anonymous
  • anonymous
how can we derive thew formula for half angle for tangent.. pls help..
Mathematics
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anonymous
  • anonymous
how can we derive thew formula for half angle for tangent.. pls help..
Mathematics
chestercat
  • chestercat
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IsTim
  • IsTim
Is this Advanced Functions?
IsTim
  • IsTim
The derivaive of TAN is cosx/sinx I think? Or is that just the reciprocal?
eyust707
  • eyust707
\[\tan \theta = {\sin \theta \over \cos \theta}\]

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eyust707
  • eyust707
The derivative of tangent is: \[\frac{ d }{ d \theta } \tan \theta = \sec^2\theta \]
anonymous
  • anonymous
I think he's looking for a proof of the tangent half angle formula, not an actual derivative.
ash2326
  • ash2326
@shun you want to derive \(\tan \frac {\theta}{2}\)??
ash2326
  • ash2326
We know that \[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\times \tan B}\] Put \[A=B=\frac {\theta}{2}\] \(\large \text{you'd get a relation of}\ \tan \theta \ \text{in terms of} \tan \frac{\theta}2\)
anonymous
  • anonymous
Here, to prove it, use the half angle formulae for sine and cosine. \[\sin(x/2) = \pm \sqrt{ (1-cosx / 2}\] \[\cos(x/2) = \pm \sqrt{ (1+cosx / 2) }\]
anonymous
  • anonymous
@ash yes,, actually,, it is,
ash2326
  • ash2326
you got it?
anonymous
  • anonymous
yes i got it,, tnx,,
anonymous
  • anonymous
\[\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}\]
anonymous
  • anonymous
@qp tnx.. i get it.. tank you.. but that it is the formulaim asking for the derivation.. ..
anonymous
  • anonymous
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.
anonymous
  • anonymous
@sat.. yes that the beginning.. did you know some restrictions?
anonymous
  • anonymous
@qp ok ok.. i get it.. tnx,,,

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