Question

When $\theta$ varies over the real numbers, the maximum value of $\cos \theta -\cos 2\theta$ is

• A. $2$
• B. $\frac{7}{8}$
• C. $\frac{-9}{8}$
• D. $\frac{5}{4}$

Answer 1

Answer: (C)

Let $y=\cos \theta -\cos 2\theta$ $\Rightarrow$ $y^{\prime }=-\sin \theta +2\sin 2\theta$
$\Rightarrow$ $y^{\prime \prime }=-\cos \theta +4\cos 2\theta$
For maxima or minima, put $y^{\prime }=0$
$\Rightarrow$ $2\sin 2\theta =\sin \theta$
$\Rightarrow$ $4sin\theta \cos \theta =\sin \theta$
$\Rightarrow$ $\cos \theta =\frac{1}{4}$ and $\sin \theta =0$
$\Rightarrow$ $\theta ={\cos }^{-1}\left(\frac{1}{4}\right),\theta =0^o$
At $\theta ={\cos }^{-1}\left(\frac{1}{4}\right),y^{\prime \prime }<0,$ maximum
$\therefore$ Maximum value, $y_{\max }={(\cos \theta -2{\cos }^2\theta +1)}_{\theta ={{\cos }^{-1}}_{(1/4)}}$
$=\frac{1}{4}-2\left(\frac{1}{16}\right)+1$
$=\frac{5}{4}-\frac{1}{8}$
$=\frac{9}{8}$