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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: kaanonball on March 07, 2009, 10:44:13 am

Title: quick question
Post by: kaanonball on March 07, 2009, 10:44:13 am: how do i find the domain and range of cos(sin^-1 x)
Title: Re: quick question
Post by: TrueTears on March 07, 2009, 11:26:49 am: $cos(sin^{-1}x)$

the way i do these kind of questions is to split them into a composite function

so let $f(x) = cosx$

and $g(x) = sin^{-1}x$

therefore $f o g (x) = cos(sin^{-1}x)$

but we know for f o g to exist ran g must be a subset of dom f

so at the moment $ran g = [\frac{-\pi}{2},\frac{\pi}{2}]$

and dom f = R

so yes ran g is a subset of dom f.

and we also know that dom f o g = dom g , hence the domain of the composite function is just the domain of $sin^{-1}x$ which is [-1,1]
Title: Re: quick question
Post by: kaanonball on March 07, 2009, 12:03:47 pm: ty trutears, but what if the domain of Cos(x) is restricted to [0,pi] and its range restricted to [-1,1]

how can you go about finding the domain and range of the composite function of cos(sin^-1 x) then?
Title: Re: quick question
Post by: TrueTears on March 07, 2009, 12:09:18 pm: ok so we have the same set up as before...

then we still find ran g which is still $[\frac{-\pi}{2},\frac{-\pi}{2}]$

but now dom f is $[0,\pi]$

which means now we have to RESTRICT ran g to be a subset of dom f. So ran g must be at least restricted to $[0,\frac{\pi}{2}]$

now it is a subset of dom f, however to restrict the ran of g we must restrict the domain of g.

so domain of g must be restricted to [0,1]. (note you can also think of it as $0 \le sin^{-1}x \le \frac{\pi}{2}$ and just solve for x)

so now we know dom f o g = dom g which is [0,1]

to work out the range is easy, just sub in [0,1] into the composite function and those are your endpoints for the range.

so f o g(0) = 1 and f o g(1) = 0

so range is [0,1]
Title: Re: quick question
Post by: kaanonball on March 07, 2009, 12:42:43 pm: ty =]
Title: Re: quick question
Post by: 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]
Title: Re: quick question
Post by: kaanonball on March 07, 2009, 10:35:58 pm: thanks Mao, that makes it heaps easier =]
Title: Re: quick question
Post by: Flaming_Arrow on March 11, 2009, 08:08:44 am: 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]

what if its something like $\tan^-1(\cos x)$ , how would you approach that problem using that method?
Title: Re: quick question
Post by: TrueTears on March 11, 2009, 01:08:08 pm: $\cos{x} = \frac{adj}{hyp} = \frac{x}{1}$

So the adjacent side is of length x and the hypotenuse is 1.

The other side is $\sqrt{1-x^2}$

So $\tan{x} = \frac{\sqrt{1-x^2}}{x}$
Title: Re: quick question
Post by: Flaming_Arrow on March 11, 2009, 03:43:19 pm: Quote from: TrueTears on March 11, 2009, 01:08:08 pm
$\cos{x} = \frac{adj}{hyp} = \frac{x}{1}$

So the adjacent side is of length x and the hypotenuse is 1.

The other side is $\sqrt{1-x^2}$

So $\tan{x} = \frac{\sqrt{1-x^2}}{x}$

yes but that doesn't give the correct domain/range
Title: Re: quick question
Post by: Mao on March 12, 2009, 12:13:21 am: Quote from: Flaming_Arrow on March 11, 2009, 03:43:19 pm
Quote from: TrueTears on March 11, 2009, 01:08:08 pm
$\cos{x} = \frac{adj}{hyp} = \frac{x}{1}$

So the adjacent side is of length x and the hypotenuse is 1.

The other side is $\sqrt{1-x^2}$

So $\tan{x} = \frac{\sqrt{1-x^2}}{x}$

yes but that doesn't give the correct domain/range

because TT found tan(cos^-1x) as opposed to what you asked.

I can't think of how to algebraically find $\tan^{-1}(\cos (x))$ right now, might have a crack at it tomorrow

Suspicion confirmed, there is no simple expression for arctan(sin(x)) or arctan(cos(x)). (fourier expansion yields infinite number of trignometric terms)