Question

How to verify this identity: $\tan \dfrac{x}{2} = \dfrac{{\sin x}}{{1 + \cos x}}?$

Answer 1

Hint: In this question we have to prove that L.H.S is equal to R.H.S. We can see that there are trigonometric ratios in the given equation. So here we will use the trigonometric double angle and half angle formulas to solve this question.
The double angle formula of sine function is
$\sin 2x = 2\sin x\cos x$
The half angle formula of cosine is:
$\cos \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 + \cos x}}{2}} $ .

Complete step by step solution:
Here we have been given the expression:
 $\tan \dfrac{x}{2} = \dfrac{{\sin x}}{{1 + \cos x}}$ .
We will try to simplify the value in the right hand side of the equation.
First of all we will write the half angle formula in simpler value. Let us first take the cosine formula i.e.
$\cos \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 + \cos x}}{2}} $ .
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:
$2{\cos ^2}\left( {\dfrac{x}{2}} \right) = 1 + \cos x$
We re arranging the terms we can write the equation also as :
$\cos (x) = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$
Now we have double angle formula i.e.
 $\sin 2x = 2\sin x\cos x$
We can see in the question that we have to solve for $\left( {\dfrac{x}{2}} \right)$
So we will replace $x = \left( {\dfrac{x}{2}} \right)$ or we can say that we will put the value in half.
Si=o it gives us new expression:
$\sin \left( {2 \times \dfrac{x}{2}} \right) = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$
$\sin (x) = 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$ .
We will now substitute these new expressions in the R.H.S of the given question:
$\dfrac{{2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{x}{2}} \right) - 1 + 1}}$
We will now simplify this,
$\dfrac{{2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}$
We can see that there is a common term in denominator and numerator, so it will get cancelled out:
$\dfrac{{\sin \left( {\dfrac{x}{2}} \right)}}{{\cos \left( {\dfrac{x}{2}} \right)}}$
Therefore it gives us value
 $\tan \left( {\dfrac{x}{2}} \right)$
Hence it is proved that $\tan \dfrac{x}{2} = \dfrac{{\sin x}}{{1 + \cos x}}$ .

Note:
We should note that the trigonometric ratio of tangent function is:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ .
So in the above solution we have
$\theta = \dfrac{x}{2}$ .
Therefore it gives us $\tan \left( {\dfrac{x}{2}} \right)$ .
We should also know the double angle formula of cosine function i.e.
$\cos 2x = {\cos ^2}x - {\sin ^2}x$ .
There are also two other values of cosine function:
$\cos 2x = 2{\cos ^2}x - 1 = 1 - 2{\sin ^2}x$