Use the half-angle identities sin2x = [1-cos(2x)]/2 and cos2x = [1+cos(2x)]/2:
sin2x·cos2x
= [1-cos(2x)]/2 * [1+cos(2x)]/2
= [1-cos2(2x)]/4
= sin2(2x)/4
= [[1-cos(2(2x))]/2]/4
= [1-cos(4x)]/8
Hope this helped!
Javian W.
asked 04/29/23Use the half-angle identities sin2x = [1-cos(2x)]/2 and cos2x = [1+cos(2x)]/2:
sin2x·cos2x
= [1-cos(2x)]/2 * [1+cos(2x)]/2
= [1-cos2(2x)]/4
= sin2(2x)/4
= [[1-cos(2(2x))]/2]/4
= [1-cos(4x)]/8
Hope this helped!
Richard C. answered 04/29/23
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AJ L.
04/29/23