Đáp án:
$2a)
A=\dfrac{\sqrt{2}}{4}\\
b)P=2\sin\alpha$
Giải thích các bước giải:
$2a)
A=\cos\dfrac{3\pi}{8}.\cos\dfrac{\pi}{8}\\
=\dfrac{1}{2}\left [ \cos\left ( \dfrac{3\pi}{8}-\dfrac{\pi}{8} \right )+\cos\left ( \dfrac{3\pi}{8}+\dfrac{\pi}{8} \right ) \right ]\\
=\dfrac{1}{2}\left [ \cos\dfrac{2\pi}{8}+\cos\dfrac{4\pi}{8} \right ]\\
=\dfrac{1}{2}\left [ \cos\dfrac{\pi}{4}+\cos\dfrac{\pi}{2} \right ]\\
=\dfrac{1}{2}\left [ \dfrac{\sqrt{2}}{2}+0 \right ]\\
=\dfrac{\sqrt{2}}{4}\\
b)P=\dfrac{\sin5\alpha\cos3\alpha-\sin3\alpha\cos5\alpha}{\cos\alpha}\\
=\dfrac{\dfrac{1}{2}\left [\sin(5\alpha-3\alpha)+\sin(5\alpha+3\alpha) \right ]-\dfrac{1}{2}\left [\sin(3\alpha-5\alpha)+\sin(3\alpha+5\alpha) \right ]}{\cos\alpha}\\
=\dfrac{\dfrac{1}{2}\left [\sin2\alpha+\sin8\alpha -\sin(-2\alpha)-\sin8\alpha \right ]}{\cos\alpha}\\
=\dfrac{\dfrac{1}{2}\left [\sin2\alpha +\sin2\alpha \right ]}{\cos\alpha}\\
=\dfrac{\dfrac{1}{2}.2.\sin2\alpha }{\cos\alpha}\\
=\dfrac{2\sin\alpha\cos\alpha }{\cos\alpha}\\
=2\sin\alpha$
