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Math Help - Inverse Trigonometry Problem - can't find correct solution and test coming up

  1. #1
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    Inverse Trigonometry Problem - can't find correct solution and test coming up

    Find the inverse function f-1 of function f. Find the range of f & the domain of f-1:

    f(x) = -2cos(3x); 0 ≤ x π/3

    I was able to easily find the inverse function which is (1/3)cos
    -1(-x/2)=y and confirmed correct by the answers in the back of my textbook. However, when it comes to finding the range of f, I am very confused. I went about finding it like so:

    f(x) = -2cos(3x)
    y = -2cos(3x) replacing f(x) with y and proceeding to solve for cos x
    (-y/2) = cos(3x) divided by -2
    (1/3)(-y/2) = cos x multiplied by 1/3 and now i have cos x isolated.

    My text gives me an example of solving a similar inequality but using sine, and when finding the range for f, they use this definition: -1 ≤ sin x ≤ 1 so I figure I should do the same thing here but using cosine (my textbook offers no examples with cosine, I hate this book)

    So because I calculated (1/3)(-y/2) = cos x, I plug in (1/3)(-y/2) for cos x in the inequality
    -1 ≤ cos x ≤ 1 and begin solving:
    -1 ≤ (1/3)(-y/2) ≤ 1
    -1 ≤ (-y/6) ≤ 1 and this is where I just stopped and was very confused. I know I'm doing something horribly wrong but I cannot figure it out. The book tells me that the range for f (and thusly the domain of f-1 is [-2,2]. I'm hoping that I'm making a simple problem solving error, but I seriously need to figure out what I'm doing wrong without wasting hours when the issue is probably apparent to someone who is better at math. Please help me. Thank you so so much.




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  2. #2
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    Re: Inverse Trigonometry Problem - can't find correct solution and test coming up

    Quote Originally Posted by sadmaths View Post
    Find the inverse function f-1 of function f. Find the range of f & the domain of f-1:

    f(x) = -2cos(3x); 0 ≤ x [FONT=arial]π/3
    $0 \leqslant x \leqslant \frac{\pi }{3} \Rightarrow \;0 \leqslant 3x \leqslant \pi \Rightarrow \;0 \leqslant \cos (3x) \leqslant 1 \Rightarrow \; - 2 \leqslant -2\cos (3x) \leqslant 0$

    Look at this graph. It is clear that the inverse exists.
    Last edited by Plato; March 22nd 2014 at 03:16 PM.
    Thanks from mash
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  3. #3
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    Re: Inverse Trigonometry Problem - can't find correct solution and test coming up

    I'm sorry but this is highly confusing. It'd be more helpful to know where I went wrong during the solving process. I successfully converted it to the inverse form, but I can't figure out HOW to obtain the range of f (domain of f-1) Basically I think I'm having trouble solving the inequality itself, or simply did not successfully set it up. Again, the inequality I'm using to attempt to find the range of f is: -1 ≤ (1/3)(-y/2) ≤ 1 but if that's not the setup I should have, it would be helpful to know which setup I should use and why.
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    Re: Inverse Trigonometry Problem - can't find correct solution and test coming up

    Quote Originally Posted by sadmaths View Post
    I'm sorry but this is highly confusing. It'd be more helpful to know where I went wrong during the solving process. I successfully converted it to the inverse form, but I can't figure out HOW to obtain the range of f (domain of f-1) Basically I think I'm having trouble solving the inequality itself, or simply did not successfully set it up. Again, the inequality I'm using to attempt to find the range of f is: -1 ≤ (1/3)(-y/2) ≤ 1 but if that's not the setup I should have, it would be helpful to know which setup I should use and why.
    Did you bother to even look at the graph I linked to?
    That is exactly what that graph provides: $[-2,2]$.

    What is your problem?
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  5. #5
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    Re: Inverse Trigonometry Problem - can't find correct solution and test coming up

    Hi,
    Your computation is fine up to this point:

    (-y/2) = cos(3x) divided by -2
    (1/3)(-y/2) = cos x multiplied by 1/3 and now i have cos x isolated.

    Apparently you think $\cos(3x)=3\cos(x)$ not true!

    Next, a typo on Plato's part may have confused you:

    $0\leq x\leq\pi/3$ implies $0\leq3x\leq\pi$

    So $\cos(0)\geq\cos(3x)\geq\cos(\pi)$, cos is decreasing on $[0,\pi]$ -- look at the graph of cos.

    Or $1\geq\cos(3x)\geq-1$ implies $-2\leq-2\cos(3x)\leq2$ is the range of the function.
    Thanks from mash
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    Re: Inverse Trigonometry Problem - can't find correct solution and test coming up

    THANK YOU that's very helpful. I really appreciate it!
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  7. #7
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    Re: Inverse Trigonometry Problem - can't find correct solution and test coming up

    @ Plato. It doesn't help me to know that from the graph, because I have to know hot to solve it without a graph. I wasn't trying to be mean, but your reply was a bit rude. If you won't accept me looking at a graph and then stating that it doesn't help my specific situation, just don't answer my posts from now on. It's no big deal that I couldn't gain anything from it, I'm really bad at math, don't take it personally please.
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