A geometry problem by prasanth pp

Given $cos^2(76) + cos^2(16) - cos(76) \times cos(16)$ = $\frac{1}{2} \times (2cos^2(76) + 2cos^2(16) - 2cos(76) \times cos(16))$ = $\frac{1}{2} \times (1+cos(152) + 1+cos(32) - cos(92) - cos(60))$ = $\frac{1}{2} \times (2-cos(60)+cos(152) +cos(32) - cos(92) )$ = $\frac{1}{2} \times (2-cos(60)+2cos(\frac{152+32}{2}) \times cos(\frac{152-32}{2}) - cos(92) )$ = $\frac{1}{2} \times (2-cos(60)+2cos(\frac{184}{2}) \times cos(\frac{120}{2}) - cos(92) )$ = $\frac{1}{2} \times (2-cos(60)+2cos(92) \times cos(60) - cos(92) )$ = $\frac{1}{2} \times (2-\frac{1}{2}+2cos(92) \times \frac{1}{2} - cos(92) )$ = $\frac{1}{2} \times (2-\frac{1}{2}+cos(92) - cos(92) )$ = $\frac{1}{2} \times (2-\frac{1}{2} )$ = $\frac{1}{2} \times (\frac{3}{2} )$ = $\frac{3}{4}$ =0.75