# How do you verify sin 2x = (2 tan x)/(1+ tan² x)?

Apr 23, 2016

Modifying only the right hand side of the equation, we should first notice that the fraction's denominator is a form of the Pythagorean Identity:

${\sin}^{2} x + {\cos}^{2} x = 1$

$\implies \text{ } {\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} \frac{x}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x$

=>" "color(blue)(barul(|color(white)(a/a)color(black)(tan^2x+1=sec^2x)color(white)(a/a)|

Thus, we see that we can rewrite the fraction:

$\frac{2 \tan x}{1 + {\tan}^{2} x} = \frac{2 \tan x}{{\sec}^{2} x}$

Recalling that " "color(green)(barul(|color(white)(a/a)color(black)(sec^2x=1/cos^2x)color(white)(a/a)|$\text{ }$, we see that

$\frac{2 \tan x}{\sec} ^ 2 x = \frac{2 \tan x}{\frac{1}{\cos} ^ 2 x}$

Now, instead of dividing by $\frac{1}{\cos} ^ 2 x$, note that this is the same as multiplying by ${\cos}^{2} \frac{x}{1}$.

$\frac{2 \tan x}{\frac{1}{\cos} ^ 2 x} = 2 \tan x \cdot {\cos}^{2} x$

Rewrite $2 \tan x$ using the identity: " "color(purple)(barul(|color(white)(a/a)color(black)(tanx=sinx/cosx)color(white)(a/a)|

$2 \tan x \cdot {\cos}^{2} x = \frac{2 \sin x}{\cos} x \cdot {\cos}^{2} x$

The $\cos x$ in the denominator will cancel with one of the $\cos x$ terms in the numerator:

$\frac{2 \sin x}{\cos} x \cdot {\cos}^{2} x = \frac{2 \sin x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\cos x}}}} \cdot {\cos}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} x = 2 \sin x \cos x$

This is the identity for $\sin 2 x$, which is the left hand side of the original equation:

color(brown)(barul(|color(white)(a/a)color(black)(2sinxcosx=sin2x)color(white)(a/a)|

Thus, the equation is verified.

Jul 8, 2016

$L H S = \sin 2 x = \frac{2 \sin x \cos x}{1}$

$= \frac{2 \sin x \cos x}{{\cos}^{2} x + {\sin}^{2} x}$

color(red)("Dividing both numerator and dinominator by "cos^2x

$= \frac{\frac{2 \sin x \cos x}{\cos} ^ 2 x}{{\cos}^{2} \frac{x}{\cos} ^ 2 x + {\sin}^{2} \frac{x}{\cos} ^ 2 x}$

$= \frac{2 \tan x}{1 + {\tan}^{2} x} = R H S$