Prove that #cos^2xsin^2x = (1 - cos(4x))/8#?

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Answer 1 · 2017-07-04T08:36:01

It looks like you're asking why

#cos^2x sin^2x = (1 - cos(4x))/8#.

Well, here's one way to verify whether this is true or not.

Recall the following identities:

Notice how the #sin^2x# corresponds to an argument of #2x# in the #cos(2x)#. We'll use this idea later.

So far... identity #(1)# gives:

#cos^2x sin^2x = (1 - sin^2x)(sin^2x)#

Using #(2)# leads to:

#= (1 - (1 - cos(2x))/2)(1 - cos(2x))/2#

Getting common denominators:

#= ((1 + cos(2x))/2)((1 - cos(2x))/2)#

Multiply through for the difference of squares:

#= (1 - cos^2(2x))/4#

Use #(1)# again:

#= (sin^2(2x))/4#

And now, recycling #(2)#, we have that for #sin^2(2x)# instead of #sin^2x#, we get #(1 - cos(4x))/2# instead of #(1 - cos(2x))/2#:

#= ((1 - cos(4x))/2)/4 = (1 - cos(4x))/2 cdot 1/4#

#= color(blue)((1 - cos(4x))/8)#

Answer 2 · 2017-07-04T10:04:28

#LHS=cos^2xsin^2x#

#=1/4(2sinxcosx)^2#

#=1/4sin^2 2x#

#=1/4xx1/2(1-cos4x)#

#=(1-cos4x)/8=RHS#