Simplify the expression

[killasnake]

How do I simplify this expression?

\(\displaystyle tan(2cos^-^1(x/3))\)

this is what I came up with but it's wrong

\(\displaystyle tan(2*arccos((1/3)*x))\)

 

[o_O]

\(\displaystyle tan(2\theta) = \frac{sin(2\theta)}{cos(2\theta)} = \frac{2sin\theta cos\theta}{2cos^{2}\theta - 1}\)

Where :

\(\displaystyle \theta = cos^{-1}\left(\frac{x}{3}\right)\)

There may be a better solution to this but I don't see any at the moment ...

 

[Loren]

I would suggest that you draw a sketch. Let theta be the angle whose cosine is x/3. In QI theta has an abscissa of x and a hypotenuse of 3. Use the Pythagorean Th. to determine the measure of the ordinate. I got \(\displaystyle \sqrt{9-x^2}\). Therefore you can determine that \(\displaystyle \tan \theta = \frac{\sqrt{9-x^2}}{x}\). Combine that with fact that \(\displaystyle \tan 2 \theta = \frac{2\tan\theta}{1-\tan^2\theta}\) and go from there.