t-results and trig identities help!

[annnabannnana]

Help is needed please!!

 

[Harry_the_cat]

Can you write an expression for tan 2A in terms of tan A ?

 

[annnabannnana]

Can you write an expression for tan 2A in terms of tan A ?

i think it has to be in terms of 't', sorry I'm not sure. thanks for commenting!

 

[nasi112]

[MATH]\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{2\sin \theta \cos \theta}{\cos^{2} \theta - \sin^2 \theta} = \frac{2\tan \theta}{1 - \tan^2 \theta}[/MATH]

[MATH]\tan \theta = \frac{\sin \theta}{\cos \theta}[/MATH]

[MATH]\sin \theta = \frac{2\tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}[/MATH]

[MATH]\cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}[/MATH]

What can you do more with these expressions?

 

[Harry_the_cat]

i think it has to be in terms of 't', sorry I'm not sure. thanks for commenting!

Yes it does but I see it as a two-step process.
First write tan 2A in terms of tan A as Nasi112 has shown in post #4.
And then write tan A in terms of tan (A/2) ie in terms of t.

 

[annnabannnana]

[MATH]\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{2\sin \theta \cos \theta}{\cos^{2} \theta - \sin^2 \theta} = \frac{2\tan \theta}{1 - \tan^2 \theta}[/MATH]

[MATH]\tan \theta = \frac{\sin \theta}{\cos \theta}[/MATH]

[MATH]\sin \theta = \frac{2\tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}[/MATH]

[MATH]\cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}[/MATH]

What can you do more with these expressions?

hi would it just end up being 2t/1-t^2

 

[Deleted member 4993]

hi would it just end up being 2t/1-t^2

No!

Please share your work.

 

[annnabannnana]

No!

Please share your work.

 

[nasi112]

hi would it just end up being 2t/1-t^2

Yeah, that is correct for [MATH]\tan \theta[/MATH], but you need to do [MATH]\tan 2\theta[/MATH]

 

[Harry_the_cat]

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

It follows that:

\(\displaystyle tan A = \frac{2 tan(A/2)}{1-tan^2 (A/2)}=\frac{2t}{1-t^2} \)

Sub the second one into the first one and tidy it up.

 

[Deleted member 4993]

tan(2x) = sin(2x) / cos(2x) = 2*tan(x) / [1 - tan^2(x)]

tan(x) = 2 * tan(x/2) / [1 - tan^2(x/2)] = 2*t/(1-t^2)

tan(2x) = [ 4*t/(1- t^2)] / [ 1 - {2*t/(1-t^2)}^2] ...... simplify....

Same as response above ... different formats ......