Question

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

Answer 1

Hint: We first find the value $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$. We use $\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1$ for the replacement of the value. We get $2{{\cos }^{2}}\dfrac{\pi }{8}-1=\dfrac{1}{\sqrt{2}}$. We solve the quadratic to find the solution of $\cos \left( \dfrac{\pi }{8} \right)$. We only take the positive value as it is in its primary domain.

Complete step by step solution:
We need to find the value of $\cos \left( \dfrac{\pi }{8} \right)$. We know that $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$.
We can express it as $\cos \left( \dfrac{\pi }{4} \right)=\cos \left( 2\times \dfrac{\pi }{8} \right)=\dfrac{1}{\sqrt{2}}$.
We know the multiple angle formula of $\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1$.
$\cos \left( \dfrac{\pi }{4} \right)=\cos \left( 2\times \dfrac{\pi }{8} \right)=2{{\cos }^{2}}\dfrac{\pi }{8}-1=\dfrac{1}{\sqrt{2}}$. The simplified form is $2{{\cos }^{2}}\dfrac{\pi }{8}-1=\dfrac{1}{\sqrt{2}}$.
\[\begin{align}
  & 2{{\cos }^{2}}\dfrac{\pi }{8}-1=\dfrac{1}{\sqrt{2}} \\
 & \Rightarrow 2{{\cos }^{2}}\dfrac{\pi }{8}=1+\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}+1}{\sqrt{2}} \\
 & \Rightarrow {{\cos }^{2}}\dfrac{\pi }{8}=\dfrac{\sqrt{2}+1}{2\sqrt{2}}=\dfrac{2+\sqrt{2}}{4} \\
 & \Rightarrow \cos \dfrac{\pi }{8}=\pm \sqrt{\dfrac{2+\sqrt{2}}{4}}=\pm \dfrac{\sqrt{2+\sqrt{2}}}{2} \\
\end{align}\]
Now the value of $\cos \left( \dfrac{\pi }{8} \right)$ is positive and that’s why \[\cos \dfrac{\pi }{8}=\dfrac{\sqrt{2+\sqrt{2}}}{2}\]

Note:
Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi \pm a$ for $\cos \left( x \right)=\cos a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.