Question

The maximum value of ${\cos }^2\left(\frac{\mathrm{\pi }}{3}-x\right)-{\cos }^2\left(\frac{\mathrm{\pi }}{3}+x\right)$ is

• A. $-\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C

$\frac{\sqrt{3}}{2}$

Explanation for the correct option:

${\cos }^2\left(\frac{\mathrm{\pi }}{3}-x\right)-{\cos }^2\left(\frac{\mathrm{\pi }}{3}+x\right)$

Let, A $=\frac{\pi }{3}+x$, B $=\frac{\pi }{3}-x$

Using the formula,

${\cos }^2\left(\mathrm{B}\right)–{\cos }^2\left(\mathrm{A}\right)=\sin \left(\frac{A–B}{2}\right)\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)$

$\Rightarrow {\cos }^2\left(\frac{\pi }{3}–x\right)–{\cos }^2\left(\frac{\pi }{3}+x\right)=\sin \left(\frac{\left(\frac{\mathrm{\pi }}{3}\right)+x–\left(\frac{\mathrm{\pi }}{3}\right)+x}{2}\right).\sin \left(\frac{\left(\frac{\mathrm{\pi }}{3}\right)+x+\left(\frac{\mathrm{\pi }}{3}\right)–x}{2}\right)$

$$\begin{gathered} =\sin \left(2x\right).\sin \left(\frac{\mathrm{\pi }}{3}\right) \\ =\frac{\sqrt{3}}{2}\sin \left(2x\right) \end{gathered}$$

We know that the maximum value of $\sin \left(2x\right)=1$

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is the correct answer.