# Trigonometry Help!

I am having extreme difficulty with my Trigonometery homework.

In particular, When I simplify the expression:
cos2xsin2x-cos2x
I come up with cos2x.

This is obviously incorrect. Where am I going wrong?

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TD
Homework Helper
Well it depends on how you want to simplify. Do you need to get rid of the double angles or just to 'shortes' possible way of expression?

$\cos{2x}$ is indeed wrong...

You could factor $\cos{2x}$, don't know if that's what you want.
You could also use the double angle formula backwards on $\cos{2x}\sin{2x}$ to get $\frac{\sin{4x}}{2}$

I should have written it:

cos(squared) x sin(squared) x - cos(squared) x

shortest way

TD
Homework Helper
What do you mean? Now you say $\cos ^2 x\sin ^2 x - \cos ^2 x$.
That's not the same as your initial expressions You mean that's what you actually meant? What did you try already then?

Still trying figure out how to get all of the expressions to translate into typed text....sorry.

I cos(sq'd)x-cos(sq'd)x
then 1-sin(sq'd)=cos(sq'd)X

this is why I came up with cos(sq'd)x

TD
Homework Helper
Well, I'd do about the same but you get something different:

$$\cos ^2 x\sin ^2 x - \cos ^2 x = \cos ^2 x\left( {1 - \cos ^2 x} \right) - \cos ^2 x = \cos ^2 x - \cos ^4 x - \cos ^2 x = - \cos ^4 x$$

I see the step I missed. Thank you very much.
School is hard when you have been out of it for 18 years. TD
Homework Helper
I can imagine you need some freshing up 