I'm having problems in proving that cos(θ+φ) cos(θ-φ) = cos²θ-sin²θ using the trigonometric identities. i'm having cos²θ-sin²φ as the answer. help!

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- Aug 15th 2010, 03:58 AMkhmstrigonometric identities problem
**I'm having problems in proving that cos(θ+φ) cos(θ-φ) = cos²θ-sin²θ using the trigonometric identities. i'm having cos²θ-sin²φ as the answer. help!**

- Aug 15th 2010, 05:01 AMUnknown008

Unless (which I don't think is true), I think there was a mistake in the question you got. - Aug 15th 2010, 05:07 AMArchie Meade
- Aug 15th 2010, 05:46 AMkhms
i know. but what i am looking for is the proof that LHS and RHS are equal. the answers i'm getting are incorrect. i need

**cos²θ-sin²θ**as the answer - not cos²θ-sin²φ. - Aug 15th 2010, 06:02 AMArchie Meade
- Aug 15th 2010, 06:12 AMkhms
- Aug 15th 2010, 06:24 AMProve It
No, what they are saying is that is only possible if ...

- Aug 17th 2010, 07:07 AMmartinbellThere is a typo in the question !
http://www.mathhelpforum.com/math-he...ed6e142d2a.png

http://www.mathhelpforum.com/math-he...afca39de5c.png

http://www.mathhelpforum.com/math-he...f7c1aa2d52.png

... from previous poster. Continue as follows:

= (1 - sin²A).(1-sin²B) - sin²A.sin²B (sorry can't use greek letters !)

= 1 -sin²A - sin²B + sin²A.sin²B - sin²A.sin²B

= 1 -sin²A - sin²B

= sin²A + cos²A - sin²A - sin²B

= cos²A - sin²B

The question as written asked for the RHS to be cos²A - sin²A which is, of course, incorrect !

Good luck,

M(Nerd)