Question

How do you find the value of $\csc \left( {\dfrac{{3\pi }}{4}} \right)$ ?

Answer 1

Hint : In order to solve this question ,find $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ as $\csc \theta = \dfrac{1}{{\sin \theta }}$ by converting $\left( {\dfrac{{3\pi }}{4}} \right)$ into some (A+B) and apply $\sin \left( {A + B} \right)$ formula and find the inverse of it to find the required answer.
Formula:
$\sin \left( {A - B} \right) = \sin \left( A \right)\cos \left( B \right) - \sin \left( B \right)\cos \left( A \right)$

Complete step-by-step answer :
In trigonometric table ,the value of $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ is not given ,so to find this we have use some other means to find the required value .We’ll use other trigonometric values given in the trigonometric table .
First ,we’ll ,convert $\left( {\dfrac{{3\pi }}{4}} \right)$ into some (A-B) form
$\sin \left( {\dfrac{{3\pi }}{4}} \right) = \sin \left( {\dfrac{\pi }{2} + \dfrac{\pi }{4}} \right)$
Now we will apply formula $\sin \left( {A + B} \right) = \sin \left( A \right)\cos \left( B \right) + \sin \left( B \right)\cos \left( A \right)$
Where , A is equal to $\dfrac{\pi }{2}$ and B is equal to $\dfrac{\pi }{4}$
$\sin \left( {\dfrac{\pi }{2} + \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{\pi }{4}} \right) + \sin \left( {\dfrac{\pi }{4}} \right)\cos \left( {\dfrac{\pi }{2}} \right)$ -(1)
From the trigonometric table
$\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\,$ ,
$\sin \left( {\dfrac{\pi }{2}} \right) = 1$
$\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\,$
$\cos \left( {\dfrac{\pi }{2}} \right) = 0$
Putting values in equation (1)
$= 1 \times \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} \times 0 \\\ = \dfrac{1}{{\sqrt 2 }} \\\$
Therefore, the value of $\sin \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
Since using identity of trigonometry $\csc \theta = \dfrac{1}{{\sin \theta }}$ ,here $\theta = \dfrac{{3\pi }}{4}$

$$\csc \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{4}} \right)}} \\\ \csc \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}} \\\ \csc \left( {\dfrac{{3\pi }}{4}} \right) = \sqrt 2 \\\$$

Hence the value of $\csc \left( {\dfrac{{3\pi }}{4}} \right)$ is equal to $\sqrt 2$ .
So, the correct answer is “ $\sqrt 2$ ”.

Note : 1. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta$ for all values of $\theta$ and n $\in$ N.
2. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta$
Therefore, $\sin \theta$ and $\tan \theta$ and their reciprocals, $\cos ec\theta$ and $\cot \theta$ are odd functions whereas $\cos \theta$ and its reciprocal $\sec \theta$ are even functions.