Question

If `sinx+siny=a` and `cosx+cosy=b` how do you find `cos(x-y)` ?

Answer 1

`cos(x-y) = (a^2+b^2-2)/2`

Explanation:

Using de Moivre's identity

`e^(i phi) = cos phi + i sin phi`

Adding term to term

#((i sin x + i sin y = i a), (cos x + cos y = b))#

giving

`e^(ix) + e^(iy) = rho e^(ialpha)`

with `rho = sqrt(a^2+b^2)`

Adding now

#((-i sin x - i sin y =-i a), (cos x + cos y = b))#

`e^(-ix) + e^(-iy) = rho e^(-ialpha)`

Multiplying term to term

`1+e^(i(x-y))+e^(-i(x-y)) + 1= rho^2` or

`2+2cos(x-y)=rho^2` and finally

`cos(x-y) = (a^2+b^2-2)/2`

of course `a,b` must obey

`abs( (a^2+b^2-2)/2) le 1`

Answer 2

`cos(x-y)=1/2(a^2+b^2-2)`

Explanation:

As `cosx+cosy=b`, `b^2=cos^2x+cos^2y+2cosxcosy`, and similarly as `sinx+siny=a`, `a^2=sin^2x+sin^2y+2sinxsiny`

Adding the two, we get

`a^2+b^2=cos^2x+cos^2y+2cosxcosy+sin^2x+sin^2y+2sinxsiny` or

`a^2+b^2=2+2×(cosxcosy+sinxsiny)` or

`a^2+b^2=2+2×cos(x-y)` or

`cos(x-y)=1/2(a^2+b^2-2)`