Verify if cos^4x + 2sin^2x - sin^4x = 1.
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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
=> 1
Therefore we have proved that cos^4x + 2sin^2x - sin^4x = 1
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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.
We'll re-write the identity, so that to get both sides differences of squares:
(cos x)^4 - (sin x)^4 = 1 - 2(sin x)^2
We'll write the difference of square from the left side, using the formula:
a^2 - b^2 = (a-b)(a+b)
(cos a)^4 - (sin a)^4 = [(cos a)^2 - (sin a)^2][(cos a)^2 + (sin a)^2]
We'll recall the fundamental formula of trigonometry:
(cos a)^2 + (sin a)^2 = 1
(cos a)^4 - (sin a)^4 = [(cos a)^2 - (sin a)^2]
We'll substitute (cos a)^2 = 1 - (sin a)^2
(cos a)^4 - (sin a)^4 = 1 - (sin a)^2 - (sin a)^2
We'll combine like terms:
(cos a)^4 - (sin a)^4 = 1 - 2(sin a)^2 q.e.d
We notice that it's no need to transform the right side, since working to the left side, we've get the same expression we have to the right side.
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