Question

How do you find $\sin (x/2),\cos (x/2)$ and $\tan (x/2)$ from the given $\cot (x) = 13$?

Answer 1

Hint: First we will mention all the terms that you need to know before solving this type of question. Then we will evaluate the values of the above mentioned terms directly. Then again mention some points to elaborate the explanation.

Complete step-by-step solution:
We have the following identities to be used for getting the results:
$
  \cos x = 2{\cos ^2}\dfrac{x}{2} - 1\,\, \\
   \Rightarrow 2{\cos ^2}\dfrac{x}{2} = 1 + \cos x\,\,\, \\
   \Rightarrow \cos \dfrac{x}{2} = \sqrt {\dfrac{{1 + \cos x}}{2}} \\
 $
Also
$
  \cos x = 1 - 2{\sin ^2}\dfrac{x}{2}\, \\
   \Rightarrow 2{\sin ^2}\dfrac{x}{2} = 1 - \cos x\,\,\, \\
   \Rightarrow \sin \dfrac{x}{2} = \sqrt {\dfrac{{1 - \cos x}}{2}} \\
 $
And
\[\tan \dfrac{x}{2} = \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} \]

Now we have given that
$\cot (x) = 13$
From basic trigonometric ratio we have,
\[\sin x = \dfrac{1}{{\sqrt {170} }}, \cos x = \dfrac{{13}}{{\sqrt {170} }}, \tan x = \dfrac{1}{{13}}\]
Now putting the values we get
$
   \Rightarrow \cos \dfrac{x}{2} = \sqrt {\dfrac{{1 + \dfrac{1}{{\sqrt {170} }}}}{2}} \\
    \\
   \Rightarrow \cos \dfrac{x}{2} = \sqrt {\dfrac{{\sqrt {170} + 1}}{{2\sqrt {170} }}} \\
 $
Also
$
   \Rightarrow \sin \dfrac{x}{2} = \sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt {170} }}}}{2}} \\
    \\
   \Rightarrow \sin \dfrac{x}{2} = \sqrt {\dfrac{{\sqrt {170} - 1}}{{2\sqrt {170} }}} \\
 $
And
\[
  \tan \dfrac{x}{2} = \sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt {170} }}}}{{1 + \dfrac{1}{{\sqrt {170} }}}}} \\
    \\
  \tan \dfrac{x}{2} = \sqrt {\dfrac{{\sqrt {170} - 1}}{{\sqrt {170} + 1}}} \\
    \\
  \tan \dfrac{x}{2} = \sqrt {\dfrac{{\sqrt {170} - 1}}{{\sqrt {170} + 1}} \times \dfrac{{\sqrt {170} - 1}}{{\sqrt {170} - 1}}} (On\,rationalising) \\
    \\
  \tan \dfrac{x}{2} = \sqrt {\dfrac{{{{\left( {\sqrt {170} - 1} \right)}^2}}}{{{{\left( {\sqrt {170} } \right)}^2} - {1^2}}}} \\
    \\
  \tan \dfrac{x}{2} = \sqrt {\dfrac{{{{\left( {\sqrt {170} - 1} \right)}^2}}}{{169}}} \\
    \\
  \tan \dfrac{x}{2} = \dfrac{{\sqrt {170} - 1}}{{13}} \\
 \]

Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities $\cos 2x = 2{\cos ^2}x - 1$. While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that$\tan x = \,\dfrac{{\sin x}}{{\cos x}}$.
While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity.