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Question

$tan(2x-\pi/4)tan(2x+\pi/4)=4cos^2x/(tanx-cotx)$

score 1 · Answer 1

ĐK sin x cos x # 0, tan x – cot x # 0, cos (2x – π/4) cos (2x + π/4) # 0

VT=sin (2x – π/4) sin (2x + π/4) / [cos (2x – π/4) cos (2x + π/4)]

<=> [cos (π/2) – cos 4x] / [cos 4x + cos (π/2)]

= -1 (*)

VP= (4cos² 2x) / (sin x / cos x – cos x / sin x)

= (4cos² 2x) sin x cos x / (sin² x – cos² x)

= (4cos² 2x) sin x cos x / (– cos 2x)

= -4 cos 2x sin x cos x

= 2 cos 2x sin 2x

= sin 4x (**)

tu (*)(**) ta co pt da cho tro thanh sin 4x = 1

Nero · Answer 2 · 6/25/2014 10:47:44 PM

Đk tự đặt nhé!

Ta có: $VT=\frac{(sin2x+cos2x)(sin2x-cos2x)}{-(sin2x-cos2x)(sin2x+cos2x)}=-1$

Ta lại có: $cotx-tanx=\frac{cos2x}{sinx.cosx}$

Pt $\Leftrightarrow cotx-tanx=4cos^22x$

$\Leftrightarrow cos2x(2cos2x.2sinxcosx-1)=0$

$\Leftrightarrow cos2x(2sin2x.cos2x-1)=0$

$\Leftrightarrow cos2x(sin4x-1)=0$

Tự làm nhé nhớ hợp nghiệm đấy

Sửa 25-06-14 10:47 PM

Nero
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Trả lời 25-06-14 10:47 PM

Nero
1 1

299K 309K

hic.đúng luk mk cug làm ra nhưng cug cảm ơn p nhiều.^_^ – white cloud 25-06-14 10:56 PM

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