Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sin a.\sin b = - \frac{1}{2}\left( {\cos \left( {a + b} \right) - \cos \left( {a - b} \right)} \right)\\
\sin \left( {\frac{\pi }{4} + x} \right).\sin \left( {\frac{\pi }{4} - x} \right)\\
= - \frac{1}{2}.\left[ {\cos \left( {\frac{\pi }{4} + x + \frac{\pi }{4} - x} \right) - \cos \left( {\frac{\pi }{4} + x - \frac{\pi }{4} + x} \right)} \right]\\
= - \frac{1}{2}\left[ {\cos \frac{\pi }{2} - \cos 2x} \right]\\
= \frac{1}{2}\cos 2x\\
b,\\
\sin x\left( {1 + \cos 2x} \right) = \sin x.\left( {1 + 2{{\cos }^2}x - 1} \right) = \sin x.2{\cos ^2}x = \left( {2\sin x.\cos x} \right).\cos x = \sin 2x.\cos x\\
c,\\
\frac{{\sin x + \sin 3x + \sin 5x}}{{\cos x + \cos 3x + \cos 5x}}\\
= \frac{{\left( {\sin x + \sin 5x} \right) + \sin 3x}}{{\left( {\cos x + \cos 5x} \right) + \cos 3x}}\\
= \frac{{2.\sin 3x.\cos 2x + \sin 3x}}{{2.\cos 3x.\cos 2x + \cos 3x}}\\
= \frac{{\sin 3x\left( {2\cos 2x + 1} \right)}}{{\cos 3x\left( {2.\cos 2x + 1} \right)}}\\
= \frac{{\sin 3x}}{{\cos 3x}} = \tan 3x
\end{array}\)
