simplifying inverse trig functions?

[twohaha]

simplify

I tried changing tan to sin/cos, then I tried using double angle and half angle formulas. In the end, I still couldn't remove the 2arccos(x/5) in the inside function...

Does anyone have any ideas on how to proceed?
Thanks!

 

[twohaha]

This is the expression:

tan(2arccos(x/5))

 

[Deleted member 4993]

This is the expression:

tan(2arccos(x/5))

Assume:

2arccos(x/5) = Θ

simplify...

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[twohaha]

Ok so...

2tan(arccos(x/5))/(1-(tan(arccos(x/5)))^2)

Could I replace arccos(x/5) with arctan(sqrt(25-x^2)/x)? Should I take the positive square root or the negative one?
Thanks!

 

[Deleted member 4993]

Ok so...

2tan(arccos(x/5))/(1-(tan(arccos(x/5)))^2)

Could I replace arccos(x/5) with arctan(sqrt(25-x^2)/x)? Should I take the positive square root or the negative one?
Thanks!

Cannot quite figure out what the heck did you do - but -

2arccos(x/5) = Θ

cos(Θ/2) = x/5

cos(Θ) = 2(x/5)² - 1

sin(Θ) = ?

tan(Θ) = ?