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.

  • 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

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

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

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


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