Ai đó giải cho em với,😂😂😂😂 bài 1 á

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
\sin x.\cos x = \frac{1}{2}\sin 2x\\
a,\\
A = \cos 36^\circ .\cos 72^\circ \\
 \Leftrightarrow A.\sin 36^\circ  = \sin 36^\circ .\cos 36^\circ .\cos 72^\circ \\
 \Leftrightarrow A.\sin 36^\circ  = \frac{1}{2}\sin 72^\circ .cos72^\circ \\
 \Leftrightarrow A.\sin 36^\circ  = \frac{1}{2}.\frac{1}{2}\sin 144^\circ \\
 \Leftrightarrow A.\sin 36^\circ  = \frac{1}{4}\sin 144^\circ \\
 \Leftrightarrow A.\sin 36^\circ  = \frac{1}{4}\sin \left( {180^\circ  - 144^\circ } \right)\\
 \Leftrightarrow A.\sin 36^\circ  = \frac{1}{4}\sin 36^\circ \\
 \Leftrightarrow A = \frac{1}{4}\\
b,\\
B = \cos \frac{\pi }{5}.\cos \frac{{2\pi }}{5}\\
 \Rightarrow B.\sin \frac{\pi }{5} = \sin \frac{\pi }{5}.\cos \frac{\pi }{5}.\cos \frac{{2\pi }}{5}\\
 \Leftrightarrow B.\sin \frac{\pi }{5} = \frac{1}{2}\sin \frac{{2\pi }}{5}.\cos \frac{{2\pi }}{5}\\
 \Leftrightarrow B.\sin \frac{\pi }{5} = \frac{1}{2}.\frac{1}{2}\sin \frac{{4\pi }}{5}\\
 \Leftrightarrow B.\sin \frac{\pi }{5} = \frac{1}{4}.\sin \left( {\pi  - \frac{{4\pi }}{5}} \right)\\
 \Leftrightarrow B.\sin \frac{\pi }{5} = \frac{1}{4}\sin \frac{\pi }{5}\\
 \Leftrightarrow B = \frac{1}{4}\\
c,\\
C = \sin \frac{\pi }{8}.\cos \frac{\pi }{8}.\cos \frac{\pi }{4} = \frac{1}{2}\sin \frac{\pi }{4}.\cos \frac{\pi }{4} = \frac{1}{2}.\frac{1}{2}\sin \frac{\pi }{2} = \frac{1}{4}\sin \frac{\pi }{2} = \frac{1}{4}\\
d,\\
D = \sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{2}.\frac{1}{2}\sin \frac{\pi }{4} = \frac{1}{4}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{8}
\end{array}\)