Trigonometry Identities Proof

**omgilymcr** · April 4th 2012, 02:18 PM

I really need help with this problem.
sin2X+sin4X tan3X
__________ + _____ =0
sin2X-sin4X tanX

they are 2 fractions added together. It all equals 0. PLEASE help!
the 2X and 4X are not exponents they are identites, for trig functions. Thanks!

**Soroban** · April 4th 2012, 07:11 PM

Hello, omgilymcr!

We will need these identities:

. . $\sin2\theta \:=\:2\sin\theta\cos\theta$

. . $\sin3\theta \:=\:3\sin\theta - 4\sin^3\theta$

. . $\cos3\theta \:=\:4\cos^3\theta - 3\cos\theta$

. . $\cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}$

$\text{Verify: }\:\frac{\sin2x + \sin4x}{\sin2x-\sin4x} + \frac{\tan3x}{\tan x} \:=\:0$

The first fraction is: . $\frac{\sin 2x + \sin4x}{\sin2x - \sin4x} \;=\;\frac{\sin2x + 2\sin2x\cos2x}{\sin2x - 2\sin2x\cos2x}$

. . . . . . . . . . $=\;\frac{\sin2x(1 + 2\cos2x)}{\sin2x(1 - 2\cos2x)} \;=\;\frac{1+2\cos2x}{1-2\cos2x}\;\;{\color{blue}[1]}$

The second fraction is: . $\frac{\tan3x}{\tan x} \;=\;\frac{\sin3x}{\cos3x}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3\sin x - 4\sin^3x}{4\cos^3x - 3\cos x}\cdot\frac{\cos x}{\sin x}$

. . . . . . . . . . $=\;\frac{\sin x(3 - 4\sin^2x)}{\cos x(4\cos^2x - 3)}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3-4\sin^2x}{4\cos^2x-3}$

. . . . . . . . . . $=\;\frac{3-4(1-\cos^2x)}{4\cos^22x-3} \;=\;\frac{4\cos^2x-1}{4\cos^2x-3}$

. . . . . . . . . . $=\;\frac{4\left(\frac{1+\cos^2x}{2}\right) - 1}{4\left(\frac{1+\cos^2x}{2}\right)-3} \;=\;\frac{1 + 2\cos2x}{2\cos2x-1}$

. . . . . . . . . . $=\;-\frac{1 + 2\cos2x}{1 - 2\cos2x}\;\;{\color{blue}[2]}$

We see that [2] is the negative of [1].

Therefore, their sum is zero.

Edit: corrected typo.
.

**Prove It** · April 5th 2012, 01:03 AM

Originally Posted by Soroban

Hello, omgilymcr!

We will need these identities:

. . $\sin2\theta \:=\:2\sin\theta\cos\theta$

. . $\sin3\theta \:=\:3\sin\theta - 4\sin^3\theta$

. . $\cos3\theta \:=\:4\cos^3\theta - 3\cos\theta$

. . $\cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}$

The first fraction is: . $\frac{\sin 2x + \sin4x}{\sin2x - \sin4x} \;=\;\frac{\sin2x + 2\sin2x\cos2x}{\sin2x - 2\sin2x\cos2x}$

. . . . . . . . . . $=\;\frac{\sin2x(1 + 2\cos2x)}{\sin2x(1 - 2\cos2x)} \;=\;\frac{1+2\cos2x}{1-2\cos2x}\;\;{\color{blue}[1]}$

The second fraction is: . $\frac{\tan3x}{\tan x} \;=\;\frac{\sin3x}{\cos3x}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3\sin x - 4\sin^3x}{4\cos^3x - 3\cos x}\cdot\frac{\cos x}{\sin x}$

. . . . . . . . . . $=\;\frac{\sin x(3 - 4\sin^2x)}{\cos x(4\cos^2x - 3)}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3-4\sin^2x}{4\cos^2x-3}$

. . . . . . . . . . $=\;\frac{3-4(1-\cos^2x)}{4\cos^22x-3} \;=\;\frac{4\cos^2x-1}{4\cos^2x-3}$

. . . . . . . . . . $=\;\frac{4\left(\frac{1+\cos^2x}{2}\right) - 1}{4\left(\frac{1+\cos^2x}{2}\right)-3} \;=\;\frac{1 + 2\cos2x}{2\cos2x-1}$

. . . . . . . . . . $=\;-\frac{1 + 2\cos2x}{1 - \cos2x}\;\;{\color{blue}[2]}$

We see that [2] is the negative of [1].

Therefore, their sum is zero.

You are missing a 2 in the denominator of [2]...

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