# trigonometric identities problem

• August 15th 2010, 03:58 AM
khms
trigonometric 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!
• August 15th 2010, 05:01 AM
Unknown008
$cos(\theta + \phi) cos(\theta - \phi) = (cos\theta cos\phi - sin\theta sin \phi)(cos\theta cos\phi + sin\theta sin \phi)$

$=cos^2\theta cos^2\phi + cos\theta sin\theta cos\phi sin\phi - cos\theta sin\theta cos\phi sin\phi - sin^2 \theta sin^2 \phi$

$=cos^2\theta cos^2\phi - sin^2 \theta sin^2 \phi$

$=cos^2\theta (1- sin^2\phi) - sin^2 \theta (1-cos^2 \phi)$

$=cos^2\theta - cos^2\theta sin^2\phi - sin^2 \theta+ sin^2\theta cos^2 \phi$

$=cos^2\theta - sin^2 \theta+ sin^2\theta cos^2 \phi - cos^2\theta sin^2\phi$

Unless $sin^2\theta cos^2 \phi - cos^2\theta sin^2\phi = 0$ (which I don't think is true), I think there was a mistake in the question you got.
• August 15th 2010, 05:07 AM
Quote:

Originally Posted by khms
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!

$Cos(A+B)=CosACosB-SinASinB$

$Cos(A-B)=Cos[A+(-B)]=CosACos(-B)-SinASin(-B)$

$Cos(-B)=CosB,\ Sin(-B)=-SinB$

$Cos(A-B)=CosACosB+SinASinB$

$[Cos(A+B)]Cos(A-B)=[CosACosB-SinASinB][CosACosB+SinASinB]$

$=Cos^2ACos^2B-Sin^2ASin^2B$

$=Cos^2A\left(1-Sin^2B\right)-Sin^2ASin^2B=Cos^2A-Sin^2B\left(Cos^2A+Sin^2A\right)=Cos^2A-Sin^2B$

which is what you got!
• August 15th 2010, 05:46 AM
khms
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.
• August 15th 2010, 06:02 AM
Quote:

Originally Posted by khms
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.

You will get that answer if $\theta=\varphi$

since $Cos(2A)=Cos^2A-Sin^2A$
• August 15th 2010, 06:12 AM
khms
Quote:

You will get that answer if $\theta=\varphi$

since $Cos(2A)=Cos^2A-Sin^2A$

so what you're saying is it is impossible to get cos²θ-sin²θ as the answer for that equation?
• August 15th 2010, 06:24 AM
Prove It
No, what they are saying is that $\cos^2{\theta} - \sin^2{\theta}$ is only possible if $\theta = \varphi$...
• August 17th 2010, 07:07 AM
martinbell
There 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)