How can you do?: #cos(x) sin(4x) - 2sin^2(2x) sin(x) = sin(5x) - sin(x)#

Question

377 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-14T14:38:35

One verifies an identity by using identities and axioms to change only one side of the equation until it is identical to the other side.

Given: #cos(x) sin(4x) - 2sin^2(2x) sin(x) = sin(5x) - sin(x)#

Use the power reduction identity #sin^2(theta) = (1-cos(2theta))/2# where #theta = 2x#:

#cos(x) sin(4x) - 2(1-cos(4x))/2 sin(x) = sin(5x) - sin(x)#

#2/2# becomes 1:

#cos(x) sin(4x) - (1-cos(4x)) sin(x) = sin(5x) - sin(x)#

Distribute the #-sin(x)# factor:

#cos(x) sin(4x) +sin(x)cos(4x) - sin(x) = sin(5x) - sin(x)#

We recognize the identity #cos(x) sin(4x) +sin(x)cos(4x) = sin(4x+x)#:

#sin(4x+x) - sin(x) = sin(5x) - sin(x)#

Substitute #4x + x = 5x#:

#sin(5x) - sin(x) = sin(5x) - sin(x)# Q.E.D.

Answer 2 · 2018-03-14T14:38:52

Kindly go through the following Explanation.

We will use the following Identities :

#(1): sin2theta=2sinthetacostheta#.

#(2): cosucosv-sinusinv=cos(u+v)#.

#(3): 2cosusinv=sin(u+v)-sin(u-v)#.

#"Now, "sin4xcosx-2sin^2 2xsinx#,

#=(2sin2xcos2x)cosx-2sin^2 2xsinx#,

#=2sin2x{cos2xcosx-sin2xsinx}#,

#=2sin2x{cos(2x+x)}#,

#=2cos3xsin2x#,

#=sin(3x+2x)-sin(3x-2x)#,

#=sin5x-sinx#, as desired!