Evaluate sin^2 31'-cos^2 51'

Dear student

Assuming the correct question as : Evaluate \sin^{2} 39 ° - \cos^{2} 51 ° = \sin^{2} 39 ° - \cos^{2} (90 ° - 39 °) = \sin^{2} 39 ° - \sin^{2} 39 ° = 0 Note : \cos (90 ° - x) = sinx

Regards