10 câu Trắc nghiệm Công thức lượng giác có đáp án (Thông hiểu)

Đáp án A

$$\begin{gathered} A={\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta .cos\left(\alpha +\beta \right) \\ ={\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta .\left(cos\alpha .cos\beta -\sin \alpha \sin \beta \right) \\ ={\sin }^2\alpha +{\sin }^2\beta -2{\sin }^2\alpha {\sin }^2\beta +2\sin \alpha \sin \beta cos\alpha .cos\beta \\ ={\sin }^2\alpha (1-{\sin }^2\beta )+{\sin }^2\beta (1-{\sin }^2\alpha )+2\sin \alpha \sin \beta cos\alpha .cos\beta \\ ={\sin }^2\alpha co{s}^2\beta +{\sin }^2\beta co{s}^2\alpha +2\sin \alpha \sin \beta cos\alpha .cos\beta \\ ={(\sin \alpha cos\beta +\sin \beta cos\alpha )}^2={\sin }^2\left(\alpha +\beta \right) \end{gathered}$$

Đáp án D

Ta có:

$$\begin{gathered} A={\left(\sin acosb+cosa\sin b\right)}^2-{\sin }^{2}a-{\sin }^{2}b \\ ={\sin }^{2}aco{s}^{2}b+2\sin a\cos a\sin b\cos b+co{s}^{2}a{\sin }^{2}b-{\sin }^{2}a-{\sin }^{2}b \\ ={\sin }^{2}a(co{s}^{2}b-1)+{\sin }^{2}b(co{s}^{2}a-1)+2\sin a\cos a\sin b\cos b \\ =2\sin a\cos a\sin b\cos b-2{\sin }^{2}a{\sin }^{2}b \\ =2\sin a\sin b(\cos a\cos b-\sin a\sin b)=2\sin a\sin bcos\left(a+b\right) \end{gathered}$$

Đáp án D

Ta có:

$$\begin{gathered} P={\left(\sin a+\sin b\right)}^2+{\left(cosa+cosb\right)}^2 \\ ={\sin }^{2}a+2\sin a\sin b+{\sin }^{2}b+co{s}^{2}a+2cosacosb+co{s}^{2}b \\ =2+2(\sin a\sin b+cosacosb) \\ =2+2cos\left(a-b\right)=2+2cos\frac{\pi }{4}=2+\sqrt{2} \end{gathered}$$

Đáp án A

$$\begin{gathered} \sin \left(\alpha +2\beta \right)=\sin \alpha .cos2\beta +cos\alpha .\sin 2\beta \\ =\sin \alpha .(1-2{\sin }^2\beta )+2cos\alpha \sin \beta cos\beta \\ =\sin \alpha +2\sin \beta (cos\alpha .cos\beta -\sin \alpha .\sin \beta ) \\ =\sin \alpha +2\sin \beta cos\left(\alpha +\beta \right)=\sin \alpha \end{gathered}$$

Đáp án B

Ta có:

$$\begin{gathered} \sin \left(a+b\right)\sin \left(a-b\right)=\frac{1}{2}(cos2b-cos2a) \\ =\frac{1}{2}\left[\left(2co{s}^{2}b-1\right)-\left(2co{s}^{2}a-1\right)\right] \\ =co{s}^{2}b-co{s}^{2}a \end{gathered}$$

Đáp án B

Ta có:

$$\begin{gathered} P=\frac{1+cos\alpha +cos2\alpha }{\sin \alpha +\sin 2\alpha }=\frac{2co{s}^2\alpha +cos\alpha }{\sin \alpha +2\sin \alpha cos\alpha } \\ =\frac{cos\alpha (1+2cos\alpha )}{\sin \alpha (1+2cos\alpha )}=\cot \alpha =\frac{1}{2} \end{gathered}$$

Đáp án A

$$\begin{gathered} A=\frac{\sin \frac{\pi }{9}+\sin \frac{5\pi }{9}}{cos\frac{\pi }{9}+cos\frac{5\pi }{9}}=\frac{2\sin \left(\frac{\pi }{3}\right)cos\left(\frac{2\pi }{9}\right)}{2cos\left(\frac{\pi }{3}\right)cos\left(\frac{2\pi }{9}\right)} \\ =\tan \left(\frac{\pi }{3}\right)=\sqrt{3} \end{gathered}$$

Đáp án A

$$\begin{gathered} A=\frac{1-cos\frac{\pi }{4}}{2}+\frac{1-cos\frac{3\pi }{4}}{2}+\frac{1-cos\frac{5\pi }{4}}{2}+\frac{1-cos\frac{7\pi }{4}}{2} \\ =2-\frac{1}{2}\left(cos\frac{\pi }{4}+cos\frac{3\pi }{4}+cos\frac{5\pi }{4}+cos\frac{7\pi }{4}\right) \\ =2-\frac{1}{2}\left(cos\frac{\pi }{4}+cos\frac{3\pi }{4}-cos\frac{3\pi }{4}-cos\frac{\pi }{4}\right)=2 \end{gathered}$$

Đáp án C

Ta có:

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

Đáp án B

Ta có:

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