Question

How do you find the exact value of \[\csc \left( \dfrac{9\pi }{2} \right)\]?

Answer 1

Hint: Divide \[9\pi \] with 2 and write the given angle as the sum of two angles in which one must be in range \[\left[ 0,\dfrac{\pi }{2} \right]\]. Now, find the sine of the simplified angle and use the property: - \[\sin \left( 2n\pi +\theta \right)=\sin \theta \], where ‘n’ is any positive integer. Now, use the relation, \[\sin \theta =\dfrac{1}{\csc \theta }\] to find the value of \[\csc \left( \dfrac{9\pi }{2} \right)\].

Complete step by step answer:
Here, we have been asked to find the exact value of the trigonometric expression: - \[\csc \left( \dfrac{9\pi }{2} \right)\].
Now, we know that \[\dfrac{9\pi }{2}\] is a large angle if converted in degrees. So, we need to find a simpler method to solve the question. On, observing the angle \[\dfrac{9\pi }{2}\] we can say that the angle \[9\pi \] is divided by 2, so by this division, we can write the given angle as: -
\[\Rightarrow \dfrac{9\pi }{2}=\dfrac{\left( 2\times 4\pi \right)+\pi }{2}\]
\[\Rightarrow \dfrac{9\pi }{2}=4\pi +\dfrac{\pi }{2}\] - (1)
We know that the sine function repeats its value after an interval of the angle \[2\pi \]. Let us draw the graph of the sine function. It can be shown as below: -

From the above graph we can clearly see that the values of the function \[\sin x\] starts repeating after an interval of \[2\pi \]. So, let us use this property to get our answer. From equation (1), we have,
\[\Rightarrow \dfrac{9\pi }{2}=4\pi +\dfrac{\pi }{2}\]
Taking sine function both the sides, we get,
\[\Rightarrow \sin \left( \dfrac{9\pi }{2} \right)=\sin \left( 4\pi +\dfrac{\pi }{2} \right)\]
Now, the value of the above expression of sine will first repeat after \[2\pi \] and the second time it will start repeating again after \[4\pi \], which can be given by the mathematical expression: - \[\sin \left( 2n\pi +\theta \right)=\sin \theta \], where n is a positive integer. So, on comparing \[\sin \left( 4\pi +\dfrac{\pi }{2} \right)\] with \[\sin \left( 2n\pi +\theta \right)\], we have,
\[\Rightarrow n=2\] and \[\theta =\dfrac{\pi }{2}\]
 \[\begin{align}
  & \Rightarrow \sin \left( 4\pi +\dfrac{\pi }{2} \right)=\sin \left( \dfrac{\pi }{2} \right) \\
 & \Rightarrow \sin \left( \dfrac{9\pi }{2} \right)=\sin \left( 4\pi +\dfrac{\pi }{2} \right)=\sin \left( \dfrac{\pi }{2} \right) \\
 & \Rightarrow \sin \left( \dfrac{9\pi }{2} \right)=1\left( \because \sin \left( \dfrac{\pi }{2} \right)=\sin {{90}^{\circ }}=1 \right) \\
\end{align}\]
Hence, the value of \[\sin \left( \dfrac{9\pi }{2} \right)\] is 1.
Now, we know that \[\csc =\dfrac{1}{\sin x}\], so we have,
\[\begin{align}
  & \Rightarrow \sin x=\dfrac{1}{\csc x} \\
 & \Rightarrow \sin \left( \dfrac{9\pi }{2} \right)=\dfrac{1}{\csc \left( \dfrac{9\pi }{2} \right)} \\
 & \Rightarrow 1=\dfrac{1}{\csc \left( \dfrac{9\pi }{2} \right)} \\
\end{align}\]
\[\Rightarrow \csc \left( \dfrac{9\pi }{2} \right)=\dfrac{1}{1}=1\], which is our answer.

Note:
 One may note that in the expression: - \[\sin \left( 2n\pi +\theta \right)=\sin \theta \] we have considered ‘n’ as any positive integer. Actually, this formula is true for any negative integer also but we do not need to remember that because we have the property of sine of an angle given as: - \[\sin \left( -x \right)=-\sin x\]. So, even if we are provided with an angle like \[\left( \dfrac{-9\pi }{2} \right)\] we can change it to \[\left( \dfrac{9\pi }{2} \right)\] and then use the formula: - \[\sin \left( 2n\pi +\theta \right)=\sin \theta \]. Remember the graph of all the trigonometric functions.