## shun Group Title how can we derive thew formula for half angle for tangent.. pls help.. 2 years ago 2 years ago

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1. IsTim Group Title

2. IsTim Group Title

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

3. eyust707 Group Title

$\tan \theta = {\sin \theta \over \cos \theta}$

4. eyust707 Group Title

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

5. qpHalcy0n Group Title

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

6. ash2326 Group Title

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

7. ash2326 Group Title

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

8. qpHalcy0n Group Title

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

9. shun Group Title

@ash yes,, actually,, it is,

10. ash2326 Group Title

you got it?

11. shun Group Title

yes i got it,, tnx,,

12. satellite73 Group Title

$\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}$

13. shun Group Title

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

14. qpHalcy0n Group Title

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.

15. shun Group Title

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

16. shun Group Title

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