Nếu sin(2alpha + beta) = 3sin beta; cos alpha khác 0; cos(alpha + beta) khác 0

* Hướng dẫn giải

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

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