Cho $\cos a = \frac{3}{5}$ với $0 < a < \frac{\pi }{2}$. Tính $\sin \left( {a + \frac{\pi }{6}} \right),cos\left( {a - \frac{\pi }{3}} \right),\tan \left( {a + \frac{\pi }{4}} \right)$.

Do $0 < a < \frac{\pi }{2}$ nên $\sin a > 0$.

Áp dụng công thức sin²a + cos²a = 1, ta có:

$si{n^2}a + {\left( {\frac{3}{5}} \right)^2} = 1$

$\Rightarrow si{n^2}a = 1 - {\left( {\frac{3}{5}} \right)^2} = 1 - \frac{9}{{25}} = \frac{{16}}{{25}}$

$\Rightarrow \sin a = \frac{4}{5}$ (do sina > 0).

Khi đó $\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{4}{5}}}{{\frac{3}{5}}} = \frac{4}{3}$.

Áp dụng công thức cộng, ta có:

• $\sin \left( {a + \frac{\pi }{6}} \right) = \sin a\cos \frac{\pi }{6} + \cos a\sin \frac{\pi }{6} = \frac{4}{5}.\frac{{\sqrt 3 }}{2} + \frac{3}{5}.\frac{1}{2} = \frac{{4\sqrt 3 + 3}}{{10}}$;

• $cos\left( {a - \frac{\pi }{3}} \right) = \cos a\,cos\frac{\pi }{3} + \sin a\sin \frac{\pi }{3} = \frac{3}{5}.\frac{1}{2} + \frac{4}{5}.\frac{{\sqrt 3 }}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}$;

• $\tan \left( {a + \frac{\pi }{4}} \right) = \frac{{\tan a + \tan \frac{\pi }{4}}}{{1 - \tan a\tan \frac{\pi }{4}}} = \frac{{\frac{4}{3} + 1}}{{1 - \frac{4}{3}.1}} = \frac{{\frac{7}{3}}}{{ - \frac{1}{3}}} = - 7$.