Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
shun
Group Title
how can we derive thew formula for half angle for tangent.. pls help..
 one year ago
 one year ago
shun Group Title
how can we derive thew formula for half angle for tangent.. pls help..
 one year ago
 one year ago

This Question is Open

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Is this Advanced Functions?
 one year ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
The derivaive of TAN is cosx/sinx I think? Or is that just the reciprocal?
 one year ago

eyust707 Group TitleBest ResponseYou've already chosen the best response.1
\[\tan \theta = {\sin \theta \over \cos \theta}\]
 one year ago

eyust707 Group TitleBest ResponseYou've already chosen the best response.1
The derivative of tangent is: \[\frac{ d }{ d \theta } \tan \theta = \sec^2\theta \]
 one year ago

qpHalcy0n Group TitleBest ResponseYou've already chosen the best response.0
I think he's looking for a proof of the tangent half angle formula, not an actual derivative.
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
@shun you want to derive \(\tan \frac {\theta}{2}\)??
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
We know that \[\tan (A+B)=\frac{\tan A+\tan B}{1\tan A\times \tan B}\] Put \[A=B=\frac {\theta}{2}\] \(\large \text{you'd get a relation of}\ \tan \theta \ \text{in terms of} \tan \frac{\theta}2\)
 one year ago

qpHalcy0n Group TitleBest ResponseYou've already chosen the best response.0
Here, to prove it, use the half angle formulae for sine and cosine. \[\sin(x/2) = \pm \sqrt{ (1cosx / 2}\] \[\cos(x/2) = \pm \sqrt{ (1+cosx / 2) }\]
 one year ago

shun Group TitleBest ResponseYou've already chosen the best response.0
@ash yes,, actually,, it is,
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
you got it?
 one year ago

shun Group TitleBest ResponseYou've already chosen the best response.0
yes i got it,, tnx,,
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}\]
 one year ago

shun Group TitleBest ResponseYou've already chosen the best response.0
@qp tnx.. i get it.. tank you.. but that it is the formulaim asking for the derivation.. ..
 one year ago

qpHalcy0n Group TitleBest ResponseYou've already chosen the best response.0
The derivation is based on those formulae. The half angle tangent is half angle sine / half angle cosine. You can prove half angle sine and cosine as well, if needs be.
 one year ago

shun Group TitleBest ResponseYou've already chosen the best response.0
@sat.. yes that the beginning.. did you know some restrictions?
 one year ago

shun Group TitleBest ResponseYou've already chosen the best response.0
@qp ok ok.. i get it.. tnx,,,
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.