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Trigonometry Help!

  • Thread starter TonyC
  • Start date
86
0
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?
 

Answers and Replies

TD
Homework Helper
1,020
0
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?

[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]
 
86
0
I should have written it:

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

shortest way
 
TD
Homework Helper
1,020
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What do you mean? Now you say [itex]\cos ^2 x\sin ^2 x - \cos ^2 x[/itex].
That's not the same as your initial expressions :confused:

You mean that's what you actually meant? What did you try already then?
 
86
0
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
1,020
0
Well, I'd do about the same but you get something different:

[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]
 
86
0
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
Homework Helper
1,020
0
I can imagine you need some freshing up :wink:
 

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