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!
$\displaystyle cos(\theta + \phi) cos(\theta - \phi) = (cos\theta cos\phi - sin\theta sin \phi)(cos\theta cos\phi + sin\theta sin \phi)$
$\displaystyle =cos^2\theta cos^2\phi + cos\theta sin\theta cos\phi sin\phi - cos\theta sin\theta cos\phi sin\phi - sin^2 \theta sin^2 \phi$
$\displaystyle =cos^2\theta cos^2\phi - sin^2 \theta sin^2 \phi$
$\displaystyle =cos^2\theta (1- sin^2\phi) - sin^2 \theta (1-cos^2 \phi)$
$\displaystyle =cos^2\theta - cos^2\theta sin^2\phi - sin^2 \theta+ sin^2\theta cos^2 \phi$
$\displaystyle =cos^2\theta - sin^2 \theta+ sin^2\theta cos^2 \phi - cos^2\theta sin^2\phi $
Unless $\displaystyle sin^2\theta cos^2 \phi - cos^2\theta sin^2\phi = 0$ (which I don't think is true), I think there was a mistake in the question you got.
$\displaystyle Cos(A+B)=CosACosB-SinASinB$
$\displaystyle Cos(A-B)=Cos[A+(-B)]=CosACos(-B)-SinASin(-B)$
$\displaystyle Cos(-B)=CosB,\ Sin(-B)=-SinB$
$\displaystyle Cos(A-B)=CosACosB+SinASinB$
$\displaystyle [Cos(A+B)]Cos(A-B)=[CosACosB-SinASinB][CosACosB+SinASinB]$
$\displaystyle =Cos^2ACos^2B-Sin^2ASin^2B$
$\displaystyle =Cos^2A\left(1-Sin^2B\right)-Sin^2ASin^2B=Cos^2A-Sin^2B\left(Cos^2A+Sin^2A\right)=Cos^2A-Sin^2B$
which is what you got!
... 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