6
$\begingroup$

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

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:

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

$\endgroup$
  • 7
    $\begingroup$ Don't you have an $=$ on your keyboard? $\endgroup$ – Git Gud Mar 29 '13 at 14:52
  • $\begingroup$ 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
  • 4
    $\begingroup$ @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
3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
7
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Would you have attempted this in a way different than I did? $\endgroup$ – seeker Mar 29 '13 at 14:59
3
$\begingroup$

Another way of looking at it:

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.