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.