Simplifying the Triple Angle of Tangent Function

[pavadrin]

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:

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

thank you,
Pavadrin

 

[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

 

[pavadrin]

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

 

[pavadrin]

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?

 

[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]<br /> <br /> 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.[/tex]

 

[pavadrin]

okay thanks for the reliy and the correction, ill try to be more careful next time