Đáp án:
$A=\tan 4x\\
B=\dfrac{1}{64}\sin8x$
Giải thích các bước giải:
$A=\dfrac{\sin3x+\sin5x}{\cos3x+\cos5x}\\
=\dfrac{2\sin\dfrac{3x+5x}{2}\cos\dfrac{3x-5x}{2}}{2\cos\dfrac{3x+5x}{2}\cos\dfrac{3x-5x}{2}}\\
=\dfrac{\sin4x\cos x}{\cos4x\cos x}\\
=\tan 4x\\
B=\dfrac{1}{8}\sin x\cos x\cos2x\cos4x\\
=\dfrac{1}{8}.\dfrac{1}{2}\sin2x\cos2x\cos4x\\
=\dfrac{1}{16}.\dfrac{1}{2}\sin4x\cos4x\\
=\dfrac{1}{32}.\dfrac{1}{2}\sin8x\\
=\dfrac{1}{64}\sin8x$
