If cos⁡α+cos⁡β=a,sin⁡α+sin⁡β=b and α−β=2θ, then cos⁡3θcos⁡θ=

From the given relations we havea2+b2=cos2⁡α+cos2⁡β+2cos⁡αcos⁡β+sin2⁡α+sin2⁡β+2sin⁡αsin⁡β=2+2cos⁡(α−β)=2+2cos⁡2θ=4cos2⁡θNow cos⁡3θcos⁡θ=4cos3⁡θ−3cos⁡θcos⁡θ=4cos2⁡θ−3=a2+b2−3.