Question

How do you evaluate $\sec \left( {\dfrac{\pi }{9}} \right)$ ?

Answer 1

Hint: Since $\sec \theta $ is $\dfrac{1}{{\cos \theta }}$ solving the given trigonometric quantity using $\cos \theta $ would be easier. Write $\sec \left( {\dfrac{\pi }{9}} \right)$ in terms of $\cos \left( {\dfrac{\pi }{3}} \right)$ and substitute it in the $\cos (3\theta )$ formula to get an equation of degree $3$. Solve it to get the values of $\cos \theta $ which should later be converted into $\sec \theta $ for the final answer.

Formula used:
$\cos (3\theta ) = 4{\cos ^3}\theta - 3\cos \theta $

Complete step-by-step answer:
Given trigonometric expression, $\sec \left( {\dfrac{\pi }{9}} \right)$
Consider, $\cos \left( {\dfrac{\pi }{3}} \right)$ whose value is $\dfrac{1}{2}$
Multiply $3$ on the numerator and denominator of the $\cos \theta $.
$ \Rightarrow \cos \left( {\dfrac{{3\pi }}{{3 \times 3}}} \right) = \dfrac{1}{2}$
$ \Rightarrow \cos \left( {\dfrac{{3\pi }}{9}} \right) = \dfrac{1}{2}$
Consider $\theta = \dfrac{\pi }{9}$, then
$ \Rightarrow \cos (3\theta ) = \dfrac{1}{2}$
The formula for $\cos (3\theta )$ is,
$ \Rightarrow $$\cos (3\theta ) = 4{\cos ^3}\theta - 3\cos \theta $
$ \Rightarrow 4{\cos ^3}\theta - 3\cos \theta = \dfrac{1}{2}$
Now put $\cos \theta = x$ for easy evaluation of the polynomial of degree $3$
$ \Rightarrow 4{x^3} - 3x = \dfrac{1}{2}$
Take the $\dfrac{1}{2}$onto the L.H.S
$ \Rightarrow 4{x^3} - 3x - \dfrac{1}{2} = 0$
Multiply the whole equation with $2$
$ \Rightarrow 8{x^3} - 6x - 1 = 0$
Solving this equation results in $3$ roots since it is a polynomial of degree $3$
By using a scientific calculator,
We get the roots as,
$ \Rightarrow {x_1} = 0.93969;{x_2} = - 0.76604;{x_3} = - 0.17364$
Since the angle is $ < 90^\circ $ and $\cos \theta $ is positive in that region so we consider the positive value only.
$\cos \theta $is positive in the first and fourth quadrant.
$ \Rightarrow \cos \theta = 0.93969$
But $\theta = \dfrac{\pi }{9}$
$ \Rightarrow \cos \left( {\dfrac{\pi }{9}} \right) = 0.93969$
Since,$\sec \theta = \dfrac{1}{{\cos \theta }}$
$ \Rightarrow \sec \left( {\dfrac{\pi }{9}} \right) = \dfrac{1}{{\cos \left( {\dfrac{\pi }{9}} \right)}}$
From the above evaluation, we know that,$\cos \left( {\dfrac{\pi }{9}} \right) = 0.93969$
$ \Rightarrow \sec \left( {\dfrac{\pi }{9}} \right) = \dfrac{1}{{0.93969}}$
On evaluating We get,
$ \Rightarrow \sec \left( {\dfrac{\pi }{9}} \right) = \dfrac{1}{{0.93969}} = 1.0641$

$\therefore \sec \left( {\dfrac{\pi }{9}} \right) = 1.0641$

Additional information: Whenever complex equations are given to solve one must always Firstly start from the complex side and then convert all the terms into $\cos \theta $ or $\sin \theta $. Then combine them into single fractions. Now it’s most likely to use Trigonometric identities for the transformations if there are any. Know when and where to apply the Subtraction-Addition formula.

Note:
Always check when the trigonometric functions are given in degrees or radians. $1^\circ \times \dfrac{\pi }{{180}} = 0.017Rad$. Express everything in $\sin \theta $ or $\cos \theta $ to easily evaluate. Always check where both the trigonometric functions become negative or positive. Most of the problems can easily be solved by memorizing Quotient identities and Subtraction-Addition identities.