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

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

Grade:8

3 Answers

Om Alok
24 Points
7 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
askIITians Faculty 907 Points
7 years ago
PFA228-2262_3.PNG
Umang Kumar
15 Points
5 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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