How do you prove #sin^2(x)cos^2(x) = (1-cos(4x))/8#?

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Answer 1 · 2015-05-14T23:05:07

Let's start form the Left Hand Side(#"LHS"#):

#=>" LHS "=sin^2xcos^2x =(2/2sinxcosx)^2#

Remember that, #2sinxcosx=sin2x#

#=>" LHS "=(1/2sin2x)^2 = 1/4sin^2(2x)#

Use the half angle identity, #cos2theta=1-2sin^2theta#

Replace #theta# by #2x# #=>cos4x=1-2sin^2(2x)#

Rearrange that #" ; "sin^2(2x)=1/2(1-cos4x)#

#=> " LHS"=1/4(1/2(1-cos4x))#

#=1/8(1-cos4)x=(1-cos(4x))/8=" RHS"#

And that's it!