Verify if cos^4x + 2sin^2x - sin^4x = 1.
We have to verify if cos^4x + 2sin^2x - sin^4x = 1.
Now cos^4x + 2sin^2x - sin^4x
=> [(cos x)^2]^2 + 2sin^2x - sin^4x
Now use the relation (cos x)^2 = 1 - (sin x)^2
=> [1 - (sin x)^2]^2 + 2sin^2x - sin^4x
Open the brackets
=> 1 + (sin x)^4 - 2( sin x)^2 +2sin^2x - sin^4x
cancel the common terms
Therefore we have proved that cos^4x + 2sin^2x - sin^4x = 1
To verify if cos^4x + 2sin^2x - sin^4x = 1.
We know that LHS : cos^4- - sin^4x = (cos^2x-sin^2x)(cos^2+sin^2x) = cos^2x-sin^2x, as cos^2+sin^2x = 1.
Therefore cos^4x-sin^4 = cos^2x-sin^2x.
cos^4x-sin^4x = (1-sin^2) -sin^2x , as ccos^2 = 1-sin^2x
=> cos^4x- sin^4x = 1-2sin^2x.
We add 2sin^2sx to both sides and get:
cos^4x+2sin^2-sin^4 x = 1-2sin^2x +2sin^2x = 1.
Therefore cos^4x+2sin^2-sin^4x = 1.