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

Question

I have a question regarding trigonometric identities. The question I am currently struggling to understand is:

Prove that $$\cos(A + B) \cos(A − B) = \cos^2A − \sin^2B$$

When approaching this problem I know that there is $$\cos2A = \cos^2A - \sin^2A$$ but how would I apply this here, or is it completely wrong way of approaching it, or should I try
$$\cos(A+B)=\cos A \cos B - \sin A \sin B$$

Many Thanks.

Well what do you get when you try those method? How far does it take you? — Divide1918, Jun 29 '20 at 09:00
https://math.stackexchange.com/questions/345703/prove-that-cos-a-b-cos-a-b-cos-2a-sin-2b — lab bhattacharjee, Jun 30 '20 at 02:21

score 1 · Answer 1 · answered Jun 29 '20 at 09:04

\begin{align*} \cos(x+y)\cdot\cos(x-y)&= (\cos x\cos y-\sin x\sin y)(\cos x\cos y+\sin x\sin y)\\ &=\cos^2 x\cos^2 y-\sin^2 x\sin^2 y\\ &=\cos^2 x(1-\sin^2 y)-(1-\cos^2 x)\sin^2 y\\ &=\cos^2 x-\cos^2 x\sin^2 y-\sin^2 y+\cos^2 x\sin^2y\\ &=\cos^2 x-\sin^2y \end{align*}

Quanto · Answer 2 · 2022-07-25T14:35:49.130

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Note

\begin{align} \cos^2A − \sin^2B =& \frac12(1+ \cos 2A )-\frac12(1-\cos2B)\\ =& \frac12( \cos 2A+\cos2B) =\cos(A + B) \cos(A − B) \end{align}

edited Jul 25 '22 at 14:35

answered Jun 29 '20 at 20:15

Quanto

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score 0 · Answer 3 · answered Jun 29 '20 at 09:13

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Hint: The only identities you should need to prove your claim are:

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

answered Jun 29 '20 at 09:13

5xum

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score 0 · Answer 4 · answered Jun 29 '20 at 09:21

If

$$\cos(a+b)\cos(a-b)=\cos^2(a)-\sin^2(b)$$ holds, then it also holds with $\dfrac\pi2-a$ instead of $a$ and

$$\sin(a-b)\sin(a+b)=\sin^2(a)-\sin^2(b).$$

Then by subtraction,

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

score 0 · Answer 5 · answered Jun 29 '20 at 09:26

You need to use the addition/subtraction formulas $$ \cos(A+B)=\cos A\cos B-\sin A\sin B\\ \cos(A-B)=\cos A\cos B+\sin A\sin B $$ Therefore the product is $$ \cos(A+B)\cos(A-B)=\cos^2A\cos^2B-\sin^2A\sin^2B $$ The identity you're due to prove only has $\cos^2A$ and $\sin^2B$: then change $\cos^2B=1-\sin^2B$ and $\sin^2B=1-\cos^2B$. Plug these in and see what happens.

score 0 · Answer 6 · answered Jun 29 '20 at 09:49

Using the formulas you mentioned, you can derive the following convenient forms (they are useful when doing integration):

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

Applying the above formulas, one easily sees that $$\cos(A+B)\cos(A-B)=\frac 12(\cos(2A)+\cos(2B))$$ $$=\frac 12(2\cos^2 A-1+1-2\sin^2 B)=\cos^2 A-\sin^2 B.$$

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

6 Answers6