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

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

Explanation

Solution

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:
cosx=2cos2x21 2cos2x2=1+cosx cosx2=1+cosx2  \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
cosx=12sin2x2 2sin2x2=1cosx sinx2=1cosx2  \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
tanx2=1cosx1+cosx\tan \dfrac{x}{2} = \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}}

Now we have given that
cot(x)=13\cot (x) = 13
From basic trigonometric ratio we have,
sinx=1170,cosx=13170,tanx=113\sin x = \dfrac{1}{{\sqrt {170} }}, \cos x = \dfrac{{13}}{{\sqrt {170} }}, \tan x = \dfrac{1}{{13}}
Now putting the values we get
cosx2=1+11702  cosx2=170+12170  \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
sinx2=111702  sinx2=17012170  \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

tanx2=111701+1170  tanx2=1701170+1  tanx2=1701170+1×17011701(Onrationalising)  tanx2=(1701)2(170)212  tanx2=(1701)2169  tanx2=170113  \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 cos2x=2cos2x1\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 thattanx=sinxcosx\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.