# How do you verify (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x?

May 1, 2018

Pretty sure the question is $\left(\sin x\right) \left(\tan x \cos x - \cot x \cos x\right) = 1 - 2 {\cos}^{2} x$ ,or else it will be not provable.

#### Explanation:

Some basic knowledge to begin with:
1. ${\sin}^{2} x + {\cos}^{2} x = 1$
2. $\tan x = \sin \frac{x}{\cos} x$
3. $\cot x = \cos \frac{x}{\sin} x$

Let's start from the left hand side
$\left(\sin x\right) \left(\tan x \cos x - \cot x \cos x\right)$

$= \sin x \tan x \cos x - \sin x \cot x \cos x$

$= \sin x \left(\sin \frac{x}{\cos} x\right) \cos x - \sin x \left(\cos \frac{x}{\sin} x\right) \cos x$

$= {\sin}^{2} x - {\cos}^{2} x$

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

$= 1 - 2 {\cos}^{2} x$

May 1, 2018

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)tanx=sinx/cosx" and "cotx=cosx/sinx

•color(white)(x)sin^2x+cos^2x=1

$\Rightarrow {\sin}^{2} x = 1 - {\cos}^{2} x$

$\text{consider the left side}$

$\text{distributing the parenthesis}$

$\sin x \times \sin \frac{x}{\cancel{\cos x}} \times \cancel{\cos x} - \cancel{\sin x} \times \cos \frac{x}{\cancel{\sin x}} \times \cos x$

$= {\sin}^{2} x - {\cos}^{2} x$

$= \left(1 - {\cos}^{2} x\right) - {\cos}^{2} x$

$= 1 - 2 {\cos}^{2} x = \text{ right side "rArr"verified}$