Bài 111314

Question

Chứng minh:
a) $\sin 5x - 2\sin x(\cos 2x+\cos 4x)=\sin x$
b) $\frac{\sin x.\sin(60^0+x)}{\sin 3x}=\frac{1}{4}$
c) $(\tan 2x-\tan x)(\sin 2x-\tan x)+1=\frac{1}{\cos^2 x} $
d) $\sin^2(\frac{\pi}{8}+\alpha )-\sin^2(\frac{\pi}{8}-\alpha )=\frac{1}{\sqrt[]{2} } \sin 2\alpha $.

hoàng anh thọ · Answer 1 · 7/11/2012 11:27:41 AM

a) $\sin 5x-2\sin x(\cos 2x+\cos 4x)=\sin 5x-2\sin x\cos 2x-2\sin x \cos 4x$
$=\sin 5x-(\sin 3x-\sin x)-(\sin 5x-\sin 3x)$$=\sin x$
b) $\sin x. \sin (60^0-x). \sin (60^0+x)=\frac{1}{2}\sin x(\cos 2x -\cos 120^0) $
$= \frac{1}{2}\sin x \cos 2x +\frac{1}{4} \sin x = \frac{1}{4}(\sin 3x-\sin x)+\frac{1}{4} \sin x =\frac{1}{4} \sin 3x $.
Suy ra đpcm.
c) $(\tan 2x-\tan x)(\sin 2x-\tan x)=\frac{\sin x}{\cos 2x \cos x}(2\sin x \cos x-\frac{\sin x}{\cos x} ) $
$= \frac{\sin^2 x(2\cos^2x-1)}{\cos 2x. \cos^2 x}=\tan^2x $
$\tan^2x=1=\frac{1}{\cos^2x} $ (đpcm)
d) Dùng công thức hạ bậc:
$\sin^2(\frac{\pi}{8}+\alpha )=\frac{1}{2}[1-\cos(\frac{\pi}{4}+2\alpha )]; \sin^2(\frac{\pi}{8}-\alpha )=\frac{1}{2}[1-\cos(\frac{\pi}{4}-2\alpha )] $
dễ dàng chứng minh

Bài 111314

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Bài 111314

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