Inverse of function - Difficulty solving

Question

$$y=\sqrt{35\tan (\frac{\pi }{180}x)}$$

I have a really hard time finding the inverse of this particular function. Can anyone shine some light on why that may be, or alternatively solve it if I'm just being silly? :-)

Thank you.

Answer 1

$$y^2=35\tan (\frac{\pi x}{180})\Rightarrow \tan(\frac{\pi x}{180})=\frac{y^2}{35}\Rightarrow \tan^{-1}(\frac{y^2}{35} )=\frac{\pi x}{180}\Rightarrow x= \frac{180}{\pi }\tan^{-1}(\frac{y^2}{35} )$$ Thus $f^{-1}(x)=\frac{180}{\pi }\tan^{-1}(\frac{x^2}{35} )$

Answer 2

Hint: Your function is composed by performing the following operations on a give value of $x$:

• multiply by $\frac{\pi}{180}$
• apply the tangent function
• multiply by 35
• apply the square root function

This yields a value $f(x)$. So, starting with a given value of $f(x)$, you need to find operations that will undo each of these operations, in reverse order.

The individual operations listed are simple enough that you should know how to undo each one separately. Then just put them together. The result is the inverse function.

(Note: I have omitted any discussion of domain and range, but you should probably determine those as well.)

Answer 3

Divide by 35 Square it Take the ArcTan Multiply by $\frac{180}{\pi}$

Answer 4

$$y=\sqrt{35\tan\left(\frac{\pi}{180}x\right)}\Longleftrightarrow$$ $$y=\sqrt{35}\sqrt{\tan\left(\frac{\pi x}{180}\right)}\Longleftrightarrow$$ $$\sqrt{\tan\left(\frac{\pi x}{180}\right)}=\frac{y}{\sqrt{35}}\Longleftrightarrow$$ $$\left(\sqrt{\tan\left(\frac{\pi x}{180}\right)}\right)^2=\left(\frac{y}{\sqrt{35}}\right)^2\Longleftrightarrow$$ $$\tan\left(\frac{\pi x}{180}\right)=\frac{y^2}{35}\Longleftrightarrow$$ $$\frac{\pi x}{180}=\arctan\left(\frac{y^2}{35}\right)+\pi n\Longleftrightarrow$$ $$x=\frac{\arctan\left(\frac{y^2}{35}\right)+\pi n}{\frac{\pi}{180}}\Longleftrightarrow$$ $$x=\frac{180\arctan\left(\frac{y^2}{35}\right)}{\pi}+180n$$

With $n\in\mathbb{Z}$