Question

If,$\alpha =22°30\text{'}$ then$(1+\cos \alpha )(1+\cos 3\alpha )(1+\cos 5\alpha )(1+\cos 7\alpha )$ equals

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1+\sqrt{2}}{2\sqrt{2}}$
• D. $\frac{\sqrt{2}-1}{\sqrt{2}+1}$

Answer 1

The correct option is A

$\frac{1}{8}$

Explanation for correct option:

Step 1: Convert the degree measurement into radian measurement.

We have given$\alpha =22°30\text{'}$

$\alpha =22°30\text{'}=22°{\left(\frac{30}{60}\right)}^{°}=22°\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)°=\left(\frac{45}{2}\right)°=\frac{\mathrm{\pi }}{8}$

Step 2: We have to find the value of$\mathbf{(}\mathbf{1}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{\alpha }\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{5}\mathit{\alpha }\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{7}\mathit{\alpha }\mathbf{)}$.

$\Rightarrow (1+\cos \alpha )(1+\cos 3\alpha )(1+\cos 5\alpha )(1+\cos 7\alpha )$

$\Rightarrow (1+\cos \frac{\mathrm{\pi }}{8})(1+\cos \frac{3\mathrm{\pi }}{8})(1+\cos \frac{5\mathrm{\pi }}{8})(1+\cos \frac{7\mathrm{\pi }}{8})$

$$\begin{gathered} \cos \frac{5\mathrm{\pi }}{8}=\cos (\mathrm{\pi }-\frac{3\mathrm{\pi }}{8})=-\cos \left(\frac{3\mathrm{\pi }}{8}\right)\because \mathrm{cos\theta }\mathrm{lies}\mathrm{in}2\mathrm{nd}\mathrm{quadrant} \\ \mathrm{and}\cos \frac{7\mathrm{\pi }}{8}=\cos (\mathrm{\pi }-\frac{\mathrm{\pi }}{8})=-\cos \left(\frac{\mathrm{\pi }}{8}\right)\because \mathrm{cos\theta }\mathrm{lies}\mathrm{in}2\mathrm{nd}\mathrm{quadrant} \end{gathered}$$

$$\begin{gathered} \Rightarrow (1+\cos \frac{\mathrm{\pi }}{8})(1+\cos \frac{3\mathrm{\pi }}{8})(1-\cos \frac{3\mathrm{\pi }}{8})(1-\cos \frac{\mathrm{\pi }}{8}) \\ \Rightarrow (1-{\cos }^2\frac{\mathrm{\pi }}{8})(1-{\cos }^2\frac{3\mathrm{\pi }}{8}) \\ \Rightarrow {\sin }^2\left(\frac{\mathrm{\pi }}{8}\right){\sin }^2\left(\frac{3\mathrm{\pi }}{8}\right) \end{gathered}$$

Step 3: Applying the formula$\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\left(x\right)\mathit{s}\mathit{i}\mathit{n}\left(y\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\left(x-y\right)\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\left(x+y\right)$

${\left[\sin \left(\frac{\mathrm{\pi }}{8}\right)\sin \left(\frac{3\mathrm{\pi }}{8}\right)\right]}^2$

$$\begin{gathered} \Rightarrow {\left[\sin \left(\frac{\mathrm{\pi }}{8}\right)\sin \left(\frac{3\mathrm{\pi }}{8}\right)\right]}^2 \\ \Rightarrow {\left[\frac{1}{2}\right\{\cos \left(\frac{\mathrm{\pi }}{8}-\frac{3\mathrm{\pi }}{8}\right)-\cos \left(\frac{\mathrm{\pi }}{8}+\frac{3\mathrm{\pi }}{8}\right)\left\}\right]}^2\{2\sin \left(x\right)\sin \left(y\right)=\cos \left(x-y\right)-\cos \left(x+y\right)\} \\ \Rightarrow \frac{1}{4}{\{\cos \left(\frac{-\mathrm{\pi }}{4}\right)-\cos \left(\frac{\mathrm{\pi }}{2}\right)\}}^2 \\ \Rightarrow \frac{1}{4}{\{-\frac{1}{\sqrt{2}}-0\}}^2 \\ \Rightarrow \frac{1}{4}{\{-\frac{1}{\sqrt{2}}\}}^2 \\ \Rightarrow \frac{1}{4}\left(\frac{1}{2}\right) \\ \Rightarrow \frac{1}{8} \end{gathered}$$

Hence, the correct option is$\left(A\right)$