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:

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