Question

What is the cosine of $\dfrac{13\pi }{6}$ and what other radian does it correspond to?

Answer 1

We solve this problem by using the standard conversion of cosine angle that is,
$\cos \left( 2\pi \pm \theta \right)=\cos \theta$
We convert the given angle in the form of $\left( 2\pi +\theta \right)$ so that we can convert the cosine of the given angle to the cosine of angle which is in standard table of trigonometric ratios. Then we get the required value by using the standard trigonometric table that is, $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$

Complete step-by-step answer:
We are asked to find the cosine of $\dfrac{13\pi }{6}$
Let us assume that the required value as,
$\Rightarrow x=\cos \left( \dfrac{13\pi }{6} \right)$
Now, let us try to convert the above angle in the form $\left( 2\pi +\theta \right)$
Here, we can see that the denominator of the angle is 6. So, let us divide the numerator as sum of two terms such that ne term is nearest 6 multiple of 13 that is,
$\Rightarrow x=\cos \left( \dfrac{12\pi +\pi }{6} \right)$
Now, by separating the terms inside the cosine ratio then we get,
$\begin{aligned} & \Rightarrow x=\cos \left( \dfrac{12\pi }{6}+\dfrac{\pi }{6} \right) \\\ & \Rightarrow x=\cos \left( 2\pi +\dfrac{\pi }{6} \right) \\\ \end{aligned}$
We know that the standard conversion of cosine ratio is given as,
$\cos \left( 2\pi \pm \theta \right)=\cos \theta$
By using this conversion in above equation then we get,
$\Rightarrow x=\cos \left( \dfrac{\pi }{6} \right)$
We know that the standard value of some angle from standard trigonometric table that is,
$\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$
By using the above value we get the required value as,
$\Rightarrow x=\dfrac{\sqrt{3}}{2}$
Therefore, the required value of cosine of $\dfrac{13\pi }{6}$ is given as,
$\therefore \cos \left( \dfrac{13\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$
We are also asked to find the other radian that represents the same value.
We know that the cosine angle is positive in fourth and first quadrants and we also know that the standard conversion $\cos \left( 2\pi \pm \theta \right)=\cos \theta$
Now, let us combine both the above statements and by generalising the conversion we get,
$\Rightarrow \cos \left( 2n\pi \pm \theta \right)=\cos \theta$ where $n=0,1,2,3,....$
Here, let us assume that $\theta =\dfrac{\pi }{6}$ then we get,
$\Rightarrow \cos \left( 2n\pi \pm \dfrac{\pi }{6} \right)=\cos \left( \dfrac{\pi }{6} \right)$ where $n=0,1,2,3,....$
Here, we can see that for all angles in the form $\left( 2n\pi \pm \dfrac{\pi }{6} \right)$ the value is some which is $\dfrac{\sqrt{3}}{2}$
Therefore, the angles that correspond to same value are given as,
$\therefore \dfrac{\pi }{6},\dfrac{11\pi }{6},\dfrac{23\pi }{6},\dfrac{25\pi }{6},.....$

Note: We need to note that there will be infinitely many numbers of angles that give the same cosine value. So, we need to take any few by generalising the angles that give the same cosine value. So, if we combine the statements that cosine is positive in first and fourth quadrants and the standard conversion $\cos \left( 2\pi \pm \theta \right)=\cos \theta$ to get the required answer.
But the main mistake that can be done is taking the odd multiples of $\pi$ also. If we take the odd multiples of $\pi$ that is $\cos \left( \pi \pm \theta \right)=-\cos \theta$ then the cosine value will be negative but we need to take only positive value because $\cos \left( \dfrac{13\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$ which is positive.