Question

For all real numbers x and y, prove that (i)sin(x+y)=sinxcosy + cosxsiny​

Answer 1

We know that,

cos (x-y) = cosx .cos y + sinx.siny ----------- (1)

Then,

from formula. --(1) we get ,

If we give x=(90-x)

So ,cos (90°-x-y)=cos (90°-x).cosy +sin (90°-x).siny

=> cos[90°-(x+y)]

=sinx.cosy+cosx.siny

{cos(90°-x)=sinx and sin(90°-x)=cosx}

=>sin(x+y)= sinx .cosy +cosx.siny

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Answer 2

We Know that,

sinx=cos(π/2-x)

sin(x+y)=cos[π/2-(x-y)]

sin(x+y)=cos[(π/2-x)-y] since,cos(x-y)=cosx cosy+sinx siny

sin(x+y)=cos(π/2-x)cosy+sin(π/2-x)siny

Sin(x+y)=sinx cosy + cosx siny.