Question

If $L={\sin }^2\frac{\mathrm{\pi }}{16}-{\sin }^2\frac{\mathrm{\pi }}{8}$ and $M={\cos }^2\frac{\mathrm{\pi }}{16}-{\sin }^2\frac{\mathrm{\pi }}{8}$, then

• A. $M=\frac{1}{2\sqrt{2}}+\frac{1}{2}\cos \frac{\mathrm{\pi }}{8}$
• B. $M=\frac{1}{4\sqrt{2}}+\frac{1}{4}\cos \frac{\mathrm{\pi }}{8}$
• C. $L=-\frac{1}{2\sqrt{2}}+\frac{1}{2}\cos \frac{\mathrm{\pi }}{8}$
• D. $L=\frac{1}{4\sqrt{2}}-\frac{1}{4}\cos \frac{\mathrm{\pi }}{8}$

Answer 1

The correct option is A

$M=\frac{1}{2\sqrt{2}}+\frac{1}{2}\cos \frac{\mathrm{\pi }}{8}$

Explanation for the correct option:

Step 1: Simplify the expression $L$

Given, $L={\sin }^2\frac{\mathrm{\pi }}{16}-{\sin }^2\frac{\mathrm{\pi }}{8}$

$$\begin{gathered} =\sin \left(\frac{\mathrm{\pi }}{16}+\frac{\mathrm{\pi }}{8}\right)\sin \left(\frac{\mathrm{\pi }}{16}-\frac{\mathrm{\pi }}{8}\right)\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{A}\mathbf{-}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{B}\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{\right]} \\ =\sin \frac{3\mathrm{\pi }}{16}\sin \frac{-\mathrm{\pi }}{16} \\ =\frac{1}{2}\left[-2\sin \frac{3\mathrm{\pi }}{16}\sin \frac{\mathrm{\pi }}{16}\right]\mathbf{\left[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{\right]} \\ =\frac{1}{2}\left[\cos \frac{4\mathrm{\pi }}{16}-\cos \frac{2\mathrm{\pi }}{16}\right]\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{A}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{B}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{\right]} \\ =\frac{1}{2}\left[\frac{1}{\sqrt{2}}-\cos \frac{\mathrm{\pi }}{8}\right]\mathbf{[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{]} \\ =\frac{1}{2\sqrt{2}}-\frac{1}{2}\cos \frac{\mathrm{\pi }}{8} \end{gathered}$$

Step 2: Simplify the expression $M$

Given, $M={\cos }^2\frac{\mathrm{\pi }}{16}-{\sin }^2\frac{\mathrm{\pi }}{8}$

$$\begin{gathered} =\cos \left(\frac{\mathrm{\pi }}{16}+\frac{\mathrm{\pi }}{8}\right)\cos \left(\frac{\mathrm{\pi }}{16}-\frac{\mathrm{\pi }}{8}\right)\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{A}\mathbf{-}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{B}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{\right]} \\ =\cos \frac{3\mathrm{\pi }}{16}\cos \frac{-\mathrm{\pi }}{16} \\ =\frac{1}{2}\left(2\cos \frac{3\mathrm{\pi }}{16}\cos \frac{\mathrm{\pi }}{16}\right)\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{\right]} \\ =\frac{1}{2}\left[\cos \left(\frac{3\mathrm{\pi }}{16}+\frac{\mathrm{\pi }}{16}\right)+\cos \left(\frac{3\mathrm{\pi }}{16}-\frac{\mathrm{\pi }}{16}\right)\right]\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{2}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{B}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{\right]} \\ =\frac{1}{2}\left(\cos \frac{4\mathrm{\pi }}{16}+\cos \frac{2\mathrm{\pi }}{16}\right) \\ =\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\cos \frac{\mathrm{\pi }}{8}\right)\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{]} \\ =\frac{1}{2\sqrt{2}}+\frac{1}{2}\cos \frac{\mathrm{\pi }}{8} \end{gathered}$$

Therefore, option (A) $M=\frac{1}{2\sqrt{2}}+\frac{1}{2}\cos \frac{\mathrm{\pi }}{8}$ is the correct answer.