# Verify the following identities: (a) 2sin^2(2x) + cos4x=1 (b) tanx+co x =2sec2x (c) tan3x =[tanx(3-tan^2x)]/ (1-3tan^2x)?

Aug 31, 2015

Verify:
1. 2sin^2 (2x) + cos 4x = 1
2. tan x + cot x = 2csc 2x

#### Explanation:

1. 2sin^2 (2x) + cos 4x = 2sin^2 (2x) + (1 - 2sin^2 (2x) = 1
Reminder of trig identity: $\cos 2 a = 1 - 2 {\sin}^{2} a$

2. $\tan x + \cot x = \sin \frac{x}{\cos} x + \cos \frac{x}{\sin} x =$
= $\frac{{\sin}^{2} x + {\cos}^{2} x}{\sin x . \cos x} = \frac{1}{\sin x . \cos x} =$
$= \frac{2}{\sin 2 x} = 2 \csc \left(2 x\right)$

Aug 31, 2015

Verify: $\tan 3 x = \frac{\left(\tan x\right) \left(3 - {\tan}^{2} x\right)}{1 - 3 {\tan}^{2} x}$

#### Explanation:

Reminder: Trig identity $\tan 2 a = \frac{2 \tan a}{1 - {\tan}^{2} a} \left(1\right)$, and
Trig Identity: $\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a . \tan b} \left(2\right)$

$\tan 3 x = \tan \left(2 x + x\right) = \frac{\tan x + \tan 2 x}{1 - \tan x . \tan 2 x} \left(3\right)$
Substitute into (3) the value of tan 2x from Identity (1)

Develop the numerator:(tan x + (2tan x)/(1 - tan^2 x)) = (3tan x - tan^3 x)/(1 - tan^2 x) =
$= \frac{\left(\tan x\right) \left(3 - {\tan}^{2} x\right)}{1 - {\tan}^{2} x}$ (4)
Develop the denominator: (1 - tan x.tan 2x) = =1 - (tan x(2tan x))/(1 - tan^2 x) =
$\frac{1 - {\tan}^{2} x - 2 {\tan}^{2} x}{1 - 2 {\tan}^{2} x} = \frac{1 - 3 {\tan}^{2} x}{1 - {\tan}^{2} x}$ (5)

$\tan 3 x = \frac{\left(4\right)}{\left(5\right)} = \frac{\tan x \left(3 - {\tan}^{2} x\right)}{1 - 3 {\tan}^{2} x}$