1) Chứng minh đẳng thức sau: $\frac{1-tanx}{1+tanx}=tan(\frac{\pi}{4}-x)$ 
 2) Rút gọn biểu thức sau: $C=\frac{cos2a-cos4a}{sin2a+sin4a}$
Câu 1.
$VT=\frac{cosx-sinx}{cosx+sinx}=\frac{\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx}{\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx}$
$=\frac{sin\frac{\pi}{4}.cosx-cos\frac{\pi}{4}.sinx}{cos\frac{\pi}{4}.cosx+sin\frac{\pi}{4}.sinx}$
$=\frac{sin(\frac{\pi}{4}-x)}{cos(\frac{\pi}{4}-x)}=tan(\frac{\pi}{4}-x)=VP$(đpcm)
Câu 2.
$C=\frac{-(2cos^22x-cos2x-1)}{sin2x+2sin2x.cos2x}=\frac{-(cos2x-1)(1+2cos2x)}{sin2x(1+2cos2x)}=\frac{1-cos2x}{sin2x}$
$=\frac{2sin^2x}{2sinx.cosx}=\frac{sinx}{cosx}=tanx$
uk. sai ^^ –  Nero 24-04-14 12:50 PM
sai rồi kìa :D... cos2x-cos4x chứ k phải cos4x-cos2x đâu –  ♂Vitamin_Tờ♫ 24-04-14 12:43 PM
1/$\tan(\frac{\pi}{4}-x)=\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}.\tan x}=\frac{1-\tan x}{1+\tan x}$ (đpcm)
$2/C=\frac{-2\sin3x.\sin(-x)}{2\sin3x.\cos(-x)}=\tan x$
Được. Ý tưởng ko trùng lặp :)) –  Nero 24-04-14 12:42 PM

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