Showing A Identity Holds

SportfreundeKeaneKent · #1 $Old$ May 22nd, 2008, 09:04 AM

I have to show that the following identity holds when 1 + cos(2Ө) does not equal 0.

tan(Ө) = sin(2Ө)/1 + cos(2Ө)

I think to do this, you'd have to use cos2Ө and sin2Ө's formulas to derive a formula for tan(Ө) or tan(2Ө). It's the (2Ө) thing that throws me off.

angel.white · #2 $Old$ May 22nd, 2008, 09:50 AM

Quote:

Originally Posted by SportfreundeKeaneKent $View Post$

I have to show that the following identity holds when 1 + cos(2Ө) does not equal 0.

tan(Ө) = sin(2Ө)/1 + cos(2Ө)

I think to do this, you'd have to use cos2Ө and sin2Ө's formulas to derive a formula for tan(Ө) or tan(2Ө). It's the (2Ө) thing that throws me off.

$tan(\theta) = \frac{sin(2\theta)}{1+cos(2\theta)}$

$= \frac{2~sin(\theta)~cos(\theta)}{1+cos(2\theta)}$

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

$= \frac{2~sin(\theta)~cos(\theta)}{2cos^2(\theta)}$

$= \frac{sin(\theta)}{cos(\theta)}$

$= tan(\theta)$

SportfreundeKeaneKent · #3 $Old$ May 22nd, 2008, 10:51 AM

Thanks. The third line is tricky.

I wanna get a forumula for tan(2Ө) in terms of tan(Ө) by using the cos(2Ө) = cos^2(Ө) - sin^2(Ө) and sin(2Ө) = 2sinӨcosӨ formulas.

Can I use tan(2Ө) = sin2Ө/cos2Ө and go on from there? And how would I get it back to tanӨ?

Moo · #4 $Old$ May 22nd, 2008, 11:42 AM

Hello,

Quote:

Originally Posted by SportfreundeKeaneKent $View Post$

Thanks. The third line is tricky.

I wanna get a forumula for tan(2Ө) in terms of tan(Ө) by using the cos(2Ө) = cos^2(Ө) - sin^2(Ө) and sin(2Ө) = 2sinӨcosӨ formulas.

Can I use tan(2Ө) = sin2Ө/cos2Ө and go on from there? And how would I get it back to tanӨ?

By dividing if necessary

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

Divide by $\cos^2 \theta$ above and below

SportfreundeKeaneKent · #5 $Old$ May 22nd, 2008, 11:52 AM

I can get

(2sinӨ/cosӨ)/1-tan^2Ө

2tanӨ/1-tan^2Ө = tan(2Ө)

Would that be the formula or could you simplify it more?

Moo · #6 $Old$ May 22nd, 2008, 11:53 AM

Quote:

Originally Posted by SportfreundeKeaneKent $View Post$

I can get

(2sinӨ/cosӨ)/1-tan^2Ө

2tanӨ/1-tan^2Ө = tan(2Ө)

Would that be the formula or could you simplify it more?

It's the formula