How can you simplify #sin^4a + cos^4a#?

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Answer 1 · 2017-07-17T15:15:02

#sin^4a + cos^4a = 1 - 2sin^2acos^2a = 1 - 1/2sin^2(2a)#.

Note that

#(sin^2a + cos^2a)^2 = sin^4a + 2sin^2acos^2a + cos^4a#

We see that #sin^2x + cos^2x = 1#, so that

#1 = sin^4a + 2sin^2acos^2a + cos^4a#

Therefore,

#sin^4a + cos^4a = 1 - 2sin^2acos^2a#

Hopefully this helps!