Giúp bài lượng giác

[motsak]

1. Rút gọn:

\[A=2\sin\left( x+\frac{\pi}{2}\right) +\sin (5\pi -x)+\sin\left(\frac{3\pi}{2}+x\right) +\cos\left(\frac{\pi}{2}+x\right)\]

\[B= \sin (3\pi - x) - 2\cos (5\pi +x) + \sin \left( \frac{7\pi}{2} - x\right) + \cos\left(\frac{9\pi}{2} + x\right) + \tan (3\pi +x ) + \cot (9\pi - x)\]

2. Biến đổi về tích

\[A= 1 - 4\sin ^2 4x\]

\[B= 1 + 2\cos 2x\]

\[C= \cos x + \cos 2x + \cos 3x + \cos 4x\]

3. CMR:

a, \[sin4x =\frac{1}{8}\cos 4x - \frac{1}{2}\cos 2x + \frac{3}{8}\]

b, \[ 4\sin^3x\cos 3x + 4\cos ^3x\sin 3x=3\sin 4x\]

 

[proboyvip1995]

chia cả 2 vế cho 4 ta có:

\[sin^{3}cos3x + cos^{3}xsin3x = \frac{3}{4}sin4x\]
\[VT= \frac{1}{4}(3sinx - sin3x)cos3x + \frac{1}{4}(3cosx - cos3x)sin3x\]
\[=\frac{3}{4}(sinx.cos3x + cosx.sin3x)=\frac{3}{4}sin4x = VP\]