# a=sinα+sinβ,b=cosα+cosβ,then  sin(α+β)=? what is answer

Om Alok
24 Points
5 years ago
2ab/a^2+b^2} read alpha =x and beta=yMultiply both a.b=sin x cos x+sin y cos y+sin x cos y + cos x sin y= sin 2x/2 + sin 2y/2 +sin (x+ y)= sin (x+y).cos(x-y) + sin(x+y)By squaring both and adding u will geta^2+b^2= 2+2cos(x-y)By solving above 2 eqn u will get the result
Harsh Patodia IIT Roorkee
5 years ago
PFA
Umang Kumar
15 Points
3 years ago
Given that:

$a=sin\alpha +sin\beta$
$b=cos\alpha +cos\beta$
$a^{2}+b^{2}=4cos^{2}\frac{\alpha -\beta}{2}$
$(a+b)^{2}=4cos^{2}\frac{\alpha-\beta}{2} + 8.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$a^{2}+b^{2}+2ab=4cos^{2}\frac{\alpha-\beta}{2} + 8.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$4cos^{2}\frac{\alpha-\beta}{2}+2ab=4cos^{2}\frac{\alpha-\beta}{2} + 8.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$2ab=2*4.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$2ab=(a^{2}+b^{2})*2sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$\frac{2ab}{a^{2}+b^{2}}=2sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$\frac{2ab}{a^{2}+b^{2}}=sin(\alpha +\beta )$
hence the value of $sin(\alpha +\beta )=\frac{2ab}{a^{2}+b^{2}}$

Thank You
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