Question

Find $ \sin \left( {\dfrac{x}{2}} \right)\,\,,\,\,\cos \left( {\dfrac{x}{2}} \right)\,\,and\,\,\tan \left( {\dfrac{x}{2}} \right) $ if $ \tan x = \dfrac{{ - 4}}{4} $ $ x \in IInd\,\,quadrant $ .

Answer 1

Hint: For this we first from the given $ \tan x $ we find value of $ \sin x $ and $ \cos x $ then using these values in half angle formula to find the value of $ \sin \dfrac{x}{2}\,\,and\,\,\cos \dfrac{x}{2} $ and then dividing result of two to get value of $ \tan \dfrac{x}{2} $ .

Complete step-by-step answer:
Given, $ \tan x = \dfrac{{ - 4}}{4} $
Or can be written as:
 $ \tan x = - 1 $
 $ \Rightarrow x = {45^0} $
Then, $ \sin x = \dfrac{1}{{\sqrt 2 }}\,\,\,and\,\,\,\cos x = \dfrac{{ - 1}}{{\sqrt 2 }} $ as $ x \in IInd\,\,quadrant $
Also, for half angles, we know that:
 $ \sin \dfrac{x}{2} = \sqrt {\dfrac{{1 - \cos x}}{2}} $
Substituting value in above we have:
 $
  \sin \dfrac{x}{2} = \sqrt {\dfrac{{1 - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}{2}} \\
   \Rightarrow \sin \dfrac{x}{2} = \sqrt {\dfrac{{1 + \dfrac{1}{{\sqrt 2 }}}}{2}} \\
   \Rightarrow \sin \dfrac{x}{2} = \sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} \\
  $
Also,
 $ \cos \dfrac{x}{2} = \sqrt {\dfrac{{1 + \cos x}}{2}} $

Substituting value in above we have:
 $
  \cos \dfrac{x}{2} = \sqrt {\dfrac{{1 + \left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}{2}} \\
   \Rightarrow cox\dfrac{x}{2} = \sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{2}} \\
   \Rightarrow \cos \dfrac{x}{2} = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\
  $
On dividing above two results. We have
 $
  \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} = \dfrac{{\sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} }}{{\sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} }} \\
  \tan \dfrac{x}{2} = \sqrt {\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}} \\
  $
Hence, values of $ \sin \dfrac{x}{2},\,\,\cos \dfrac{x}{2}\,\,and\,\,\tan \dfrac{x}{2} $ are $ \sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} $ , $ \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} $ and $ \sqrt {\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}} $ respectively.

Note: For trigonometric function problem when solving according to quadrant one must take care of positive and negative signs according to quadrant and also to find solution of trigonometric function one should choose correct trigonometric formula to find correct solution of the problem.