1/ $\frac{cos(x-\frac{\pi }{4}).cos(\frac{\pi }{4}-x)}{sin^2(x+\frac{\pi }{4})}=1$

2/ $\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}=2cosx$
câu 1: tử biến đỏi cos thành sin là racâu 2:biến đổi cos2x và cos3x ra cosx là ra rùi –  mr_chuyen9x 14-06-13 02:47 PM
1/ chứng minh bằng gì thế? –  alxu_s2_vuvu 24-04-13 08:47 PM
2/
$\frac{1 + cosx + cos2x + cos3x}{2cos^{2}x + cosx - 1}$ 
$(biện luận,nhân cả tử và mẫu với sin\frac{x}{2})$
= $\frac{sin\frac{x}{2} + \frac{1}{2}( sin\frac{3x}{2} - sin\frac{x}{2} + sin\frac{5x}{2} - sin\frac{3x}{2} + sin\frac{7x}{2} - sin\frac{5x}{2}) }{ sin\frac{5x}{2} - sin\frac{3x}{2} + sin\frac{3}{x} - sin\frac{x}{2}}$  
=$\frac{sin\frac{x}{2} + sin\frac{7x}{2}} {2( sin \frac{5x}{2} - sin\frac{x}{2})}$
=$\frac{2sin4x.cos3x}{4cos3x.sin2x}$
=$\frac{1}{2}(2.cos2x)$
= cos2x

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