Prove that sin^4 x+cos^4 x+(sin^2 2x)/2=1
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We have to prove that (sin x)^4 + (cos x)^4 + (sin 2x)^2/2 = 1
(sin x)^4 + (cos x)^4 + (sin 2x)^2/2
=> (sin x)^4 + (cos x)^4 + (2*sin x * cos x)^2/2
=> (sin x)^4 + (cos x)^4 + 4*(sin x)^2*(cos x)^2/2
=> (sin x)^4 + (cos x)^4 + 2*(sin x)^2*(cos x)^2
=> [(sin x)^2 + (cos x)^2]^2
=> 1^2
=> 1
This proves that (sin x)^4 + (cos x)^4 + (sin 2x)^2/2 = 1
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We'll use the Pythagorean identity to solve the problem.
(sin x)^2 + (cos x)^2 = 1
If we'll raise to square both sides, we'll get:
[(sin x)^2 + (cos x)^2]^2 = 1^2
(sin x)^4 + (cos x)^4 + 2 (sin x)^2 *(cos x)^2 = 1
We'll keep the sum (sin x)^4 + (cos x)^4 to the left:
(sin x)^4 + (cos x)^4 = 1 - 2 (sin x)^2 *(cos x)^2
We'll also apply the double angle identity:
sin 2x = 2 sin x*cos x
We'll raise to square both sides:
(sin 2x)^2 = 4 (sin x)^2 *(cos x)^2
We'll divide by 2:
[(sin 2x)^2]/2 =2 (sin x)^2 *(cos x)^2
We'll re-write the identity to be proved:
1 - 2 (sin x)^2 *(cos x)^2 + 2 (sin x)^2 *(cos x)^2 = 1
We'll eliminate like terms:
1 = 1
We'll get equal values both sides, therefore the identity (sin x)^4 + (cos x)^4 + [(sin 2x)^2]/2 = 1 is verified.
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