verifying trig identities

**robasc** · November 18th 2008, 07:16 AM

Could someone please explain to me how to get through to the next step of verifying the identity?

$ cos^3 x sin^2 x = (sin^2 x - sin^4 x) cos x $

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

$ cos x sin^2 x - sin^4 x cos x = $

What Next?

**Jhevon** · November 18th 2008, 07:29 AM

Originally Posted by robasc

I am tyring to prove that:

$ cos^3 x sin^2 x $

Equals:
$ (sin^2 x - sin^4 x) cos x $

Prove $\cos^3 x \sin^2 x \equiv (\sin^2 x - \sin^4 x) \cos x$

Consider the LHS

$\cos^3 x \sin^2 x = \cos x \cos^2 x \sin^2 x$

$= \cos x (1 - \sin^2 x ) \sin^2 x$

$= \cdots$

**robasc** · November 18th 2008, 07:45 AM

Prove

Consider the LHS

__________________

So the next move would be:

$ cos x(1 - sin^2 x) sin^2 x = $

$ cos x(sin^2 x - sin^4 x) $

and that's it!

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