# Double angle formula

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• Mar 11th 2012, 10:01 PM
Hooperoo
Double angle formula
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 :)
• Mar 11th 2012, 10:19 PM
princeps
Re: Double angle formula
Quote:

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 :)

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

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

$\displaystyle =\frac{2\sin^2 \theta}{2\sin \theta \cdot \cos \theta}=\frac{\sin \theta}{\cos \theta}=\tan \theta$
• Mar 11th 2012, 10:37 PM
Hooperoo
Re: Double angle formula
Dang. You make it seem so easy! I had an attempt at it, and i got kinda close. But thank you@!
• Mar 12th 2012, 04:44 AM
Soroban
Re: Double angle formula
Hello, Hooperoo!

You are expected to know these identities:

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

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

Quote:

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

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

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