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I was working through some trig exercises when I stumbled upon the following problem:

Prove that: $ \cos(A+B) \cdot \cos(A-B)=\cos^2A- \sin^2B$.

I started out by expanding it such that $$ \cos(A+B) \cdot \cos(A-B)=(\cos A \cos B-\sin A \sin B) \cdot (\cos A \cos B+ \sin A \sin B),$$ which simplifies to: $$ \cos^2 A \cos^2 B- \sin^2 A \sin^2 B .$$ However, I don't know how to proceed from here. Does anyone have any suggestions on how to continue.

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2 Answers 2

up vote 4 down vote accepted

You're on the right track! From where you left off:

$\cos^2A\cos^2B-\sin^2A\sin^2B = \cos^2A(1-\sin^2B)-\sin^2A\sin^2B$ $ = \cos^2A - \sin^2B(\cos^2A+\sin^2A)=\cos^2A-\sin^2B$

Note: using \sin and \cos results in prettier TeX formatting.

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How did you get from $\cos^2 A \cos^2 B=\cos^2 A (1-\sin^2 B)$? –  E.O. Dec 17 '11 at 2:49
    
From $\sin^2B + \cos^2B = 1$. –  Bruno Joyal Dec 17 '11 at 2:50
    
Ahh! Thanks a lot! –  E.O. Dec 17 '11 at 2:51
    
You are most welcome Emile. :) –  Bruno Joyal Dec 17 '11 at 2:53

The identities

$$\cos(\theta) = \frac{e^{i \theta}+e^{- i \theta}}{2}$$ $$\sin(\theta) = \frac{e^{i \theta}-e^{- i \theta}}{2i}$$

can reduce a trigonometric identity to a identity of polynomials. Let's see how this works in your example:

$$\cos(A+B) \cos(A-B)=\cos(A)^2-\sin(B)^2$$

is rewritten into:

$$\frac{e^{i (A+B)}+e^{- i (A+B)}}{2} \frac{e^{i (A-B)}+e^{- i (A-B)}}{2}=\left(\frac{e^{i A}+e^{- i A}}{2}\right)^2-\left(\frac{e^{i B}-e^{- i B}}{2i}\right)^2$$

now we change it into a rational function of $X = e^{i A}$ and $Y = e^{i B}$:

$$\frac{(X Y + \frac{1}{X Y})(\frac{X}{Y} + \frac{Y}{X})}{4} = \frac{(X + \frac{1}{X})^2 + (Y - \frac{1}{Y})^2}{4}$$

and you can simply multiply out both sides to see that they are both $\frac{1}{4}\left(X^2 + \frac{1}{X^2} + Y^2 + \frac{1}{Y^2}\right)$ which proves the trigonometric equality.

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