My Math Forum Analytic Trig problem

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 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
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Quote:
 Originally Posted by p41s3r need some help
Do you need help with obtaining the domain and range for the inverse?

November 27th, 2014, 12:57 PM   #4
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Quote:
 Originally Posted by skipjack Do you need help with obtaining the domain and range for the inverse?
No, I know how to do those, but if its a tricky way to get them, I'll repost and I'll ask again, thank you and happy thanksgiving

November 27th, 2014, 01:07 PM   #5
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 Originally Posted by greg1313 Your result is correct but your working is confused. $\displaystyle y=-2\cos(3x)$ $\displaystyle -\frac{y}{2}=\cos(3x)$ $\displaystyle \cos^{-1}\left(-\frac{y}{2}\right)=3x$ $\displaystyle x=\frac{\cos^{-1}\left(-\frac{y}{2}\right)}{3}$ $\displaystyle y^{-1}=\frac{\cos^{-1}\left(-\frac{x}{2}\right)}{3}$
Do you think, you can help me understand why I couldn't figure out the answer? If possible? Thanks for the reply, it really helps! Also, can you explain how your answer is y^-1=cos^-1(-x/2)/3 different from, y=1/3cos^-1(-x/2)? The answer I got, and I checked is from answers from the back, but I don't quite understand because I am not fluent in math!

November 27th, 2014, 01:13 PM   #6
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 Originally Posted by skipjack Do you need help with obtaining the domain and range for the inverse?
You know what, I need clear instructions, because I'm not getting it, I don't have a strong background in math so I think I need help, so thank you so much!

November 27th, 2014, 01:21 PM   #7
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Quote:
 Originally Posted by p41s3r . . . can you explain how your answer is y^-1=cos^-1(-x/2)/3 different from, y=1/3cos^-1(-x/2)?
You can ignore the "-1" and just use y. The "-1" was intended to show that the function is the inverse of the original function. So read it as:

$\displaystyle \text{The inverse of }-2\cos(3x)\text{ is }\frac{\cos^{-1}\left(-\frac{x}{2}\right)}{3}$

In short, they are the same thing.

Last edited by greg1313; November 27th, 2014 at 01:26 PM.

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