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