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:

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

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

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