Question

$co{s}^2{73}^{\circ }+co{s}^2{47}^{\circ }+cos{73}^{\circ }cos{47}^{\circ }=\frac{3}{4}$

Answer 1

L.H.S. = $\frac{1}{2}$ [1 + cos ${146}^{\circ }$ + 1 + 1 + cos ${94}^{\circ }$ + cos ${120}^{\circ }$ + cos ${26}^{\circ }$]
$=\frac{1}{2}[2-\frac{1}{2}+cos{146}^{\circ }+cos{94}^{\circ }+cos{26}^{\circ }]$
$4\because cos{120}^{\circ }=-\frac{1}{2}$
$=\frac{1}{2}[\frac{3}{2}+cos{146}^{\circ }+2cos{60}^{\circ }cos{34}^{\circ }]$
$=\frac{1}{2}[\frac{3}{2}-cos{34}^{\circ }+cos{34}^{\circ }]=\frac{3}{4}.$