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If $\cos A+\cos B=m$ and $\sin A+\sin B=n$, where $\mathrm{m}, n \neq 0$,
then what is $\sin (A+B)$ equal to ?
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then what is $\sin (A+B)$ equal to ?
Solution:
1685 Upvotes
Verified Answer
The correct answer is:
$\frac{2 m n}{m^{2}+n^{2}}$
Let $\cos A+\cos B=m$
and $\sin A+\sin B=n$
Consider $\sin (A+B)=\frac{\left(m^{2}+n^{2}\right) \sin (A+B)}{m^{2}+n^{2}}$
$=\frac{[2+2 \cos (A-B)] \sin (A+B)}{2+2 \cos (A-B)}$
(from i and ii)
$=\frac{2 \sin (\mathrm{A}+\mathrm{B})+2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})}{1+1+2 \cos (\mathrm{A}-\mathrm{B})}$
$=\frac{2 \sin (A+B)+\sin (A+B+A-B)+\sin (A+B-A+B)}{1+1+2 \cos (A-B)}$
$=\frac{2 \mathrm{mn}}{\mathrm{m}^{2}+\mathrm{n}^{2}}($ from (i) and (ii))
Hence, $\sin (\mathrm{A}+\mathrm{B})=\frac{2 \mathrm{mn}}{\mathrm{m}^{2}+\mathrm{n}^{2}}$
and $\sin A+\sin B=n$
Consider $\sin (A+B)=\frac{\left(m^{2}+n^{2}\right) \sin (A+B)}{m^{2}+n^{2}}$
$=\frac{[2+2 \cos (A-B)] \sin (A+B)}{2+2 \cos (A-B)}$
(from i and ii)
$=\frac{2 \sin (\mathrm{A}+\mathrm{B})+2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})}{1+1+2 \cos (\mathrm{A}-\mathrm{B})}$
$=\frac{2 \sin (A+B)+\sin (A+B+A-B)+\sin (A+B-A+B)}{1+1+2 \cos (A-B)}$
$=\frac{2 \mathrm{mn}}{\mathrm{m}^{2}+\mathrm{n}^{2}}($ from (i) and (ii))
Hence, $\sin (\mathrm{A}+\mathrm{B})=\frac{2 \mathrm{mn}}{\mathrm{m}^{2}+\mathrm{n}^{2}}$
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