SOLUTION: hello once again.....Could someone please establish this identity for me ? sin^3xcos^2x =sinx(cos^2x-cos^4x) thank you in advance !!

Algebra ->  Finance -> SOLUTION: hello once again.....Could someone please establish this identity for me ? sin^3xcos^2x =sinx(cos^2x-cos^4x) thank you in advance !!      Log On


   

Question 927725: hello once again.....Could someone please establish this identity for me ? sin^3xcos^2x =sinx(cos^2x-cos^4x) thank you in advance !!
Found 2 solutions by brysca, MathLover1:
Answer by brysca(112) About Me  (Show Source):
You can put this solution on YOUR website!
solution posted below
.
hope this helps!

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

sin%5E3%28x%29cos%5E2%28x%29+=sin%28x%29%28cos%5E2%28x%29-cos%5E4%28x%29%29
start with right side and prove it's equal to left side:
sin%28x%29%28cos%5E2%28x%29-cos%5E4%28x%29%29
=sin%28x%29+cos%5E2%28x+%29%281-cos%5E2%28x%29%29+
=sin%28x%29+cos%5E2%28x%29+%28sin%5E2%28x%29%29+
=sin%5E3%28x%29+cos%5E2%28x%29