Question

If \[\sin \alpha +\sin \beta =a\]and \[\cos \alpha +\cos \beta =b\], show that \[\sin (\alpha +\beta )=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}\].

Answer 1

Hint: By applying the formula sum and difference of sines and cosines and then dividing the equations and applying the formula \[\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\],\[\cos \alpha +\cos \beta =2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\] will arrive to the final solution.

Complete step-by-step answer:
Given
\[\sin \alpha +\sin \beta =a\] . . . . . . . . . . . . . . . . . . . . . . . . . (1)
\[\cos \alpha +\cos \beta =b\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
By applying the formula to \[\sin \alpha +\sin \beta =a\]
The formula is \[\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\]
\[\cos \alpha +\cos \beta =2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\]
 From the formula we can write (1) and (2) as follows,
\[\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\]\[=a\]
\[\cos \alpha +\cos \beta =2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\]\[=b\]
\[2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\]\[=a\]. . . . . . . . . . . . . . . . . . (3)
\[2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)\]\[=b\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)

Dividing 3 with 4, (3\[\div \]4) we get
\[\tan \left( \dfrac{\alpha +\beta }{2} \right)=\dfrac{a}{b}\]
Now applying the formula we get
\[\sin (\alpha +\beta )=\dfrac{2\tan \left( \dfrac{\alpha +\beta }{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{\alpha +\beta }{2} \right)}\]
Now substituting the value of \[\tan \left( \dfrac{\alpha +\beta }{2} \right)\]in the above equation we get,
\[\sin (\alpha +\beta )=\dfrac{2\left( \dfrac{a}{b} \right)}{1+{{\left( \dfrac{a}{b} \right)}^{2}}}\]
\[=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}\]

Hence showed that \[\sin (\alpha +\beta )=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}\].

Note: To solve such types of problems all the related formulas should be handy. Also be careful about the signs in all the formulas. In the above equation division was made to simplify the terms and lead to the final answer. So smart moves lead to easy completion of answers.