SOLUTION: sin^2x - sin^4x = (cos^2)(sin^2x) sin^2x - (sin^2x)(sin^2x) = (cos^2x)(sin^2x) sin^2x - (sin^2x)(1/csc^2x) = (cos^2x)(1/csc^2x) sin^2x - (sin^2x/csc^2x) = (cos^2x / csc^2x)

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: sin^2x - sin^4x = (cos^2)(sin^2x) sin^2x - (sin^2x)(sin^2x) = (cos^2x)(sin^2x) sin^2x - (sin^2x)(1/csc^2x) = (cos^2x)(1/csc^2x) sin^2x - (sin^2x/csc^2x) = (cos^2x / csc^2x)       Log On

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Question 117947: sin^2x - sin^4x = (cos^2)(sin^2x)
sin^2x - (sin^2x)(sin^2x) = (cos^2x)(sin^2x)
sin^2x - (sin^2x)(1/csc^2x) = (cos^2x)(1/csc^2x)
sin^2x - (sin^2x/csc^2x) = (cos^2x / csc^2x)
cos^2x [sin^2x - (sin^2x / csc^2x) = (cos^2x / csc^2x)]
(sin^2x)(csc^2x) - sin^2x = cos^2x
(sin^2x)(csc^2x) = cos^2x + sin^2x
(sin^2x)(csc^2x) = 1
(1/csc^2x)(csc^2x/1) = 1
1 = 1
Can I prove the above identity this way? Or have I taken a leap of faith somewhere along the line?
Thank you very much.
Answer by jim_thompson5910(14954) About Me  (Show Source):
You can put this solution on YOUR website!
When you prove identities, you only manipulate one side. So I'm going to change the left side into the right side


sin^2x - sin^4x = (cos^2x)(sin^2x).....Start with the given equation


sin^2x - (sin^2x)(sin^2x) = (cos^2x)(sin^2x).....Rewrite sin^4x as (sin^2x)(sin^2x)


sin^2x(1 - sin^2x) = (cos^2x)(sin^2x).....Factor out sin^2x


sin^2x(cos^2x) = (cos^2x)(sin^2x).....Replace 1-sin^2x with cos^2x



(cos^2x)(sin^2x) = (cos^2x)(sin^2x).....Rearrange the left side



So this verifies the identity