Question

How do you find the value of $\cos \left( \dfrac{16\pi }{3} \right)$ ?

Answer 1

Hint: This is a little tricky question. We have to express $\dfrac{16\pi }{3}$ in such a way that we get its reducible form. We rewrite $\dfrac{16\pi }{3}$ as $\dfrac{18\pi }{3}-\dfrac{2\pi }{3}$ . Now we use the properties of cosine which is, $\cos \left( x+2n\pi \right)=x$ and then after this simplification we can easily evaluate the value of the expression by using the trigonometric tables.

Complete step-by-step solution:
The given trigonometric expression is, $\cos \left( \dfrac{16\pi }{3} \right)$
Now we rewrite this expression or expand in such a way that we get a term which is periodicity of the trigonometric function which can be later used to simplify easily,
$\Rightarrow \cos \left( \dfrac{18\pi }{3}-\dfrac{2\pi }{3} \right)$
Now on evaluating we get,
$\Rightarrow \cos \left( 6\pi -\dfrac{2\pi }{3} \right)$
Since the periodicity of cosine is $2\pi$,hence we remove three full $2\pi$ rotations till we get the angle between $0$ and $2\pi$
$\Rightarrow \cos \left( -\dfrac{2\pi }{3} \right)$
And cosine is negative in the third quadrant so we can now remove the negation and then evaluate the angle.
$\Rightarrow \cos \left( \dfrac{2\pi }{3} \right)$
From the trigonometric tables, we know the value of $\cos \dfrac{2\pi }{3}$ as $-\dfrac{1}{2}$
The detailed explanation is that,
The expression, $\cos x=\dfrac{-1}{2}$ is because $\cos x$ is negative in ${{2}^{nd}}$ and ${{3}^{rd}}$ Quadrant.
So, the values of x should lie in ${{2}^{nd}},{{3}^{rd}}$ Quadrant in the interval $[0,2\pi ]$ . There are two values of $\cos x$ in this range which are,
$\Rightarrow x={{\cos }^{-1}}\left( \dfrac{-1}{2} \right)$
$\Rightarrow x={{\cos }^{-1}}\left( x\pm \dfrac{\pi }{3} \right)$
Hence, we can rewrite our expression as,
$\Rightarrow \cos \left( \dfrac{\pi }{3}+\dfrac{\pi }{3} \right)=-\dfrac{1}{2}$
Therefore, the evaluated value of $\cos \left( \dfrac{16\pi }{3} \right)$ is $-\dfrac{1}{2}$

Note: Always check when the trigonometric functions are given in degrees or radians. There is a lot of difference between both $1{}^\circ \times \dfrac{\pi }{180}=0.017Rad$.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 the Quotient identities and the Subtraction-Addition identities.