Question

The range of $cos(x+\frac{\pi }{6})cos(x-\frac{\pi }{6})$ is

• A. $[-\frac{1}{4},\frac{1}{4}]$
• B. $[-\frac{3}{4},\frac{3}{4}]$
• C. $[-\frac{1}{4},\frac{3}{4}]$
• D. $[-\frac{1}{2},\frac{1}{2}]$

Answer 1

The correct option is C $[-\frac{1}{4},\frac{3}{4}]$

Given,

$\cos (x+\frac{\pi }{6})\left(\cos (x-\frac{\pi }{6})\right)$

$\therefore$ we know,

$2\cos A\cos B=\cos (A+B)+\cos (A-B)$

Here,

$A=x+\frac{\pi }{6}$

$B=x-\frac{\pi }{6}$

Now,

$\Rightarrow \frac{1}{2}\left(2\cos (x+\frac{\pi }{6})\cos (x-\frac{\pi }{6})\right)$

$\Rightarrow \frac{1}{2}(\cos (x+\mathrm{x/}+x-\frac{x}{6})+\cos (x+\frac{\pi }{6}-x+\frac{\pi }{6}))$

$\Rightarrow \frac{1}{2}\left(\cos \right(2x)+\cos \frac{\pi }{3})$

$\Rightarrow \frac{\cos 2x}{2}+\frac{1}{2\times 2}$

$\because -1⩽\cos x⩽1\text{or}$

$-1⩽\cos 2x⩽1$

$\frac{-1}{2}⩽\frac{\mathrm{los}2x}{2}⩽\frac{1}{2}$

$-\frac{1}{2}⩽\frac{\cos 2x}{2}⩽\frac{1}{2}$

$\frac{-1}{2}+\frac{1}{4}⩽\frac{\cos 2x}{2}+\frac{1}{4}\le \frac{1}{2}+\frac{1}{4}$

$-\frac{1}{4}⩽\frac{\cos 2x}{2}+\frac{1}{4}\le \frac{3}{4}$

Option C