Question

The value of ${\cos }^2{10}^{\circ }-\cos {10}^{\circ }\cos {50}^{\circ }+{\cos }^2{50}^{\circ }$ is equal to

• A. $\frac{4}{3}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $3$

Answer 1

The correct option is C $\frac{3}{4}$

${\cos }^2{10}^{\circ }-\cos {10}^{\circ }\cos {50}^{\circ }+{\cos }^2{50}^{\circ }=\frac{1}{2}[2{\cos }^2{10}^{\circ }-2\cos {10}^{\circ }\cos {50}^{\circ }+2{\cos }^2{50}^{\circ }]=\frac{1}{2}[1+\cos {20}^{\circ }-(\cos {60}^{\circ }+\cos {40}^{\circ })+1+\cos {100}^{\circ }]$
$=\frac{1}{2}[2+\cos {20}^{\circ }-\frac{1}{2}-\cos {40}^{\circ }+\cos {100}^{\circ }]=\frac{1}{2}[\frac{3}{2}+\cos {20}^{\circ }+\cos {100}^{\circ }-\cos {40}^{\circ }]=\frac{1}{2}[\frac{3}{2}+2\cos {60}^{\circ }\cos {40}^{\circ }-\cos {40}^{\circ }]=\frac{1}{2}[\frac{3}{2}+\cos {40}^{\circ }-\cos {40}^{\circ }]=\frac{3}{4}$