prove that cos 4x= 8sin^2 x cos^2x
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jay61:
i think the question is incorrect... on the RHS, it should be 1 - 8Sin²xCos²x
no. it is right
But the question is Cos4x = 8Sin²xCos²x.. and the result comes out to be 1 - 8Sin²xCos²x
Answers
Answered by
9
LHS
= cos 4x
= cos 2(2x)
= 1 - 2sin² 2x ∵ cos 2A = 1 - sin²A
= 1 - 2(2sinx cosx )² ∵ sin 2A = 2sinA cosB
= 1 - 8 sin²xcos²x
= RHS
= cos 4x
= cos 2(2x)
= 1 - 2sin² 2x ∵ cos 2A = 1 - sin²A
= 1 - 2(2sinx cosx )² ∵ sin 2A = 2sinA cosB
= 1 - 8 sin²xcos²x
= RHS
Answered by
0
cos4x can be written as
1 - 2sin^2 2x
(cos 2x = 1 - 2 sin^2x)
1 - 2 sin^2 2x = 1 - 2(sin^2 2x)
= 1 - 2(2sin x cos x)^2
= 1 - 2(4 sin^2 x cos^2 x)
= 1 - 8 sin^2 x cos^2 x
Hope This Helps You!
1 - 2sin^2 2x
(cos 2x = 1 - 2 sin^2x)
1 - 2 sin^2 2x = 1 - 2(sin^2 2x)
= 1 - 2(2sin x cos x)^2
= 1 - 2(4 sin^2 x cos^2 x)
= 1 - 8 sin^2 x cos^2 x
Hope This Helps You!
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