Question

Evaluate $\cos \left( {\dfrac{{13\pi }}{8}} \right)$.

Answer 1

Consider one of the basic trigonometric identities $\;\cos 2\theta = 2{\cos ^2}\theta - 1$. In order to solve this question we can use the above mentioned identity. For that we have to convert our question in such a way that it can be expressed in the form of the above given identity, and then we can solve it to get the value.

Complete step by step answer:
Given, $\cos \left( {\dfrac{{13\pi }}{8}} \right)..................................................\left( i \right)$
Now let’s assume $\cos \left( {\dfrac{{13\pi }}{8}} \right) = \cos a......................\left( {ii} \right)$
$\Rightarrow \cos 2a = \cos \left( {\dfrac{{26\pi }}{8}} \right)$
We have to find the value of $\cos \left( {\dfrac{{26\pi }}{8}} \right)$ such that by using the identity we can then solve the question using the given identity $\;\cos 2\theta = 2{\cos ^2}\theta - 1$.
So finding the value of $\cos \left( {\dfrac{{26\pi }}{8}} \right)$:
We know that $\cos \left( {\dfrac{{26\pi }}{8}} \right)$ can be written as
$\cos \left( {\dfrac{{12\left( {2\pi } \right)}}{8} + \dfrac{{2\pi }}{8}} \right) = \cos \left( {3\pi + \dfrac{{2\pi }}{8}} \right) \\\$
$\Rightarrow \cos \left( {3\pi + \dfrac{{2\pi }}{8}} \right) = \cos \left( {3\pi + \dfrac{\pi }{4}} \right).................(iii)$
So from (iii) we know that $\cos \left( {3\pi + \dfrac{\pi }{4}} \right)$ would be in the III Quadrant where cosine is negative.Such that:
$\operatorname{c} \cos \left( {3\pi + \dfrac{\pi }{4}} \right) = - \cos \left( {\dfrac{\pi }{4}} \right)..................(iv)$
Also we know $- \cos \left( {\dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}....................(v)$
Now by using the identity $\;\cos 2\theta = 2{\cos ^2}\theta - 1$ we get
\Rightarrow 2{\cos ^2}a = 1 + \cos 2a \\\
Also we know from (v) $\cos 2a = - \cos \left( {\dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow 2{\cos ^2}a = 1 + - \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow 2{\cos ^2}a = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }} \\\ \Rightarrow 2{\cos ^2}a = \left( {\dfrac{{\left( {\sqrt 2 - 1} \right) \times \sqrt 2 }}{{\left( {\sqrt 2 } \right) \times \sqrt 2 }}} \right) = \dfrac{{2 - \sqrt 2 }}{2} \\\ \Rightarrow {\cos ^2}a = \dfrac{{2 - \sqrt 2 }}{4} \\\$
From (i) $\cos \left( {\dfrac{{13\pi }}{8}} \right) = \cos a$

\Rightarrow \cos a = \dfrac{{\sqrt {2 - \sqrt 2 } }}{2} \\\ \Rightarrow \cos \left( {\dfrac{{13\pi }}{8}} \right) = \dfrac{{\sqrt {2 - \sqrt 2 } }}{2} \\\ $$ **Therefore $\cos \left( {\dfrac{{13\pi }}{8}} \right)$ is $$\dfrac{{\sqrt {2 - \sqrt 2 } }}{2}$$.** **Note:** Some other equations needed for solving these types of problem are: $$\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right) \\\ \Rightarrow\cos \left( {2\theta } \right) = {\cos ^2}\left( \theta \right)-{\sin ^2}\left( \theta \right) = 1-2{\text{ }}{\sin ^2}\left( \theta \right) = 2{\text{ }}{\cos ^2}\left( \theta \right)-1$$ Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.