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

Answer: (C)

$\frac{3}{4}$

${\cos }^2-\cos 10\cos 50+{\cos }^{2}50$

$={\cos }^{2}10-\cos 10\cos (60-10)+{\cos }^2(60-10)$

$={\cos }^{2}10-\cos 10(\cos 60\cos 10+\sin 60\sin 10)+(\cos 60\cos 10+\sin 60\sin 10{)}^2$

$={\cos }^{2}10-\frac{{\cos }^{2}10}{2}-\frac{\sqrt{3}\cos 10\sin 10}{2}+\frac{{\cos }^{2}10}{4}+\frac{3{\sin }^{2}10}{4}+\frac{\sqrt{3}\sin 10\cos 10}{2}$

$={\cos }^{2}10-\frac{{\cos }^{2}10}{2}+\frac{{\cos }^{2}10}{4}+\frac{3(1-{\cos }^{2}10)}{4}+\frac{3}{4}$