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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TonyHem on March 20, 2009, 08:03:29 pm

Title: Inverse circular functions
Post by: TonyHem on March 20, 2009, 08:03:29 pm: Quick question:
Given that the domain of sin x and cos x are restricted to $[\frac{-\pi}{2}.\frac{\pi}{2}]$ and $[0,\pi]$ respectively, define the implied domain and range of each of the following where y is equal to:

1. So say you have something like $cos(sin^{-1} x)$ are you meant to find the domain of the outside sign, the range of the inside value and combine them for the implied domain and sub in those values to get the range?

2. If you have tan involved $(eg: cos(tan^{-1} x) )$ I have $tan 0\le x \le tan\frac{\pi}{2}$ and the left will give me 0 and the right gives me a domain error. If it gives a domain error does it mean that the domain would be [0,infinite)? The back of the book has [0,infinite).

3. If this is wrong can someone give me a brief explanation on working these type of Q's out. I've got a rough idea, I can kind of work out the answer after I get it wrong but I kinda wanna get a better understanding.

Thanks: (PS: latex is so hard to use -_-")
Title: Re: Inverse circular functions
Post by: TrueTears on March 20, 2009, 08:12:14 pm: I've answered a question similar to this: http://vcenotes.com/forum/index.php/topic,11793.0.html

Same principle, if you're not sure, I'll be happy to set out my working
Title: Re: Inverse circular functions
Post by: TrueTears on March 20, 2009, 08:30:34 pm: Actually, I will just show you my working for the second question.

Let $cos(x) = f(x)$ and $tan^{-1}(x) = g(x)$

so $f o g (x) = cos(tan^{-1}(x))$ (composite functions)

so now we know that for f o g to exist $ran g \subset dom f$

$ran g = (\frac{-\pi}{2} , \frac{\pi}{2})$

$dom f = [0,\pi]$ (stated in the question)

therefore ran g is not a subset of dom f, so we have to restrict ran g to at least $[0, \frac{\pi}{2})$

to do that we must restrict the domain of g to $[0, \infty)$

Hence we know that $dom f o g(x) = dom g = [0, \infty)$

EDIT: to work out range just sub in the end points of the domain of $tan^{-1}(x)$ , which then pumps out the end points of the range of $cos(x)$ , so we have the domain of $tan^{-1}(x) = [0, \infty)$ $tan^{-1}(0) = 0 \implies cos(0) = 1$ but we also know that as x approaches $\infty$ , the output of $tan^{-1}(\infty)$ approaches $\frac{\pi}{2}$ , so sub $\frac{\pi}{2}$ into $cos(x)$ and this yields 0, therefore range of $f o g(x) = (0,1]$ (notice open bracket for 0 because it can't equal 0 since the output of $tan^{-1}(\infty)$ APPROACHES $\frac{\pi}{2}$ )

Exactly the same principle for Q 1, and it's much easier :)
Title: Re: Inverse circular functions
Post by: TonyHem on March 20, 2009, 08:39:29 pm: Thanks, I get your explanation. Is the thing I said about the tan resulting in a domain error right? or did I stuff it up?

Quote from: Mao on March 07, 2009, 06:08:29 pm
Another way to think about it:

construct a right-angled triangle, with an angle $\theta$ and sides $H=1,\; O=x,\; \implies A=\sqrt{1-x^2}$

$\implies \cos(\sin^{-1}x) = \cos \theta =\frac{A}{H} = \frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}$
which is the equation for the upper semicircle of the unit circle. Domain [-1,1], range [0,1]

Probably a really stupid question, but how do you tell the domain/range from that equation?
And would this work for all of these equations or is it limited?
EDIT:
Nvm I get it, thanks
Title: Re: Inverse circular functions
Post by: TrueTears on March 20, 2009, 08:40:23 pm: Quote from: TonyHem on March 20, 2009, 08:39:29 pm
Thanks, I get your explanation. Is the thing I said about the tan resulting in a domain error right? or did I stuff it up?

Quote from: Mao on March 07, 2009, 06:08:29 pm
Another way to think about it:

construct a right-angled triangle, with an angle $\theta$ and sides $H=1,\; O=x,\; \implies A=\sqrt{1-x^2}$

$\implies \cos(\sin^{-1}x) = \cos \theta =\frac{A}{H} = \frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}$
which is the equation for the upper semicircle of the unit circle. Domain [-1,1], range [0,1]

Probably a really stupid question, but how do you tell the domain/range from that equation?
And would this work for all of these equations or is it limited?

Just draw a quick sketch of that and work it out :)
Title: Re: Inverse circular functions
Post by: Flaming_Arrow on March 20, 2009, 08:41:46 pm: Quote from: TonyHem on March 20, 2009, 08:39:29 pm
Thanks, I get your explanation. Is the thing I said about the tan resulting in a domain error right? or did I stuff it up?

Quote from: Mao on March 07, 2009, 06:08:29 pm
Another way to think about it:

construct a right-angled triangle, with an angle $\theta$ and sides $H=1,\; O=x,\; \implies A=\sqrt{1-x^2}$

$\implies \cos(\sin^{-1}x) = \cos \theta =\frac{A}{H} = \frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}$
which is the equation for the upper semicircle of the unit circle. Domain [-1,1], range [0,1]

Probably a really stupid question, but how do you tell the domain/range from that equation?
And would this work for all of these equations or is it limited?

i dont think it works for ones ranges/domain with $\pi$