Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
A = \sin \left( {21\pi - x} \right).\cos \left( {\frac{\pi }{2} + x} \right) - \cos \left( {2017\pi + x} \right).\sin \left( {x - \frac{{5\pi }}{2}} \right)\\
= \sin \left( {\left( {\pi - x} \right) + 10.2\pi } \right).sin\left( {\frac{\pi }{2} - \left( {\frac{\pi }{2} + x} \right)} \right) - \cos \left( {\left( {x + \pi } \right) + 2.1008\pi } \right).\sin \left( {\left( {x - \frac{\pi }{2}} \right) - 2\pi } \right)\\
= \sin \left( {\pi - x} \right).\sin \left( { - x} \right) - \cos \left( {x + \pi } \right).\sin \left( {x - \frac{\pi }{2}} \right)\\
= \sin x.\left( { - \sin x} \right) - \left( { - \cos \left( {\pi - x - \pi } \right)} \right).cos\left( {\frac{\pi }{2} - \left( {x - \frac{\pi }{2}} \right)} \right)\\
= - {\sin ^2}x + \cos \left( { - x} \right).\cos \left( {\pi - x} \right)\\
= - {\sin ^2}x + \cos x.\left( { - \cos x} \right)\\
= - \left( {{{\sin }^2}x + {{\cos }^2}x} \right) = - 1\\
b,\\
\cot x = \frac{3}{4} \Leftrightarrow \frac{{\cos x}}{{\sin x}} = \frac{3}{4} \Leftrightarrow \cos x = \frac{3}{4}\sin x\\
A = \frac{{{{\sin }^3}x - {{\cos }^3}x}}{{\sin x.{{\cos }^2}x + \cos x.si{n^2}x}}\\
= \frac{{{{\sin }^3}x - {{\left( {\frac{3}{4}\sin x} \right)}^3}}}{{\sin x.{{\left( {\frac{3}{4}\sin x} \right)}^2} + \left( {\frac{3}{4}\sin x} \right).{{\sin }^2}x}}\\
= \frac{{\frac{{37}}{{64}}{{\sin }^3}x}}{{\frac{{21}}{{16}}{{\sin }^3}x}} = \frac{{37}}{{84}}
\end{array}\)
