How do you prove that? Sin4x=4 cos2x sinx cosx

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Answer 1 · 2017-01-07T17:19:53

Use algebra and trigonometric identities to change only one side of the equation, until it looks the same as the other side.

Prove: $sin(4x) = 4cos(2x)sin(x)cos(x)$

I will only change the left side:

Substitute $sin(2x + 2x)" for "sin(4x)$ :

$sin(2x + 2x) = 4cos(2x)sin(x)cos(x)$

Use the identity $sin(A + B) = sin(A)cos(B) + cos(A)sin(B)$ where $A = B = 2x$ :

$sin(2x)cos(2x) + cos(2x)sin(2x) = 4cos(2x)sin(x)cos(x)$

Combine the two terms on the left:

$2cos(2x)sin(2x) = 4cos(2x)sin(x)cos(x)$

Use the identity sin(2x) = 2sin(x)cos(x) on the left:

$2cos(2x)2sin(x)cos(x) = 4cos(2x)sin(x)cos(x)$

Multiply the 2s on the left:

$4cos(2x)sin(x)cos(x) = 4cos(2x)sin(x)cos(x)$

The left side looks the same as the right side. Q.E.D.

Answer 2 · 2017-01-07T18:15:35

See the Proof in Explanation.

We have, $sin2theta=2sinthetacostheta$

Letting $theta=2x,$ we get, $sin(2(2x))=2sin2xcos2x,$ i.e.,

$sin4x=2cos2xsin2x$ , and, using the same identity for $sin2x$ ,

$sin4x=(2cos2x)(2sinxcosx)=4cos2xsinxcosx$ .