Double angle formula

**Hooperoo** · March 11th 2012, 10:01 PM

Hi ya'll. Quick question

using double angle formula, verify that

csc(2theta) - cot (2theta) = tan theta

im so very confused.

Working would be helpful to thanks

**princeps** · March 11th 2012, 10:19 PM

Originally Posted by Hooperoo

Hi ya'll. Quick question

using double angle formula, verify that

csc(2theta) - cot (2theta) = tan theta

im so very confused.

Working would be helpful to thanks

$\frac{1}{\sin 2\theta}-\frac{\cos 2\theta}{\sin 2\theta}=\frac{1-(cos^2 \theta-\sin^2 \theta)}{2\sin \theta \cdot \cos \theta}=$

$=\frac{\sin^2 \theta + \cos^2 \theta -\cos^2 \theta +\sin^2 \theta}{2\sin \theta \cdot \cos \theta}=$

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

**Hooperoo** · March 11th 2012, 10:37 PM

Dang. You make it seem so easy! I had an attempt at it, and i got kinda close. But thank you@!

**Soroban** · March 12th 2012, 04:44 AM

Hello, Hooperoo!

You are expected to know these identities:

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

. . $\sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 1 - \cos2\theta \:=\:2\sin^2\!\theta$

Using double-angle formulas, verify that: . $\csc2\theta - \cot2\theta \:=\:\tan\theta$

We have: . $\csc2\theta - \cot2\theta \;=\;\frac{1}{\sin2\theta} - \frac{\cos2\theta}{\sin2\theta} \;=\;\frac{1 - \cos2\theta}{\sin2\theta}$

. . . . . . . . . . . . . . . . . . . $\;=\;\frac{2\sin^2\!\theta}{2\sin\theta\cos\theta} \;=\;\frac{\sin\tjeta}{\cos\theta} \;=\;\tan\theta$

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