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

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)

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(35256) (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

RELATED QUESTIONS
Prove: 1-cot^2x/1+cot^2x=2cos^2-1 I tried this: LHS... (answered by Alan3354)

prove: 2-cos^2x =... (answered by stanbon)

√3 sin 2x + 2 cos^2x =... (answered by greenestamps)

sin^2x ? 1-cos... (answered by Alan3354)

sin x = cos 2x -... (answered by jsmallt9)

sin(2x)+cos(x)=1 (answered by AnlytcPhil)

Verify each identity:... (answered by Alan3354)

please prove sin^2x + cos^4x = cos^2x +... (answered by stanbon)

Simplify: cos(4x) cos(2x) - sin(2x)... (answered by ikleyn,solve_for_x)