# Triple angle of tangent

1. Sep 8, 2006

hey
sorry to distrub you, but i was surfing the net for the triple angle of tangent trig function but could not find it so i decided to use infomation i knew to solve it. i would like to know if what i have done is correct, and if it can be simplified further, thanks. What i have done is as follows:

$$\ \begin{array}{c} \tan 3A = \tan \left( {2A + A} \right) \\ = \frac{{\tan 2A + \tan A}}{{1 - \tan 2A\tan A}} \\ = \frac{{\left( {\frac{{2\tan A}}{{1 - \tan ^2 A}}} \right) + \tan A}}{{1 - \left( {\frac{{2\tan A}}{{1 - \tan ^2 A}}} \right) \cdot \tan A}} \\ = \frac{{\frac{{2\tan A + \left( {\tan A\left( {1 - \tan ^2 A} \right)} \right)}}{{1 - \tan ^2 A}}}}{{\frac{{\left( {1 - \tan ^2 A} \right) - 2\tan ^2 A}}{{1 - \tan ^2 A}}}} \\ = \frac{{2\tan A + \left( {\tan A\left( {1 - \tan ^2 A} \right)} \right)}}{{\left( {1 - \tan ^2 A} \right) - 2\tan ^2 A}} \\ = \frac{{2\tan A + \tan A - \tan ^3 A}}{{1 - \tan A - 2\tan ^2 A}} \\ = \frac{{3\tan A - \tan ^3 A}}{{1 - \tan A - 2\tan ^2 A}} \\ \end{array} \$$

thank you,

2. Sep 9, 2006

### vsage

I randomly chose A=23 and your formula didn't produce the same value as tan(69). On your 4th to 5th step you dropped the ^2 term on one of the tangents in the denominator

3. Sep 9, 2006

okay thanks for checking. ill try top fix the problem and re-post

4. Sep 9, 2006

okay i finally got some time to solve it again. is the triple angle of tangent equal to this:

$$\frac{{\frac{{2\tan A}}{{1 - \tan ^2 A}} + \tan A}}{{1 - \frac{{2\tan ^2 A}}{{1 - \tan ^2 A}}}}$$

and is that as simple as what i can get it?

5. Sep 9, 2006

### HallsofIvy

Multiply both numerator and denominator by 1- tan2A to get
$$\frac{2tanA+ tanA(1- tan^2 A)}{1- tan^2 A- 2tan^2A}$$
That's essentially what you have in the fifth line of your original calculation. Then
[tex]tan 3A= \frac{3tan A- tan^3 A}{1- 3tan^2 A}[/itex]

As vsage said (although it was between your fifth and sixth lines by my count, not fourth and fifth) one tan2 A accidently became tan A.

6. Sep 9, 2006