How do you verify the identity of: #cos^3x sin^2x=(sin^2x-sin^4x)cosx#?

Question

13603 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-22T17:10:51

From basic definitions and the Pythagorean Theorem
#cos^2(x)+sin^2(x)=1#
or
#cos^2(x) = 1-sin^2(x)#

First consider
#(sin^2(x)-sin^4(x)#

#=(sin^2(x))*(1-sin^2(x))#

#=underbrace(sin^2(x)cos^2(x))_(used below)#

So
#cos^3(x)sin^2(x)#

#=(cos(x)) * [underbrace((cos^2(x)sin^2(x)))_(as above)]#

#= (cos(x))*(sin^2(x)-sin^4(x))#

#= (sin^2(x)-sin^4(x))cos(x)#