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how can we derive thew formula for half angle for tangent.. pls help..
 one year ago
 one year ago
how can we derive thew formula for half angle for tangent.. pls help..
 one year ago
 one year ago

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

IsTimBest 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

eyust707Best ResponseYou've already chosen the best response.1
\[\tan \theta = {\sin \theta \over \cos \theta}\]
 one year ago

eyust707Best 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

qpHalcy0nBest 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

ash2326Best ResponseYou've already chosen the best response.0
@shun you want to derive \(\tan \frac {\theta}{2}\)??
 one year ago

ash2326Best 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

qpHalcy0nBest 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

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

satellite73Best ResponseYou've already chosen the best response.0
\[\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}\]
 one year ago

shunBest 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

qpHalcy0nBest 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

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

shunBest ResponseYou've already chosen the best response.0
@qp ok ok.. i get it.. tnx,,,
 one year ago
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