Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
cos30^\circ = 2{\cos ^2}15^\circ - 1\\
\Leftrightarrow \frac{{\sqrt 3 }}{2} = 2{\cos ^2}15^\circ - 1\\
\Leftrightarrow {\cos ^2}15^\circ = \frac{{\sqrt 3 + 2}}{4}\\
\Leftrightarrow \cos 15^\circ = \sqrt {\frac{{\sqrt 3 + 2}}{4}} \,\,\,\,\,\,\,\,\,\,\,\,\left( {\cos 15^\circ > 0} \right)\\
\Leftrightarrow \cos 15^\circ = \frac{{\sqrt 6 + \sqrt 2 }}{4}\\
b,\\
\cos \frac{\pi }{4} = 1 - 2{\sin ^2}\frac{\pi }{8}\\
\Leftrightarrow \frac{{\sqrt 2 }}{2} = 1 - 2{\sin ^2}\frac{\pi }{8}\\
\Leftrightarrow {\sin ^2}\frac{\pi }{8} = \frac{{2 - \sqrt 2 }}{4}\\
\Leftrightarrow \sin \frac{\pi }{8} = \sqrt {\frac{{2 - \sqrt 2 }}{4}} \,\,\,\,\,\,\,\,\,\,\left( {\sin \frac{\pi }{8} > 0} \right)\\
\Leftrightarrow \sin \frac{\pi }{8} = \frac{{\sqrt {2 - \sqrt 2 } }}{2}\\
c,\\
\cos 150^\circ = 2{\cos ^2}75^\circ - 1\\
\Leftrightarrow - \frac{{\sqrt 3 }}{2} = 2{\cos ^2}75^\circ - 1\\
\Leftrightarrow {\cos ^2}75^\circ = \frac{{2 - \sqrt 3 }}{4}\\
\Leftrightarrow \cos 75^\circ = \frac{{\sqrt 6 - \sqrt 2 }}{4}\\
\Rightarrow \sin 75^\circ = \sqrt {1 - {{\cos }^2}75^\circ } = \frac{{\sqrt 6 + \sqrt 2 }}{4}\\
\cot 75^\circ = \frac{{\cos 75^\circ }}{{\sin 75^\circ }} = \frac{{\sqrt 6 - \sqrt 2 }}{{\sqrt 6 + \sqrt 2 }} = 2 - \sqrt 3 \\
2,\\
\frac{\pi }{2} < \alpha < \pi \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha < 0
\end{array} \right.\\
\cot \alpha = - 3 \Leftrightarrow \frac{{\cos \alpha }}{{\sin \alpha }} = - 3 \Leftrightarrow \cos \alpha = - 3\sin \alpha \\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( { - 3\sin \alpha } \right)^2} + {\sin ^2}\alpha = 1\\
\Leftrightarrow {\sin ^2}\alpha = \frac{1}{{10}}\\
\sin \alpha > 0 \Rightarrow \sin \alpha = \frac{1}{{\sqrt {10} }}\\
\cos \alpha = - 3\sin \alpha = - \frac{3}{{\sqrt {10} }}\\
P = \cos \left( {\alpha - \frac{\pi }{4}} \right) = \cos \alpha .\cos \frac{\pi }{4} + \sin \alpha .\sin \frac{\pi }{4}\\
= \frac{{ - 3}}{{\sqrt {10} }}.\frac{{\sqrt 2 }}{2} + \frac{1}{{\sqrt {10} }}.\frac{{\sqrt 2 }}{2} = - \frac{{\sqrt 5 }}{5}
\end{array}\)
