Question

Cho \(\cos \alpha  = \frac{2}{3}\).

Tính \(B = \cos \frac{{3\alpha }}{2}.\cos \frac{\alpha }{2}\)

Answer 1

Ta có :

\(\begin{array}{l}B = \cos \frac{{3\alpha }}{2}.\cos \frac{\alpha }{2} = \frac{1}{2}\left[ {\cos \left( {\frac{{3\alpha }}{2} + \frac{\alpha }{2}} \right) + \cos \left( {\frac{{3\alpha }}{2} - \frac{\alpha }{2}} \right)} \right]\\ = \frac{1}{2}\left[ {\cos (2\alpha ) + \cos \alpha } \right] = \frac{1}{2}\left[ {2.{{\cos }^2}\alpha  - 1 + \cos \alpha } \right] = \frac{1}{2}\left[ {2.{{\left( {\frac{2}{3}} \right)}^2} - 1 + \frac{2}{3}} \right] = \frac{5}{{18}}\end{array}\)