Question

If $\cos \theta, \cos \left(60^{\circ}-\theta\right), \cos \left(60^{\circ}+\theta\right)$ in order are three consecutive terms of A.P, then the value of $\cos 2 \theta$ is equal to

• A. $\frac{11}{12}$
• B. $\frac{12}{13}$
• C. $\frac{13}{14}$
• D. $\frac{14}{15}$

Answer 1

Answer: (C)

$2 \cos (60-\theta)=\cos \theta+\cos (60+\theta)$
$2\left[\frac{\cos \theta}{2}+\frac{\sqrt{3}}{2} \sin \theta\right]=\cos \theta+\frac{\cos \theta}{2}-\frac{\sqrt{3}}{2} \sin \theta$
$\Rightarrow \tan \theta=\frac{1}{3 \sqrt{3}}$
$\Rightarrow \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\frac{27-1}{27+1}=\frac{13}{14} $