5
$\begingroup$

I have seen geometric proof of identities $$\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$$ and $$\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}$$

By adding two equation, $$2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$$.

But how to prove this by geometry?

Thank you.

$\endgroup$
1
  • 1
    $\begingroup$ If $\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$ and $\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}$ were proven geometrically, doesn't that mean you have already shown $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$ geometrically? $\endgroup$
    – graydad
    Feb 25 '15 at 16:52
6
$\begingroup$

enter image description here

$$\begin{align} 2 \cos A \cos B &= \cos(A-B)+\cos(A+B) \\[6pt] 2 \sin A \,\sin B &= \cos(A-B)-\cos(A+B) \end{align}$$

Note. Although not labeled (yet), these identities are also evident:

$$\begin{align} 2 \,\sin A \cos B &= \sin(A+B)+\sin(A-B) \\[6pt] 2 \cos A \,\sin B &= \sin(A+B)-\sin(A-B) \end{align}$$

$\endgroup$

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.