How do you verify $sin^2 theta cos ^2 theta = 1/8 [1-cos(4 theta)]$ ?

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Answer 1 · 2015-05-25T15:49:22

Use trig identities: $2sin a.cos a = sin 2a$
and $2sin^2 a = 1 - cos 2a$

$sin^2 a.cos^2 a$ = $(1/4)(sin^2 2a)$ = $(1/8)(2sin^2 2a)$ =

$= (1/8)(1 - cos 4a)$

How do you verify sin^2 theta cos ^2 theta = 1/8 [1-cos(4 theta)]sin^2 theta cos ^2 theta = 1/8 [1-cos(4 theta)]?