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

2 Answers
Cesareo R.
Sep 15, 2016

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

Shwetank Mauria
Sep 16, 2016

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)

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