# trig identity

• Jan 30th 2009, 06:18 AM
vungaralas
trig identity
pls solve this
cot(4/5)-A = cos 2A/(1-sin 2A)
• Jan 30th 2009, 09:50 AM
Trigonometry
Hello vungaralas
Quote:

Originally Posted by vungaralas
pls solve this
cot(4/5)-A = cos 2A/(1-sin 2A)

I'm not quite sure what you mean by cot(4/5)-A. But if it's any help, you can express the right-hand side in terms of tan A, using the half-angle formulae, or from the basic double-angle formulae, like this:

$\displaystyle \frac{\cos 2A}{1-\sin 2A} = \frac{\cos^2A - \sin^2A}{1 - 2\sin A \cos A}$

$\displaystyle = \frac{1 - \tan^2A}{\sec^2A - 2\tan A}$

$\displaystyle = \frac{1 - \tan^2A}{1 + \tan^2A - 2\tan A}$

$\displaystyle = \frac{(1 - \tan A)(1+\tan A)}{(1-\tan A)^2}$

$\displaystyle = \frac{1 + \tan A}{1 - \tan A}$

Is that any help?

• Jan 30th 2009, 02:02 PM
Soroban
Hello, vungaralas!

. . I took a guess at the expression on the left side.

Quote:

$\displaystyle \cot(4/5)-A \:= \:\frac{\cos2A}{1-\sin2A}$

If that statement is supposed to be an identity,
. . I would guess that it should look like this:

. . $\displaystyle \cot\left(\frac{\pi}{4} - A\right) \:=\:\frac{\cos2A}{1-\sin2A}$

We have: .$\displaystyle \cot(P - Q) \:=\:\frac{1+\tan P\tan Q}{\tan P - \tan Q}$

So the left side is: .$\displaystyle \cot\left(\frac{\pi}{4} - A\right) \;=\;\frac{1 + \tan\frac{\pi}{4}\tan A}{\tan\frac{\pi}{4} - \tan A} \;=\;\frac{1 + \tan A}{1 - \tan A} \;=\;\frac{1 + \frac{\sin A}{\cos A}}{1 - \frac{\sin A}{\cos A}}$

Multiply by $\displaystyle \frac{\cos A}{\cos A}\!:\;\;\frac{\cos A + \sin A}{\cos A - \sin A}$

Multiply by $\displaystyle \frac{\cos A - \sin A}{\cos A-\sin A}\!:\;\;\frac{\cos A + \sin A}{\cos A - \sin A}\cdot\frac{\cos A - \sin A}{\cos A - \sin A}$

. . $\displaystyle = \;\frac{\overbrace{\cos^2\!A - \sin^2\!A}^{\text{This is }\cos2A}}{\underbrace{\cos^2\!A + \sin^2\!A}_{\text{This is 1}} - \underbrace{2\sin A\cos A}_{\text{This is }\sin2A}} \;=\;\frac{\cos2A}{1-\sin2A}$

• Jan 30th 2009, 04:12 PM
vungaralas
thank u so much