Question

How do I find the value of $\csc \dfrac{{11\pi }}{2}$?

Answer 1

Hint: In this question, we need to find the value of the given cosecant function. We make use of the fact that $\csc x = \dfrac{1}{{\sin x}}$. So firstly, we need to find the value of $\sin \dfrac{{11\pi }}{2}$ and then just we need to find its reciprocal. We will rewrite the angle given in the expression, as a sum of two angles in radians by applying the properties of trigonometric function. Then we make use of the sum formula of sine of the trigonometric function and simplify the equation. And obtain the desired result.

Complete step-by-step answer:
In this problem we are asked to find the value of $\csc \dfrac{{11\pi }}{2}$.
We know that cosecant is the reciprocal of the sine function. So we make use of this idea to solve the given problem.
So we have, $\csc x = \dfrac{1}{{\sin x}}$
Here $x = \dfrac{{11\pi }}{2}$. So we have, $\csc \dfrac{{11\pi }}{2} = \dfrac{1}{{\sin \dfrac{{11\pi }}{2}}}$
So firstly we find the value in terms of sine and then just take the reciprocal of it.
Thus, we find the value $\sin \dfrac{{11\pi }}{2}$.
We can write the angle $\dfrac{{11\pi }}{2}$ as follows.
$\dfrac{{11\pi }}{2} = \dfrac{{3\pi }}{2} + \dfrac{{8\pi }}{2}$
$ \Rightarrow \dfrac{{5\pi }}{3} = \dfrac{{3\pi }}{2} + 4\pi $
Now, $\sin \left( {\dfrac{{11\pi }}{2}} \right) = \sin \left( {\dfrac{{3\pi }}{2} + 4\pi } \right)$
We use the formula $\sin (A + B) = \sin A\cos B + \cos A\sin B$
Here we have $A = \dfrac{{3\pi }}{2}$ and $B = 4\pi $
Putting the values in the formula, we get,
$ \Rightarrow \sin \left( {\dfrac{{3\pi }}{2} + 4\pi } \right) = \sin \dfrac{{3\pi }}{2}\cos 4\pi - \cos \dfrac{{3\pi }}{2}\sin 4\pi $
We know that the values of $\sin \dfrac{{3\pi }}{2} = - 1$, $\cos 4\pi = 1$, $\cos \dfrac{{3\pi }}{2} = 0$ and $\sin 4\pi = 0$
Substituting this we get,
$ \Rightarrow \sin \left( {\dfrac{{3\pi }}{2} + 4\pi } \right) = ( - 1) \times 1 - 0 \times 0$
Simplifying this we get,
$ \Rightarrow \sin \left( {\dfrac{{3\pi }}{2} + 4\pi } \right) = - 1 - 0$
$ \Rightarrow \sin \left( {\dfrac{{3\pi }}{2} + 4\pi } \right) = - 1$
Hence we get $\sin \dfrac{{11\pi }}{2} = - 1$.
Since we have $\csc \dfrac{{11\pi }}{2} = \dfrac{1}{{\sin \dfrac{{11\pi }}{2}}}$, we get,
$ \Rightarrow \csc \dfrac{{11\pi }}{2} = \dfrac{1}{{ - 1}} = - 1$
Therefore, the value of $\csc \dfrac{{11\pi }}{2}$ is -1.

Note:
Students must know the basic properties of trigonometric functions and also in which quadrant which function is positive or negative.
As in the first quadrant all the six trigonometric functions are positive. In the second quadrant only the sine and cosec functions are positive, rest of all are negative. In the third quadrant, only the tan and cot functions are positive and all the other functions are negative. In the fourth quadrant only the cosine and secant are positive.
The sum and difference formula related to sine and cosine are given below.
(1) $\sin (A + B) = \sin A\cos B + \cos A\sin B$
(2) $\sin (A - B) = \sin A\cos B - \cos A\sin B$
(3) $\cos (A + B) = \cos A\cos B - \sin A\sin B$
(4) $\cos (A - B) = \cos A\cos B + \sin A\sin B$