a. Cho \(\sin x = \frac{3}{5}\) với \(\frac{\pi }{2} var DO...

* Hướng dẫn giải

a) Ta có \({\sin ^2}x + {\cos ^2}x = 1 \Rightarrow \cos x =  \pm \sqrt {1 - {{\sin }^2}x}  =  \pm \sqrt {1 - \frac{9}{{25}}}  =  \pm \frac{4}{5}\)

Vì \(\frac{\pi }{2} < x < \pi \) nên \(\cos x =  - \frac{4}{5} \Rightarrow \tan x =  - \frac{3}{4}\)

Ta có: \(\tan \left( {x + \frac{\pi }{4}} \right) = \frac{{\tan x + \tan \frac{\pi }{4}}}{{1 - \tan x.\tan \frac{\pi }{4}}} = \frac{{ - \frac{3}{4} + 1}}{{1 + \frac{3}{4}}} = \frac{1}{7}\)

b) Ta có \(\sin \left( {a + \frac{\pi }{4}} \right).\sin \left( {a - \frac{\pi }{4}} \right) = \frac{1}{2}\left[ {\cos \frac{\pi }{2} - \cos 2a} \right] =  - \frac{1}{2}\cos 2a\)