Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$

Question

Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$

I have got as far as showing that:

$\cos(x+y)\cos(x-y) = \cos^2x\cos^2y -\sin^2x\sin^2y$

and

$\sin(x+y)\sin(x-y) = \sin^2x\cos^2y - \cos^2x\sin^2y$

I get stuck at showing:

$\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y - \cos^2x\sin^2y = \cos^2x - \sin^2x$

I know that $\sin^2x + \cos^2x = 1$ and I have tried rearranging this identity in various ways, but this has not helped me so far.

I believe one of your minus signs where you are stuck is incorrect. — JB King, Jul 15 '13 at 18:58
Specifically, the minus sign before $\cos^2x\sin^2y$ should be a plus. — egreg, Jul 15 '13 at 19:00
Hint: $\sin^2 x\sin^2 y + \sin^2 x\cos^2 y = \sin^2 x(\sin^2 y + \cos^2 y) = \sin^2 x$. And yes, as pointed out by JB King, you've got one sign wrong. — Peter Košinár, Jul 15 '13 at 19:00

score 5 · Answer 1 · edited Jul 15 '13 at 19:14

5

Hint: $$\cos(a+b)=\cos a \cos b-\sin a \sin b$$$$\cos(2a)=\cos^2 a -\sin ^2a $$

$$\begin{align}\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) &= \cos((x+y)+(x-y))\\&=\cos2x\\&=\cos^2x - \sin^2x\end{align}$$

edited Jul 15 '13 at 19:14

Pedro

122,002

answered Jul 15 '13 at 19:03

M.H

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Very nice +1 ${}{}{}{}{}$ – Pedro Jul 15 '13 at 19:13
@Peter Tamaroff:thanks peter – M.H Jul 15 '13 at 19:17
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It’s really a matter of pattern recognition rather than of mathematics. – Lubin Jul 15 '13 at 19:20
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@Lubin How is "pattern recognition" dissociated from "mathematics"? – Pedro Jul 15 '13 at 19:34
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Although I upvoted your comment, @PeterTamaroff, there are many cases of pattern recognition that have nothing to do with mathematics, unless one force an unnaturally broad definition on mathematics; and there are many mathematical ideas and techniques that have nothing to do with pattern recognition. – Lubin Jul 16 '13 at 15:05

score 2 · Accepted Answer · edited Jul 16 '13 at 15:09

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Looking at $\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y + \cos^2x\sin^2y$ (The last is a negative times a negative that forms a positive I believe.

You could pull $\sin^2x$ from the middle two terms to get this expression: $ -\sin^2x\sin^2y - \sin^2x\cos^2y = -\sin^2x(\sin^2y+\cos^2y) = -\sin^2x $

There is a similar reduction with the first and last terms around $\cos^2x$ that should make this appear easier.

edited Jul 16 '13 at 15:09

Aang

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answered Jul 15 '13 at 19:02

JB King

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The correction of my wrong minus sign was very helpful. Thank you. – mikoyan Jul 15 '13 at 19:18

Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$

2 Answers2