45}\\ t=\frac{1}{9}\end{array}\right.$

$$\begin{gathered} + t=\frac{13}{45}\Rightarrow A=\frac{607}{405} \\ + t=\frac{1}{9}\Rightarrow A=\frac{107}{81} \end{gathered}$$

Đáp án A

$$\begin{gathered} \frac{1}{{\sin }^2\alpha }=1+{\cot }^2\alpha =1+18=19 \\ \Rightarrow {\sin }^2\alpha =\frac{1}{19}\Rightarrow \sin \alpha =\pm \frac{1}{\sqrt{19}} \end{gathered}$$

Vì $\frac{\pi }{2}<\alpha <\pi \Rightarrow \sin \alpha >0\Rightarrow \sin \alpha =\frac{1}{\sqrt{19}}$

Suy ra $\tan \frac{\alpha }{2}+\cot \frac{\alpha }{2}=\frac{{\sin }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}}{\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}=\frac{2}{\sin \alpha }=2\sqrt{19}$

Đáp án D

A đúng vì $VT=\frac{\tan x+\tan y}{\frac{1}{\tan x}+\frac{1}{\tan y}}=\tan x.\tan y=VP$

B đúng vì 

$$\begin{gathered} VT=\frac{1+\sin a}{1-\sin a}+\frac{1-\sin a}{1+\sin a}-2 \\ =\frac{{\left(1+\sin a\right)}^2+{\left(1-\sin a\right)}^2}{1-{\sin }^{2}a}-2 \\ =\frac{2+2{\sin }^{2}a}{{\cos }^{2}a}-2=4{\tan }^{2}a=VP \end{gathered}$$

C đúng vì 

$$\begin{gathered} VT=\frac{-{\sin }^2\alpha -{\cos }^2\alpha }{{\cos }^2\alpha -{\sin }^2\alpha } \\ =\frac{{\sin }^2\alpha +{\cos }^2\alpha }{{\sin }^2\alpha -{\cos }^2\alpha }=\frac{1+{\cot }^2\alpha }{1-{\cot }^2\alpha }=VP \end{gathered}$$