How to derive the formula cos(A+B) from the formula cos(A-B)?

How do I derive the formula:

cos(A+B)=cosAcosB-sinAsinB

from the formula:

cos(A-B)=cosAcosB+sinAsinB?

The only difference that I noticed is the negative and positive sign. I was thinking that first, I replace B with (-B), but then after that how does cos(-B) turn to cos(B), and sin(-B) turn to -sin(B)?

Thank you, can someone please explain to me. I hope my question was not too confusing.

• Use the fact that sine is an odd function and cosine is an even function. – Oiler Feb 20 '17 at 5:10
• Rewrite $\cos(A+B)$ as $\cos(A-(-B))$ and then use the suggestion by @Oiler – John Wayland Bales Feb 20 '17 at 5:12
• Let $B=-B$ and you get $\cos A \cos B-\sin A \sin B=\cos (A+B)$ – Nick Pavini Feb 20 '17 at 5:39
• Its a property that $$\cos(-\theta)=\cos\theta$$ and $$\sin(-\theta)=-\sin\theta.$$ – Juniven Feb 20 '17 at 5:40
• If this helped, please upvote and mark as the answer – Nick Pavini Feb 20 '17 at 17:43

$$\cos (A-(-B)) = \cos A \cos -B + \sin A \sin -B = \cos A \cos B - \sin A \sin B= \cos (A+B)$$ Based on the even odd properties of $\sin$ and $\cos$