November 26th, 2014, 09:21 PM  #1 
Newbie Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0  Analytic Trig problem
Hi, need some help from the math gods: find the inverse function f^1 of each function f. Find the range of f and domain and range of f^1. f(x) = 2cos(3x); 0≤x≤π/3 far as steps goes, I made it this far on my own y = 2cos(3x) x = 2cos(3y) x/2 = 2cos(3y)/2 (x/2)^3 = (cos(3y))^3 then I dragged the 3 out of make 1/3 (or square root it) 1/3(x/2) = cosy cosy = 1/3(x/2) y = 1/3cos(x/2) Is this right? If not, can you have a step by step process where I can understand easily? Thank you. Last edited by skipjack; November 27th, 2014 at 01:55 AM. 
November 26th, 2014, 09:51 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 
Your result is correct but your working is confused. $\displaystyle y=2\cos(3x)$ $\displaystyle \frac{y}{2}=\cos(3x)$ $\displaystyle \cos^{1}\left(\frac{y}{2}\right)=3x$ $\displaystyle x=\frac{\cos^{1}\left(\frac{y}{2}\right)}{3}$ $\displaystyle y^{1}=\frac{\cos^{1}\left(\frac{x}{2}\right)}{3}$ 
November 27th, 2014, 02:02 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,978 Thanks: 2229  
November 27th, 2014, 12:57 PM  #4 
Newbie Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0  
November 27th, 2014, 01:07 PM  #5  
Newbie Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0  Quote:
 
November 27th, 2014, 01:13 PM  #6 
Newbie Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0  
November 27th, 2014, 01:21 PM  #7  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond  Quote:
$\displaystyle \text{The inverse of }2\cos(3x)\text{ is }\frac{\cos^{1}\left(\frac{x}{2}\right)}{3}$ In short, they are the same thing. Last edited by greg1313; November 27th, 2014 at 01:26 PM.  

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