Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\cos 2x = 2{\cos ^2}x - 1 \Leftrightarrow 2{\cos ^2}x = \cos 2x + 1 \Leftrightarrow 4{\cos ^2}x = 2\cos 2x + 1\\
A = \sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + 2\cos x} } } } \\
= \sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {4{{\cos }^2}\frac{x}{2}} } } } \\
= \sqrt {2 + \sqrt {2 + \sqrt {2 + 2\cos \frac{x}{2}} } } \\
= \sqrt {2 + \sqrt {2 + \sqrt {4{{\cos }^2}\frac{x}{4}} } } \\
= \sqrt {2 + \sqrt {2 + 2\cos \frac{x}{4}} } \\
= \sqrt {2 + \sqrt {4{{\cos }^2}\frac{x}{8}} } \\
= \sqrt {2 + 2\cos \frac{x}{8}} \\
= \sqrt {4{{\cos }^2}\frac{x}{{16}}} \\
= 2\cos \frac{x}{{16}}\\
b,\\
\cos 2x = 2{\cos ^2}x - 1 \Leftrightarrow 2{\cos ^2}x = \cos 2x + 1 \Leftrightarrow {\cos ^2}x = \frac{1}{2}\cos 2x + \frac{1}{2}\\
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow 2{\sin ^2}x = 1 - \cos 2x \Leftrightarrow {\sin ^2}x = \frac{1}{2} - \frac{1}{2}\cos 2x\\
B = \sqrt {\frac{1}{2} - \frac{1}{2}\sqrt {\frac{1}{2} - \frac{1}{2}\sqrt {\frac{1}{2} + \cos x} } } \\
= \sqrt {\frac{1}{2} - \frac{1}{2}\sqrt {\frac{1}{2} - \frac{1}{2}\sqrt {{{\cos }^2}\frac{x}{2}} } } \\
= \sqrt {\frac{1}{2} - \frac{1}{2}\sqrt {\frac{1}{2} - \frac{1}{2}\cos \frac{x}{2}} } \\
= \sqrt {\frac{1}{2} - \frac{1}{2}\sqrt {{{\sin }^2}\frac{x}{4}} } \\
= \sqrt {\frac{1}{2} - \frac{1}{2}\sin \frac{x}{4}}
\end{array}\)
