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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.

  • 3
    Use the fact that sine is an odd function and cosine is an even function. – 2017-02-20
  • 2
    Rewrite $\cos(A+B)$ as $\cos(A-(-B))$ and then use the suggestion by @Oiler – 2017-02-20
  • 0
    Let $B=-B$ and you get $\cos A \cos B-\sin A \sin B=\cos (A+B)$ – 2017-02-20
  • 1
    Its a property that $$\cos(-\theta)=\cos\theta$$ and $$\sin(-\theta)=-\sin\theta.$$ – 2017-02-20
  • 0
    If this helped, please upvote and mark as the answer – 2017-02-20

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