Nếu sin(2 alpha + bêta) = 3sin bêta; cos alpha khác 0; cos (alpha + bêta) khác 0 thì tan (alpha + bêta) bằng:

Ta có:

$\sin \left(2\alpha +\beta \right)=3\sin \beta$

$\Rightarrow \sin 2\alpha \cos \beta +\cos 2\alpha \sin \beta =3\sin \beta$

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

$\Rightarrow 2\sin \alpha \cos \alpha \cos \beta +2{\cos }^2\alpha \sin \beta =4\sin \beta$

$\Rightarrow 2\cos \alpha \left(sin\alpha \cos \beta +\sin \beta \cos \alpha \right)=4\sin \beta$

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