Why $\cos a\cdot \cos b+\sin a\cdot \sin b = \cos (b-a)$

Question

I do not get why

$$\cos a\cdot \cos b+\sin a\cdot \sin b = \cos (b-a)$$

Thank you for your help

Various proofs here: http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html. — Martin R, Nov 11 '18 at 13:45
Idea: $\cos{x}=\frac{e^{ix}+e^{-ix}}{2}$ and $\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$. — Samvel Safaryan, Nov 11 '18 at 13:51

score 2 · Accepted Answer · answered Nov 11 '18 at 13:46

2

The following diagram by Blue demonstrates this perfectly:

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Hence, we conclude

$$\cos(a\pm b) = \cos a\cos b\mp \sin a\sin b$$

You may want to check the following link.

answered Nov 11 '18 at 13:46

KM101

1

Thanks for giving me credit (you're learning! ;) ... but the proper thing to do in this case (for the sake of the Math.SE community) is to vote to close the question as a duplicate of the other, since the other one has lots of different answers. – Blue Nov 11 '18 at 14:03
Haha, I guess that was the better option of the two. :-) Will do! – KM101 Nov 11 '18 at 14:05

1 Answers1