Question

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

Answer 1

Is this Advanced Functions?

Answer 2

The derivaive of TAN is cosx/sinx I think? Or is that just the reciprocal?

Answer 3

\[\tan \theta = {\sin \theta \over \cos \theta}\]

Answer 4

The derivative of tangent is:
\[\frac{ d }{ d \theta } \tan \theta = \sec^2\theta \]

Answer 5

I think he's looking for a proof of the tangent half angle formula, not an actual derivative.

Answer 6

@shun you want to derive \(\tan \frac {\theta}{2}\)??

Answer 7

We know that
\[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\times \tan B}\]
Put
\[A=B=\frac {\theta}{2}\]
\(\large \text{you'd get a relation of}\ \tan \theta \ \text{in terms of} \tan \frac{\theta}2\)

Answer 8

Here, to prove it, use the half angle formulae for sine and cosine.
\[\sin(x/2) = \pm \sqrt{ (1-cosx / 2}\]
\[\cos(x/2) = \pm \sqrt{ (1+cosx / 2) }\]

Answer 9

@ash yes,, actually,, it is,

Answer 10

you got it?

Answer 11

yes i got it,, tnx,,

Answer 12

\[\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}\]

Answer 13

@qp tnx.. i get it.. tank you.. but that it is the formulaim asking for the derivation..
..

Answer 14

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.

Answer 15

@sat.. yes that the beginning.. did you know some restrictions?

Answer 16

@qp ok ok.. i get it.. tnx,,,