Cho cos alpha = 3/4; sin alpha >0; sin bêta = 3/4; cos bêta <0. Tính cos(alpha + bêta)

* Hướng dẫn giải

Ta có: $\cos \alpha =\frac{3}{4};\sin \alpha >0$

$\Rightarrow {\sin }^2\alpha =1-\frac{9}{16}=\frac{7}{16}$

$\Rightarrow \sin \alpha =\frac{\sqrt{7}}{4}$

Ta có: $\sin \beta =\frac{3}{4};\cos \beta <0$

$\Rightarrow {\cos }^2\beta =1-\frac{9}{16}=\frac{7}{16}$

$\Rightarrow \cos \beta =-\frac{\sqrt{7}}{4}$

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

$\Rightarrow \cos \left(\alpha +\beta \right)=\frac{3}{4}.\left(-\frac{\sqrt{7}}{4}\right)-\frac{3}{4}.\left(\frac{\sqrt{7}}{4}\right)$

$\Rightarrow \cos \left(\alpha +\beta \right)=-\frac{3\sqrt{7}}{8}$