trig identity (sin²x-sin²xcos²x)=sin²x(1-cos²x)

[flyingfreedom]

trig identity (sin²x-sin²xcos²x)=sin²x(1-cos²x)

how do you solve
(sin²x-sin²xcos²x)=sin²x(1-cos²x)

I don't see any way of going foward.

 

[soroban]

Re: trig identity (sin²x-sin²xcos²x)=sin²x(1-cos²x)

Hello, flyingfreedom!

\(\displaystyle \text{How do you solve: }\;\sin^2\!x - \sin^2\!x\cos^2\!x \;=\;\sin^2\!x(1-\cos^2\!x)\)

Are you sure it says "solve"? . . . It's an identity.


If we're supposed to prove it's an identity, just factor the left side!

\(\displaystyle \text{Then the left side is: }\;\sin^2\!x - \sin^2\!x\cos\!x \;=\;\sin^2\!x(1-\cos^2\!x) \quad\hdots\quad There!\)

 

[flyingfreedom]

Re: trig identity (sin²x-sin²xcos²x)=sin²x(1-cos²x)

your right... I missed it... sometimes I just don't see the obvious