If 0<a,b<π and cos(a) + cos(b) - cos(a+b) = 3/2 then show that a=b=π/3

[cooldudeachyut]

If 0<a,b<π and cos(a) + cos(b) - cos(a+b) = 3/2 then show that a=b=π/3

Question: If 0<a,b<π and cos(a) + cos(b) - cos(a+b) = 3/2 then show that a=b=π/3

My attempt:
2cos((a+b)/2)cos((a-b)/2) - 2cos²((a+b)/2) + 1=3/2

4cos((a+b)/2)cos((a-b)/2) - 4cos²((a+b)/2) =1

4cos((a+b)/2)(cos((a-b)/2) - cos((a+b)/2)) =1

8cos((a+b)/2)(sin(a/2)sin(b/2)) =1

That probably doesn't lead anywhere. So, how to proceed in this question?

 

[pka]

I get, if a=b=π/3, 1/2+1/2-sqrt(3)/2=(1-sqrt(3))/2 - not 3/2.

You have made a sign error.
\(\displaystyle \cos\left(\frac{\pi}{3}\right)+\cos\left(\frac{\pi}{3}\right)-\cos\left(\frac{2\pi}{3}\right)=\frac{1}{2}+\frac{1}{2}-(-\frac{1}{2})=\frac{3}{2}\)