Trigonometry Help!

[TonyC]

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?

[TD]

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}

[TonyC]

I should have written it:

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

shortest way

[TD]

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 :confused:

You mean that's what you actually meant? What did you try already then?

[TonyC]

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]

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

[TonyC]

I see the step I missed. Thank you very much.
School is hard when you have been out of it for 18 years. :smile:

[TD]

I can imagine you need some freshing up :wink: