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'=-\sin \theta +2\sin 2\theta $
$ \Rightarrow $ $ y''=-\cos \theta +4\cos \,2\theta $
For maxima or minima, put $ y'=0 $
$ \Rightarrow $ $ 2\sin \,2\theta =\sin \theta $
$ \Rightarrow $ $ 4\,sin\,\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''< 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} $