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

I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.

The way I did it:

$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$

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    $\begingroup$ Don't you have an $=$ on your keyboard? $\endgroup$ – Git Gud Mar 29 '13 at 14:52
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    $\begingroup$ The book I'm using saying you should use the "identical to" sign when dealing with identities, is this not the case? $\endgroup$ – seeker Mar 29 '13 at 14:53
  • $\begingroup$ What? That's probably a convention used only by geometers. Makes no sense if you ask me. What does $\equiv$ even mean here? How does it differ from $=$? It's just my opinion, though. $\endgroup$ – Git Gud Mar 29 '13 at 14:55
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    $\begingroup$ @GitGud: $$\cos(A+B)\cos(A-B)\equiv\cos^2A-\sin^2B$$ is an abbreviation for $$\forall A\forall B\bigl[\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B\bigr].$$ In contrast, while there exist $C$ such that $\sin C=\cos C$, we would never say $\sin C\equiv\cos C$. $\endgroup$ – Cameron Buie Mar 29 '13 at 15:37
  • $\begingroup$ @CameronBuie Thanks for clarifying. $\endgroup$ – Git Gud Mar 29 '13 at 15:38
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You can use the addition theorem which states that $$\cos(\alpha+\beta)=\cos(\alpha)\cdot \cos(\beta) -\sin(\alpha)\sin(\beta)$$

$$\cos(\alpha + \beta) \cdot \cos(\alpha -\beta)= ( \cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha) \cdot \cos(-\beta)-\sin(\alpha)\cdot \sin(-\beta))$$ As $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ this is equal to $$(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta))$$ With the third binom you get $$\cos^2(\alpha) \cos^2(\beta) -\sin^2(\alpha)\sin^2(\beta)$$ As $\cos^2(\beta)+\sin^2(\beta)=1$ we have $$\cos^2(\alpha)(1-\sin^2(\beta))-(1-\cos^2(\alpha))\sin^2(\beta)$$ Multplying it out gives you $$\cos^2(\alpha) -\sin^2(\beta)\cos^2(\alpha)-\sin^2(\beta)+\cos^2(\alpha)\sin^2 (\beta)$$ And this is $$\cos^2(\alpha)-\sin^2(\beta)$$

Another way: $$2\cos(\alpha)\cdot \cos(\beta)=\cos(\alpha+\beta) + \cos(\alpha-\beta)$$ Using this in your formula gives us $$\cos(A+B)\cdot \cos(A-B)=\frac{1}{2}\left( \cos(2A) + \cos(-2B)\right)=\frac{1}{2} \left(\cos(2 A)+ \cos(2 B)\right)$$ As $\cos(2 A)=1-2\sin^2(A)$ and $\cos(2 B) = 1-2 \sin^2 (B)$ this is equal to $$1-\sin^2(A) -\sin^2 (B)=\cos^2(A)-\sin^2(B)$$

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    $\begingroup$ This seems like the way I did it $\endgroup$ – seeker Mar 29 '13 at 15:06
  • $\begingroup$ @Assad when I wrote it I didn't know you already finished it on your own, (nor which way you have chosen) I will think of another way $\endgroup$ – Dominic Michaelis Mar 29 '13 at 15:07
  • $\begingroup$ I'm sorry, I only remembered to include my answer after. Thank you for your help! $\endgroup$ – seeker Mar 29 '13 at 15:09
  • $\begingroup$ @Assad posted another one which is shorter $\endgroup$ – Dominic Michaelis Mar 29 '13 at 15:17
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For a slightly different solution,

$2\cos^2 A-2\sin^2 B$

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

$=\cos 2A+\cos 2B$

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

and halve each side.

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Another way of looking at it:

$$\begin{align*} \cos(A+B)\cos(A-B) &= \frac14\left(e^{i(A+B)}+\frac{1}{e^{i(A+B)}}\right)\left(e^{i(A-B)}+\frac{1}{e^{i(A-B)}}\right) \\ &= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\ &= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2 - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\ &= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2\right] + \frac14\left[ - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\ &= \left(\frac{e^{iA} + e^{-iA}}{2}\right)^2 - \left( \frac{e^{iB}-e^{-iB}}{2i}\right)^2 \\ &= \cos^2 A - \sin^2 B. \end{align*} $$

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Maybe you didn't see you can use the formula $(x+y)(x-y)=x^2-y^2$. Other than that, I don't see any other simpler method.

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  • $\begingroup$ Would you have attempted this in a way different than I did? $\endgroup$ – seeker Mar 29 '13 at 14:59
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Here is an interesting different way: Let

$$x = \cos(A+B)\cos(A-B) \\ y = \sin(A+B)\sin(A-B)$$

Then,

$$x+y = \cos(A+B-A+B) = \cos(2B) \\ x-y = \cos(A+B+A-B) = \cos(2A)$$

You can add these to get $2x$ or subtract them to get $2y$. Then expand using the double-angle formulas. This gives you two trig product formulas at the same time.

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