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Given $ \sin A + \sin B = a$ and $ \cos A + \cos B = b$

where $a$ and $b$ are acute angles

find the value of $\cos(A+B)$

This is my approach

$ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A+B}{2} $

$ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A+B}{2} $

dividing yield $\tan\frac{A+B}{2}=\frac{a}{b}$

$$\tan(A+B) = \frac {2\frac{a}{b}}{1-\left(\frac{a}{b}\right)^2} = \frac{2ab}{b^2-a^2}$$

and

$$\cos(A+B)=\sqrt{\frac{1}{1+\left(\frac{2ab}{b^2-a^2}\right)^2}} = \frac{b^2-a^2}{b^2+a^2}$$

I wonder if

$$\cos(A+B)= \frac{a^2-b^2}{b^2+a^2}$$

was right, if not, why is it?

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    Note that $$\sin A + \sin B = 2\sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$$ and $$\cos A + \cos B = 2\cos \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$$2012-10-04
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    i don't get it, please elaborate, thanks2012-10-04

1 Answers 1