Nếu \(\cos a = \frac{1}{3},\,\,\sin b = \frac{{ - 2}}{3}\) thì giá trị cos(a + b) cos(a − b) bằng:

A. \( - \frac{2}{3}\).

B. \(\frac{1}{3}\).

C. \(\frac{2}{3}\).

D. \( - \frac{1}{3}\).

Đáp án đúng là: D

Ta có cos(a + b) cos(a − b) \( = \frac{1}{2}\left[ {\cos \left( {a + b + a - b} \right) + \cos \left( {a + b - a + b} \right)} \right]\)

\( = \frac{1}{2}\left( {\cos 2a + \cos 2b} \right)\)

\( = \frac{1}{2}\left[ {\left( {2{{\cos }^2}a - 1} \right) + \left( {1 - 2{{\sin }^2}b} \right)} \right]\)

\( = \frac{1}{2}.\left[ {2.{{\left( {\frac{1}{3}} \right)}^2} - 2.{{\left( { - \frac{2}{3}} \right)}^2}} \right] = - \frac{1}{3}\).