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

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} cos\alpha =\frac{3}{4};\sin \alpha >0 \\ \Rightarrow {\sin }^2\alpha =1-\frac{9}{16}=\frac{7}{16}\Rightarrow \sin \alpha =\frac{\sqrt{7}}{4} \\ \sin \beta =\frac{3}{4};cos\beta <0 \\ \Rightarrow co{s}^2\beta =1-\frac{9}{16}=\frac{7}{16}\Rightarrow cos\beta =-\frac{\sqrt{7}}{4} \\ \cos (\alpha + \beta ) = \cos \alpha \cos \beta – \sin \alpha \sin \beta \\ =\frac{3}{4}.\left(-\frac{\sqrt{7}}{4}\right)-\frac{3}{4}.\left(\frac{\sqrt{7}}{4}\right)=-\frac{3\sqrt{7}}{8} \end{gathered}$$