- #1

TonyC

- 86

- 0

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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- Thread starter TonyC
- Start date

- #1

TonyC

- 86

- 0

In particular, When I simplify the expression:

cos2xsin2x-cos2x

I come up with cos2x.

This is obviously incorrect. Where am I going wrong?

- #2

TD

Homework Helper

- 1,022

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[itex]\cos{2x}[/itex] is indeed wrong...

You could factor [itex]\cos{2x}[/itex], don't know if that's what you want.

You could also use the double angle formula backwards on [itex]\cos{2x}\sin{2x}[/itex] to get [itex]\frac{\sin{4x}}{2}[/itex]

- #3

TonyC

- 86

- 0

I should have written it:

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

shortest way

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

shortest way

- #4

TD

Homework Helper

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That's not the same as your initial expressions

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

- #5

TonyC

- 86

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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

- #6

TD

Homework Helper

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[tex]\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[/tex]

- #7

TonyC

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School is hard when you have been out of it for 18 years.

- #8

TD

Homework Helper

- 1,022

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I can imagine you need some freshing up

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