Question

How do you prove $\cot \dfrac{x}{2} = \dfrac{{1 + \cos x}}{{\sin x}}$ ?

Answer 1

In this question we will try to simplify the expression in the right hand side of the equation to get the left hand side value. We can see that in the left hand side of the equation, we have a half angle i.e. $\dfrac{x}{2}$ . So we will try to use the half angle formulas to solve this question.

Formulas used:
The formula that we are using:
$2{\cos ^2}\left( {\dfrac{x}{2}} \right) = 1 + \cos x$
$\sin (x) = 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$ .

Complete step by step solution:
According to question, we have
$\cot \dfrac{x}{2} = \dfrac{{1 + \cos x}}{{\sin x}}$ .
We should know that the original half angle formula of cosine function is
$\cos \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 + \cos x}}{2}}$ .
We will derive the formula that will help us solve this question, so we will square both the sides of the equation:
${\left( {\cos \dfrac{x}{2}} \right)^2} = \pm {\left( {\sqrt {\dfrac{{1 + \cos x}}{2}} } \right)^2}$
It gives the value
${\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 + \cos x}}{2}$
We can take $2$ from the denominator from the R.H.S to the Left hand side of the equation, and therefore it gives us:
$2{\cos ^2}\left( {\dfrac{x}{2}} \right) = 1 + \cos x$
Now, Let us take the right hand side of the equation i.e.
$\dfrac{{1 + \cos x}}{{\sin x}}$ .
Now in the numerator, we can substitute the formula :
$2{\cos ^2}\left( {\dfrac{x}{2}} \right) = 1 + \cos x$
Similarly in the denominator, we can also substitute the formula mentioned in the hint,
So we have
$\dfrac{{2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$
We can cancel out the similar terms from the numerator and denominator of the dfraction, i.e.
$\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}}$
Now we know that cotangent is the ratio of cosine and sine function, represented as”
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ , here we have $\theta = \dfrac{x}{2}$
So by applying this, it gives us
$\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}} = \cot \dfrac{x}{2}$
Hence it is proved that $\cot \dfrac{x}{2} = \dfrac{{1 + \cos x}}{{\sin x}}$ .

Additional Information:
We should note that the formula that we have used is derived from double angle formula i.e.
$\sin 2x = 2\sin x\cos x$
Since we have to solve for $\left( {\dfrac{x}{2}} \right)$, the half angle formula, we will replace $x = \left( {\dfrac{x}{2}} \right)$ or we can say that we will put the value in half.
Therefore it gives us:
$\sin (x) = 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$ .

Note:
Alternate method:
We should know another formula that we can apply in this question, i.e.
$\cos 2x = {\cos ^2}x - {\sin ^2}x$ .
Again, we will replace $x = \left( {\dfrac{x}{2}} \right)$ , and it gives us new formula i.e.
$\cos x = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}$
By putting this value in the numerator of the R.H.S, we get:
$\dfrac{{1 + {{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$
Now we know that
${\sin ^2}x + {\cos ^2}x = 1$
Or, $1 - {\sin ^2}x = {\cos ^2}x$
So by putting this we have:
$\dfrac{{{{\cos }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}} = \dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$
By cancelling out the common terms, it gives us value
$\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}} = \cot \dfrac{x}{2}$ .