cos(x)+cos(y)+cos(z)=0 and sin(x)+sin(y)+sin(z)=0. prove that cos(x-y)+cos(y-z)+cos(z-x)=-3/2

Hi Aritra,

Let's see how to solve this easily using Complex Numbers:

Let z1=cosx+isinx, z2=cosy+isiny, z3=cosz+isinz.

As Re(z1+z2+z3)=0 and Img(z1+z2+z3)=0,

we have z1+z2+z3=0.

Also as |z|=1 for all complex numbers, 1/z = z' (where ' denotes the conjugate, which is z bar).

so 1/z1 = z1'........ 1/z2 = z2'......... 1/z3 = z3'.

Next cos(x-y)+cos(y-z)+(z-x) = Re(z1/z2 + z2/z3 + z3/z1) = Re (z1z2' + z2z3' + z3z1')------(1).

Re(z) = 1/2(z+z')

So (1) = (1/2)*(z1z2'+z2z3'+z3z1'+z1'z2+z2'z3+z3'z1)

= (1/2)*[z2'(z1+z3) + z3'(z1+z2) + z1'(z2+z3)]

= (1/2)*[z2'(-z2) + z3'(-z3) + z1'(-z1)]

= (1/2)*[-|z2|²-|z3|²-|z1|²]

= -3/2.

And hence proved. Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).