In △ABC, if sin A=sin^2 B then prove 4 cos 2A−4 cos 2B=1−cos 4B


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Question Number 37914 by gunawan last updated on 19/Jun/18

$In △ A B C, if \sin A = \sin^{2} B$ $then prove$ $4 \cos 2 A - 4 \cos 2 B = 1 - \cos 4 B$
	Answered by $@ty@m last updated on 19/Jun/18
	$LHS= 4(cos 2A−cos 2B) =4.2sin (A+B)sin (B−A) =8(sin Bcos A+cos Bsin A). (sin Bcos A−cos Bsin A). =8(sin^2 Bcos^2 A−cos^2 Bsin^2 A) =8{sin^2 B(1−sin^2 A)−(1−sin^2 B)sin^2 A} =8{sinA(1−sin^2 A)−(1−sinA)sin^2 A} =8(sin A−sin^3 A−sin^2 A+sin^3 A) =8(sin^2 B−sin^4 B) =8sin^2 Bcos^2 B =2.(2sin Bcos B)^2 =2sin^2 2B =1−cos 4B =RHS$
	$L H S =$ $4 (\cos 2 A - \cos 2 B)$ $= 4 . 2 \sin (A + B) \sin (B - A)$ $= 8 (\sin B \cos A + \cos B \sin A) .$ $(\sin B \cos A - \cos B \sin A) .$ $= 8 (\sin^{2} B \cos^{2} A - \cos^{2} B \sin^{2} A)$ $= 8 {\sin^{2} B (1 - \sin^{2} A) - (1 - \sin^{2} B) \sin^{2} A}$ $= 8 {\sin A (1 - \sin^{2} A) - (1 - \sin A) \sin^{2} A}$ $= 8 (\sin A - \sin^{3} A - \sin^{2} A + \sin^{3} A)$ $= 8 (\sin^{2} B - \sin^{4} B)$ $= 8 \sin^{2} B \cos^{2} B$ $= 2 . {(2 \sin B \cos B)}^{2}$ $= 2 \sin^{2} 2 B$ $= 1 - \cos 4 B$ $= R H S$
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