# How do you prove (tanx+sinx)/(2tanx)=cos^2(x/2)?

Apr 28, 2018

We'll need these two identities to complete the proof:

$\tan x = \sin \frac{x}{\cos} x$

$\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos x}{2}}$

I'll start with the right side, then manipulate it until it looks like the left side:

$R H S = {\cos}^{2} \left(\frac{x}{2}\right)$

$\textcolor{w h i t e}{R H S} = {\left(\cos \left(\frac{x}{2}\right)\right)}^{2}$

$\textcolor{w h i t e}{R H S} = {\left(\pm \sqrt{\frac{1 + \cos x}{2}}\right)}^{2}$

$\textcolor{w h i t e}{R H S} = \frac{1 + \cos x}{2}$

$\textcolor{w h i t e}{R H S} = \frac{1 + \cos x}{2} \textcolor{red}{\cdot \sin \frac{x}{\sin} x}$

$\textcolor{w h i t e}{R H S} = \frac{\sin x + \sin x \cos x}{2 \sin x}$

$\textcolor{w h i t e}{R H S} = \frac{\sin x + \sin x \cos x}{2 \sin x} \textcolor{red}{\cdot \frac{\frac{1}{\cos} x}{\frac{1}{\cos} x}}$

$\textcolor{w h i t e}{R H S} = \frac{\sin \frac{x}{\cos} x + \frac{\sin x \cos x}{\cos} x}{2 \sin \frac{x}{\cos} x}$

$\textcolor{w h i t e}{R H S} = \frac{\tan x + \sin x}{2 \tan x}$

$\textcolor{w h i t e}{R H S} = L H S$

That's the proof. Hope this helped!

Apr 28, 2018

We seek to prove the identity:

$\frac{\tan x + \sin x}{2 \tan x} \equiv {\cos}^{2} \left(\frac{x}{2}\right)$

Consider the LHS of the expression, and use the definition of tangent:

$L H S = \frac{\tan x + \sin x}{2 \tan x}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\sin \frac{x}{\cos} x + \sin x}{2 \left(\sin \frac{x}{\cos} x\right)}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\cos \frac{x}{\sin} x\right) \left(\frac{\sin \frac{x}{\cos} x + \sin x}{2}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\cos \frac{x}{\sin} x \cdot \sin \frac{x}{\cos} x + \cos \frac{x}{\sin} x \cdot \sin x}{2}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1 + \cos x}{2}$

Now, Consider the RHS, and use the identity:

$\cos 2 A \equiv 2 {\cos}^{2} A - 1$

Giving us:

$\cos x \equiv 2 {\cos}^{2} \left(\frac{x}{2}\right) - 1 \implies 1 + \cos x \equiv 2 {\cos}^{2} \left(\frac{x}{2}\right)$

$\therefore {\cos}^{2} \left(\frac{x}{2}\right) = \frac{1 + \cos x}{2} = R H S$

Thus:

$L H S = R H S \implies \frac{\tan x + \sin x}{2 \tan x} \equiv {\cos}^{2} \left(\frac{x}{2}\right) \setminus \setminus \setminus$ QED

May 6, 2018

$L H S = \frac{\tan x + \sin x}{2 \tan x}$

$= \frac{\cancel{\tan x} \left(1 + \sin \frac{x}{\tan} x\right)}{2 \cancel{\tan x}}$

$= \frac{1 + \cos x}{2} = \frac{2 {\cos}^{2} \left(\frac{x}{2}\right)}{2} = {\cos}^{2} \left(\frac{x}{2}\right) = R H S$