Question

Find the value of \[\cos \dfrac{\pi }{8}\]
(a) \[\sqrt{\dfrac{\sqrt{2}-1}{2\sqrt{2}}}\]
(b) \[\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\]
(c) \[\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{3}}}\]
(d) \[\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}}}\]

Answer 1

Hint: We solve this problem by using the half-angle formula of the cosine ratio.
The half angle formula of cosine ratio is given as
\[\cos \theta =2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)-1\]
We assume some value for \[\theta \] so that we can get \[\dfrac{\pi }{8}\] on either side of the equation.
We use the standard result of cosine ratio that is
\[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]

Complete step by step answer:
We are asked to find the value of \[\cos \dfrac{\pi }{8}\]
Let us use the half-angle formula for solving this problem.
We know that the half angle formula of cosine ratio is given as
\[\cos \theta =2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)-1\]
Let us assume that the value of \[\theta =\dfrac{\pi }{4}\]
By substituting the required value in the half angle formula we get
\[\Rightarrow \cos \dfrac{\pi }{4}=2{{\cos }^{2}}\left( \dfrac{\pi }{8} \right)-1\]
We know that the standard result of cosine ratio that is
\[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
By using this result in above equation then we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\sqrt{2}}=2{{\cos }^{2}}\left( \dfrac{\pi }{8} \right)-1 \\
 & \Rightarrow 2{{\cos }^{2}}\left( \dfrac{\pi }{8} \right)=\dfrac{\sqrt{2}+1}{\sqrt{2}} \\
 & \Rightarrow {{\cos }^{2}}\left( \dfrac{\pi }{8} \right)=\dfrac{\sqrt{2}+1}{2\sqrt{2}} \\
\end{align}\]
Now, by applying the square root on both sides then we get
\[\Rightarrow \cos \left( \dfrac{\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\]
Therefore, we can conclude that value of \[\cos \dfrac{\pi }{8}\] is given as
\[\therefore \cos \left( \dfrac{\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\]
So, option (b) is the correct answer..

Note:
We can solve this problem in another method also.
We have the half angle formula for cosine ratio as
\[\cos \theta =1-2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)\]
Let us assume that the value of \[\theta =\dfrac{3\pi }{4}\]
By substituting the required value in the half angle formula we get
\[\Rightarrow \cos \dfrac{3\pi }{4}=1-2{{\sin }^{2}}\left( \dfrac{3\pi }{8} \right)\]
We know that the standard result of cosine ratio that is
\[\cos \dfrac{3\pi }{4}=-\dfrac{1}{\sqrt{2}}\]
By using this result in above equation then we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\sqrt{2}}=1-2{{\sin }^{2}}\left( \dfrac{3\pi }{8} \right) \\
 & \Rightarrow 2{{\sin }^{2}}\left( \dfrac{3\pi }{8} \right)=\dfrac{\sqrt{2}+1}{\sqrt{2}} \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{3\pi }{8} \right)=\dfrac{\sqrt{2}+1}{2\sqrt{2}} \\
\end{align}\]
Now, by applying the square root on both sides then we get
\[\Rightarrow \sin \left( \dfrac{3\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\]
We know that the conversion of sine to cosine as
\[\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)\]
By using this conversion to above equation then we get
\[\begin{align}
  & \Rightarrow \sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}} \\
 & \Rightarrow \cos \left( \dfrac{\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}} \\
\end{align}\]
Therefore, we can conclude that value of \[\cos \dfrac{\pi }{8}\] is given as
\[\therefore \cos \left( \dfrac{\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\]
So, option (b) is correct answer.