Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
{\sin ^2}x + {\cos ^2}x = 1\\
2,\\
\sin a + \cos a = \frac{1}{2}\\
\Leftrightarrow {\left( {\sin a + \cos a} \right)^2} = \frac{1}{4}\\
\Leftrightarrow {\sin ^2}a + 2\sin a.\cos a + {\cos ^2}a = \frac{1}{4}\\
\Leftrightarrow 1 + 2\sin a.\cos a = \frac{1}{4}\\
\Leftrightarrow 1 + \sin 2a = \frac{1}{4}\\
\Leftrightarrow \sin 2a = - \frac{3}{4}\\
3,\\
a \in \left( {\pi ;\frac{{3\pi }}{2}} \right) \Rightarrow \left\{ \begin{array}{l}
\sin a < 0\\
\cos a < 0
\end{array} \right.\\
\cos a < 0 \Rightarrow \cos a = - \sqrt {1 - {{\sin }^2}a} = - 0,8\\
\sin 2a = 2\sin a.\cos a = 2.\left( { - 0,6} \right).\left( { - 0,8} \right) = 0,96\\
4,\\
\frac{{2\pi }}{3} = \frac{{2.180^\circ }}{3} = 120^\circ \\
5,\\
\cos \frac{\pi }{4} = 2{\cos ^2}\frac{\pi }{8} - 1 \Leftrightarrow \frac{{\sqrt 2 }}{2} = 2{\cos ^2}\frac{\pi }{8} - 1\\
\Leftrightarrow 2{\cos ^2}\frac{\pi }{8} = \frac{{\sqrt 2 + 2}}{2}\\
\Leftrightarrow \cos \frac{\pi }{8} = \frac{{\sqrt {\sqrt 2 + 2} }}{2}\\
6,\\
\tan \frac{{25\pi }}{4} = \tan \left( {6\pi + \frac{\pi }{4}} \right) = \tan \frac{\pi }{4} = 1\\
7,\\
\tan \frac{{13\pi }}{{12}} = \tan \left( {\pi + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{{12}}\\
\cos \frac{\pi }{6} = 2{\cos ^2}\frac{\pi }{{12}} - 1 = 1 - 2{\sin ^2}\frac{\pi }{{12}}\\
\Rightarrow \tan \frac{\pi }{{12}} = \frac{{\sin \frac{\pi }{{12}}}}{{\cos \frac{\pi }{{12}}}} = 2 - \sqrt 3 \\
8,\\
\cos x = 1 \Rightarrow x = \frac{\pi }{2} + k2\pi
\end{array}\)
