Question

What is the value of $\cos \left(\frac{\mathrm{\pi }}{8}\right)$?

Answer 1

Find the value of $\cos \left(\frac{\mathrm{\pi }}{8}\right)$.

In the formula $\cos 2A=2{\cos }^{2}A-1$,put $A=\frac{\mathrm{\pi }}{8}$:

$\begin{array}{rcl}\cos \left[2\left(\frac{\mathrm{\pi }}{8}\right)\right]& =& 2{\cos }^2\left(\frac{\mathrm{\pi }}{8}\right)-1\\ & \Rightarrow & \cos \left(\frac{\mathrm{\pi }}{4}\right)=2{\cos }^2\left(\frac{\mathrm{\pi }}{8}\right)-1\\ & \Rightarrow & \frac{1}{\sqrt{2}}=2{\cos }^2\left(\frac{\mathrm{\pi }}{8}\right)-1\left[\because cos\frac{\pi }{4}=cos45°=\frac{1}{\sqrt{2}}\right]\\ & \Rightarrow & {\cos }^2\left(\frac{\mathrm{\pi }}{8}\right)=\frac{1+{\displaystyle \frac{1}{\sqrt{2}}}}{2}\\ & \Rightarrow & {\cos }^2\left(\frac{\mathrm{\pi }}{8}\right)=\frac{\sqrt{2}+1}{2\sqrt{2}}\\ & \Rightarrow & \cos \left(\frac{\mathrm{\pi }}{8}\right)=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}\end{array}$

Now rationalize the above expression.

$$\begin{gathered} \cos \frac{\mathrm{\pi }}{8}=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}} \\ \Rightarrow =\sqrt{\frac{2+\sqrt{2}}{4}} \\ \Rightarrow =\frac{\sqrt{2+\sqrt{2}}}{2} \end{gathered}$$

Hence, the value of $\begin{array}{l}\cos \left(\frac{\mathrm{\pi }}{8}\right)\end{array}$ is $\frac{\sqrt{2+\sqrt{2}}}{2}$.