Question

sin x +sin y = a, cos x +cos y= b then sin(x+y) =​

Answer 1

Answer:-

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Answer is 2ab/a^2+b^2✌✌⏩

let us solve:-

Identities used:-

⏩sin(2x)=>2sinx.cosx

⏩sin(x+y)=>sinx.cosy + siny.cosx

⏩sinx+siny =>2sin[(x+y)/2]cos[(x+y)/2]

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Now,According to question

ab=>(sinx+siny)(cosx+cosy)

=>sinx.cosx +(sinx.cosy +siny.cosx)+siny.cosy

=>1/2sin(2x)+1/2sin(2y)+sin(x+y)

=>sin(x+y)+1/2[sin(2x)+sin(2y)]

=>sin(x+y)[1+cos(x-y)]. -(eq1)

a^2=>(sinx+siny)^2=>sinx^2+siny^2+2sinx.siny

b^2=>(cosx+cosy)^2=>cosx^2+cosy^2+2cosx.cosy

a^2+b^2=>2 + 2cos(x-y)=>2[1+cos(x-y)]. (eq2)

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sin(x+y)=>2ab/a^2+b^2(from eq1 and eq2)

______________hope it helps uh✌✌✌❤⏩

Answer 2

Answer:

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