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Question: If \[\sin \alpha +\sin \beta =a\]and \[\cos \alpha +\cos \beta =b\], show that \[\sin (\alpha +\beta...

If sin⁡α+sin⁡β=a\sin \alpha +\sin \beta =aand cos⁡α+cos⁡β=b\cos \alpha +\cos \beta =b, show that sin⁡(α+β)=2aba2+b2\sin (\alpha +\beta )=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}.

Explanation

Solution

Hint: By applying the formula sum and difference of sines and cosines and then dividing the equations and applying the formula sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2)\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right),cos⁡α+cos⁡β=2cos⁡(α+β2)cos⁡(α−β2)\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⁡α+sin⁡β=a\sin \alpha +\sin \beta =a . . . . . . . . . . . . . . . . . . . . . . . . . (1)
cos⁡α+cos⁡β=b\cos \alpha +\cos \beta =b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
By applying the formula to sin⁡α+sin⁡β=a\sin \alpha +\sin \beta =a
The formula is sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2)\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)
cos⁡α+cos⁡β=2cos⁡(α+β2)cos⁡(α−β2)\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⁡(α+β2)=ab\tan \left( \dfrac{\alpha +\beta }{2} \right)=\dfrac{a}{b}
Now applying the formula we get
sin⁡(α+β)=2tan⁡(α+β2)1+tan⁡2(α+β2)\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⁡(α+β2)\tan \left( \dfrac{\alpha +\beta }{2} \right)in the above equation we get,
sin⁡(α+β)=2(ab)1+(ab)2\sin (\alpha +\beta )=\dfrac{2\left( \dfrac{a}{b} \right)}{1+{{\left( \dfrac{a}{b} \right)}^{2}}}
=2aba2+b2=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}

Hence showed that sin⁡(α+β)=2aba2+b2\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.