# How to derive formula for $\sin(A-B)$ from formula for $\sin(A+B)$?

It has been a while since I have done any maths and am struggling with this question -

Using the addition law $$\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$$ and the fact that $\sin(-\theta) = -\sin(\theta)$ and $\cos(-\theta) = \cos(\theta)$, show that $$\sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B)$$

I am really confused as to where to even start, so any help is appreciated!!

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Hint: Substitute $-B$ into your expression for sin(A+B) and use the identities. – Amzoti Mar 6 '13 at 4:27
Let $C=-B$. Then $\sin(A-B)=\sin(A+C)$. See if you can take it from there. – crf Mar 6 '13 at 4:28

## 2 Answers

Hint: $$\quad A-B=A+(-B)\quad$$

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There's hardly anything to do: \begin{align} \sin(A-B) & = \sin(A+(-B)) = \sin A\cos(-B) + \cos A \sin(-B) \\[6pt] & = \sin A\cos B + (\cos A)(-\sin B) \end{align} followed by one trivial step.

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