Question

prove geometrically that cos (x+y)=cosx cosy-sinxsiny

Answer 1

Answer:

We have to prove that

cos(x+y)= cosx*cosy-sinx*siny

From the figure,

The angle of the upper triangle i.e. opposite side of the length C is x-y.

Now by cosine law,

C2 = 12 + 12 - 2*1*1*cos(x-y)

=> C2 = 1 +1 - 2*cos(x-y)

=> C2 = 2 - 2*cos(x-y)

Agian the side of length C joins the points (cosy, siny) and (cosx, sinx), So from Pythagorus theorem

C2 = (cosy - cosx)2 + (siny - sinx)2

=> C2 = cos2 y + cos2 x - 2*cosx*cosy + sin2 x + sin2 y - 2*sinxsiny

=> C2 = (cos2 y + sin2 y) + (cos2 x + sin2 x) - 2(cosx*cosy - sinxsiny)

=> 2 - 2*cos(x-y) = 1 + 1 - 2(cosx*cosy - sinxsiny) (since cos2 θ + sin2 θ = 1, C2 = 2 - 2*cos(x-y) )

=> 2 - 2*cos(x-y) = 2 - 2(cosx*cosy - sinxsiny)

=> - 2*cos(x-y) = - 2(cosx*cosy - sinxsiny)

=> cos(x-y) = cosx*cosy - sinxsiny

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