Question

Write the value of $co{s}^2{76}^{\circ }+co{s}^2{16}^{\circ }-cos{76}^{\circ }cos{16}^{\circ }$

Answer 1

$co{s}^2{76}^{\circ }+co{s}^2{16}^{\circ }-cos{76}^{\circ }cos{16}^{\circ }$

$=\frac{1}{2}[1+cos2(76{)}^{\circ }+1+cos2(16{)}^{\circ }-cos(76+16{)}^{\circ }-cos(76-16{)}^{\circ }]=\frac{1}{2}[2-\frac{1}{2}+cos{152}^{\circ }+cos{32}^{\circ }-cos{92}^{\circ }]=\frac{1}{2}[\frac{3}{2}+cos{152}^{\circ }+2sin\frac{92+32}{2}sin\frac{92-32}{2}]=\frac{1}{2}[\frac{3}{2}+cos{152}^{\circ }+2sin\frac{{124}^{\circ }}{2}sin\frac{{60}^{\circ }}{2}]=\frac{1}{2}[\frac{3}{2}+cos{152}^{\circ }+2sin{62}^{\circ }sin{30}^{\circ }]$

$=\frac{1}{2}[\frac{3}{2}+cos{152}^{\circ }+sin{62}^{\circ }]=\frac{1}{2}[\frac{3}{2}+cos{152}^{\circ }+sin(90-28{)}^{\circ }]=\frac{1}{2}[\frac{3}{2}+cos{152}^{\circ }+cos{28}^{\circ }]=\frac{1}{2}[\frac{3}{2}+2cos\frac{152+28}{2}cos\frac{152-28}{2}]=\frac{1}{2}[\frac{3}{2}+2cos{90}^{\circ }cos\frac{{124}^{\circ }}{2}]=\frac{1}{2}[\frac{3}{2}+2(0\left)cos{62}^{\circ }\right]=\frac{3}{4}$

$co{s}^2{76}^{\circ }+co{s}^2{16}^{\circ }-cos{76}^{\circ }cos{16}^{\circ }=\frac{3}{4}$