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

• Mar 22nd 2014, 02:12 PM
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.

• Mar 22nd 2014, 03:02 PM
Plato
Re: Inverse Trigonometry Problem - can't find correct solution and test coming up
Quote:

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.
• Mar 22nd 2014, 03:25 PM
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.
• Mar 22nd 2014, 04:06 PM
Plato
Re: Inverse Trigonometry Problem - can't find correct solution and test coming up
Quote:

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: $[-2,2]$.

• Mar 22nd 2014, 06:38 PM
johng
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.
• Mar 26th 2014, 11:05 AM