Question

How do I evaluate $\cos \left( \dfrac{\pi }{5} \right)$ without using a calculator.

Answer 1

Hint: Now first let us consider $\theta =\dfrac{\pi }{10}$ . Now we will use the relation $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \left( \theta \right)$ we will write the form an equation using $\cos 3\theta $. Now we know that $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ and $\sin 2\theta =2\sin \theta \cos \theta $. Hence using this we will simplify the equation and form a quadratic in $\sin \theta $ using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Now using the formula for roots of the quadratic equation we will find the value of $\sin \theta $. Now we will find the value of $\cos \left( \dfrac{\pi }{5} \right)$ by substituting the value of $\sin \theta $ obtained in the equation $\cos 2\theta =1-2{{\sin }^{2}}\theta $.

Complete step-by-step solution:
Now first let us consider $\theta =\dfrac{\pi }{10}$.
Now we want to find the value of $\cos 2\theta $ .
Now first consider the value of $\cos 3\theta $
$\begin{align}
  & \Rightarrow \cos 3\theta =\cos \dfrac{3\pi }{10} \\
 & \Rightarrow \cos 3\theta =\cos \left( \dfrac{\pi }{2}+\dfrac{3\pi }{10}-\dfrac{\pi }{2} \right) \\
 & \Rightarrow \cos 3\theta =\cos \left( \dfrac{\pi }{2}+\dfrac{3\pi -5\pi }{10} \right) \\
 & \Rightarrow \cos 3\theta =\cos \left( \dfrac{\pi }{2}-\dfrac{2\pi }{10} \right) \\
 & \Rightarrow \cos 3\theta =\cos \left( \dfrac{\pi }{2}-\dfrac{\pi }{5} \right) \\
\end{align}$
Now we know that $\cos \left( \dfrac{\pi }{2}-\alpha \right)=\sin \alpha $.
Hence we get,
$\begin{align}
  & \Rightarrow \cos 3\theta =\sin \left( \dfrac{\pi }{5} \right) \\
 & \Rightarrow \cos 3\theta =\sin 2\theta \\
\end{align}$
Now we have $\cos 3\theta =\sin 2\theta $.
We know that $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ and $\sin 2\theta =2\sin \theta \cos \theta $
Hence substituting the formulas we get, ‘
$\Rightarrow 4{{\cos }^{3}}\theta -3\cos \theta =2\sin \theta \cos \theta $
Now dividing the above equation by $\cos \theta $ we get,
$\Rightarrow 4{{\cos }^{2}}\theta -3=2\sin \theta $
Now we know that $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $ Hence using this we get,
$\begin{align}
  & \Rightarrow 4\left( 1-{{\sin }^{2}}\theta \right)-3=2\sin \theta \\
 & \Rightarrow 4-4{{\sin }^{2}}\theta -3=2\sin \theta \\
 & \Rightarrow 4{{\sin }^{2}}\theta +2\sin \theta -1=0 \\
\end{align}$
Now the given equation is a quadratic equation in $\sin \theta $. We know that the solution of the quadratic equation of the form $a{{x}^{2}}+bx+c$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Hence the solution to the given equation is
$\begin{align}
  & \Rightarrow \sin \theta =\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( 4 \right)\left( -1 \right)}}{2\left( 4 \right)} \\
 & \Rightarrow \sin \theta =\dfrac{-2\pm \sqrt{4+16}}{8} \\
 & \Rightarrow \sin \theta =\dfrac{-2\pm \sqrt{20}}{8} \\
 & \Rightarrow \sin \theta =\dfrac{-2\pm 2\sqrt{5}}{8} \\
 & \Rightarrow \sin \theta =\dfrac{-1\pm \sqrt{5}}{4} \\
\end{align}$
Now $\cos 2\theta $ is nothing but $1-2{{\sin }^{2}}\theta $. Hence we get,
$\begin{align}
  & \Rightarrow \cos 2\theta =1-2{{\left( \dfrac{-1\pm \sqrt{5}}{4} \right)}^{2}} \\
 & \Rightarrow \cos 2\theta =1-2\dfrac{\left( 1+5\pm 2\sqrt{5} \right)}{16} \\
 & \Rightarrow \cos 2\theta =\dfrac{8-6\pm 2\sqrt{5}}{8} \\
 & \Rightarrow \cos 2\theta =\dfrac{1\pm \sqrt{5}}{4} \\
\end{align}$
Now since $\theta =\dfrac{\pi }{10}$ we have
$\Rightarrow \cos \left( \dfrac{\pi }{5} \right)=\dfrac{1\pm \sqrt{5}}{4}$

Note: Now note that to find the solution of such equation we will try to write a trigonometric equation which is satisfied by the given $\theta $. Then we will try to convert the equation such that we have just one trigonometric function in the equation. Hence we will solve the equation to find the value. Now we can easily use different identities to find the trigonometric ratio required from the trigonometric ratio obtained.