Question

How do you find the exact value of \[\csc \left( {\dfrac{{17\pi }}{6}} \right)\]?

Answer 1

Hint: In this question we have to find the exact value of the trigonometric function, first convert the argument of the function into principal argument by converting the angle into difference of two angles, whose trigonometric angle ratios we know and apply sin on both sides , use the fact that \[\sin \left( {3\pi - x} \right) = \sin x\] and now use the trigonometric identity \[\csc x = \dfrac{1}{{\sin x}}\], and by substituting the angle in the identity and further simplification we will get the required answer.

Complete step by step solution:
Given trigonometric function is \[\csc \left( {\dfrac{{17\pi }}{6}} \right)\],
Now rewrite the angle \[\dfrac{{17\pi }}{6}\]as the difference of two angles i.e.,
\[ \Rightarrow \dfrac{{17\pi }}{6} = 3\pi - \dfrac{\pi }{6}\],
Now applying sin on both sides we get,
\[ \Rightarrow \sin \dfrac{{17\pi }}{6} = \sin \left( {3\pi - \dfrac{\pi }{6}} \right)\],
Now we know that \[\sin \left( {3\pi - x} \right) = \sin x\], applying the identity we get,
Here \[x = \dfrac{\pi }{6}\], now substituting the value in the identity we get,
\[ \Rightarrow \sin \dfrac{{17\pi }}{6} = \sin \left( {\dfrac{\pi }{6}} \right)\],
Now using the identity, \[\csc x = \dfrac{1}{{\sin x}}\], we get,
\[ \Rightarrow \csc \dfrac{{17\pi }}{6} = \dfrac{1}{{\sin \dfrac{{17\pi }}{6}}} = \dfrac{1}{{\sin \dfrac{\pi }{6}}}\]
Now we know that \[\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}\],
\[ \Rightarrow \csc \dfrac{{17\pi }}{6} = \dfrac{1}{{\sin \dfrac{\pi }{6}}}\],
Substituting the value we get,
\[ \Rightarrow \csc \dfrac{{17\pi }}{6} = \dfrac{1}{{\dfrac{1}{2}}}\],
Now simplifying we get,
\[ \Rightarrow \csc \dfrac{{17\pi }}{6} = 1 \times \dfrac{2}{1}\].
Now simplifying we get,
\[ \Rightarrow \csc \dfrac{{17\pi }}{6} = 2\],
So, the value of \[\csc \dfrac{{17\pi }}{6}\]is 2.


\[\therefore \]The exact value of the trigonometric function \[\csc \dfrac{{17\pi }}{6}\] will be equal to 2.

Note:
We have reduced the given angle into principal argument, as students will learn the only some trigonometric values of the principal angles only. And reducing the arguments will help in solving these types of questions. Some useful facts are: The values of sine, cosine and tan are positive in the first quadrant, the values of sin only positive in the second quadrant, the values of tan are only positive in the third quadrant and the values of cosine are only positive in the fourth quadrant.