Can someone help me solve sin^3xcos^2x=cos^2xsinx-cos^4xsinx? I need to transform one side into the other
2006-11-08 13:02:03 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
sin^3xcos^2x=cos^2xsinx-cos^4xsinx
2006-11-08 13:18:48 · update #1
sin^3xcos^2x=cos^2xsinx-cos^4xsinx can you help me solve?
2006-11-08 13:22:17 · update #2
at the end it is coz^4xsinx but it wouldnt show up
2006-11-08 13:24:55 · update #3
Did you get the question correct?
Start from right-hand side:
cos^2x sinx - cos^4x
= cos^2x [sinx - cos^2x]
= cos^2x [sinx - (1 - sin^2x)]
= cos^2x [sin^2x + sinx -1]
You can't transform (sin^2x + sinx - 1) into sin^3x.
2006-11-08 13:12:07 · answer #1 · answered by orhhai 2 · 0⤊ 0⤋
Well lets see. The first thing I would do is get a list of all the trig identities that I know. Then, I'd guess and try to visualize every property I could apply.
I'd work with the right side. Factor out cos^2x. It cancels out from both sides. Work from there.
I plugged in x=8 for the equation and one side equal .0204 and the other .0205 but that could've been my calculator or I misread the question. Are those coefficients or exponents?
2006-11-08 21:16:01 · answer #2 · answered by McKevel 2 · 0⤊ 0⤋