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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: /0 on December 23, 2008, 09:51:12 am

Title: composite trig
Post by: /0 on December 23, 2008, 09:51:12 am: Is there a way to algebraically find compositions such as

$\cos^{-1} ( \tan{x})$ or $\cos(\tan^{-1}x)$ ?

I can't find any identities relating tan to either sin or cos, which would allow it to do something like this:

$\sin^{-1} (\cos{\frac{\pi}{6}})=\sin^{-1}(\sin{(\frac{\pi}{2}-\frac{\pi}{6}})}) = \frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$

Also I don't understand why $\sin^{-1}(\sin \frac{5\pi}{6}) = \frac{\pi}{6}$ and not $\frac{5\pi}{6}$ .

I would have thought that since we can define $f(x) = \sin{x}$ , it would be more sensible to have $f^{-1}(f(\frac{5\pi}{6})) = \frac{5\pi}{6}$

OR even better: $f^{-1}(f(\frac{5\pi}{6})) = n\pi + (-1)^n\frac{\pi}{6}, \ n \in Z$

Thanks

(Oh also one more thing, when we sketch $y = \sin^{-1}x$ and $y = \cos^{-1}x$ and $y = \tan^{-1}x$ is it implied that the domain and range are:
$[-1, 1], [-\frac{\pi}{2}, \frac{\pi}{2}]$ ; $[-1,1], [0, \pi]$ ; and $R, (-\frac{\pi}{2}, \frac{\pi}{2})$ respectively?
Why doesn't $y = cos^{-1}x$ have range $[\frac{-\pi}{2}, \frac{\pi}{2}]$ ? )
Title: Re: composite trig
Post by: Mao on December 23, 2008, 12:30:43 pm: I shall answer questions in no particular order =]

1. why does $\sin^{-1} \left( \sin \frac{5\pi}{6}\right)=\frac{\pi}{6}$

This is because $\sin^{-1}(x)$ has a range of $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (it is defined to that).

2. why are the domains and ranges of inverse circular functions what they are?

the circular functions are periodic, hence they repeat themselves every so often. The problem with this is that it won't be one-to-one if we didn't restrict the domain. Sketch the three circular functions to see for yourself. When you deal with the inverse circular functions, remember that they are ONLY what they are defined to be, nothing more.

3. how to do $\cos (\tan^{-1} x)$ and things similar to this?
construct a triangle, first concerning tan(x) [the opposite is x, adjacent is 1, using pytha to find the hypotenuse to be $\sqrt{x^2 + 1}$
now, we take the inverse tan of x, i.e. we get the angle between adj and hyp, taking the cos of that, we use $\frac{adj}{hyp}=\frac{1}{\sqrt{x^1+1}}$

4. how to do $f_1 ^{-1} (f_2 x)$ ?

firstly, unless both $f_1$ and $f_2$ are tan, $f_2$ cannot be tan. This is because [and you have learnt composite functions] the range of inner must be a subset of the domain of the outer, which tan is not, R does not fit in [-1,1].
and your method dealing with this when they are sin/cos are quite correct.
Title: Re: composite trig
Post by: /0 on December 23, 2008, 01:20:40 pm: Ah, thanks Mao! Now i think i understand :D
Title: Re: composite trig
Post by: Mao on December 23, 2008, 01:30:27 pm: Another more algebraic method for number 3, use the pythag identity $\sin^2(x)+\cos^2(x)=1,\; \tan^2(x)+1=\sec^2(x)$ :

e.g. $\cos (\sin^{-1}(x))=\sqrt{1-(\sin (\sin^{-1}(x)))^2}=\sqrt{1-x^2}$

e.g. $\cos (\tan^{-1}(x))$

$(\tan(\tan^{-1}(x)))^2+1=\frac{1}{(\cos (\tan^{-1}(x)))^2}$

$\implies \frac{1}{x^2+1}=(\cos (\tan^{-1}(x)))^2$

$\implies \cos (\tan^{-1}(x))=\frac{1}{\sqrt{x^2+1}}$ (negative solution discarded since inverse tan's values lies in first and fourth quadrant.

:)
Title: Re: composite trig
Post by: /0 on December 23, 2008, 01:54:20 pm: Cool! Substituting $\theta = \tan^{-1} x$ is quite inspired ;D
Title: Re: composite trig
Post by: /0 on December 23, 2008, 02:20:59 pm: So, leading on from the range of $\sin^{-1}(x)$ being defined as $\left[ \frac{-\pi}{2}, \frac{\pi}{2} \right]$ ,

If you're asked to find the inverse of $f: \left[\frac{\pi}{2}, \frac{3\pi}{2} \right] \rightarrow R, \ f(x) = \sin{x}$ ,

Would $f^{-1}: [-1,1] \rightarrow \left[\frac{\pi}{2}, \frac{3\pi}{2} \right], \ f^{-1}(x) = \sin^{-1}(x)$ technically be incorrect? I guess I am messing with predefined ranges after all...

The answers gave it as:

$f^{-1}: [-1, 1] \rightarrow R,\ f^{-1}(x) = y$ where $\sin{y} = x, \ y \in \left[\frac{\pi}{2}, \frac{3\pi}{2} \right]$

But it seems all drawn-out and complicated - is it all necessary?

Thanks again :P
Title: Re: composite trig
Post by: humph on December 23, 2008, 02:37:42 pm: Quote from: DivideBy0 on December 23, 2008, 02:20:59 pm
So, leading on from the range of $\sin^{-1}(x)$ being defined as $\left[ \frac{-\pi}{2}, \frac{\pi}{2} \right]$ ,

If you're asked to find the inverse of $f: \left[\frac{\pi}{2}, \frac{3\pi}{2} \right] \rightarrow R, \ f(x) = \sin{x}$ ,

Would $f^{-1}: [-1,1] \rightarrow \left[\frac{\pi}{2}, \frac{3\pi}{2} \right], \ f^{-1}(x) = \sin^{-1}(x)$ technically be incorrect? I guess I am messing with predefined ranges after all...

The answers gave it as:

$f^{-1}: [-1, 1] \rightarrow R,\ f^{-1}(x) = y$ where $\sin{y} = x, \ y \in \left[\frac{\pi}{2}, \frac{3\pi}{2} \right]$

But it seems all drawn-out and complicated - is it all necessary?

Thanks again :P
It'd be better to say $f^{-1}(x) = y$ where $\sin{y} = x, \ y \in \left[\frac{\pi}{2}, \frac{3\pi}{2} \right]$ , than say $f^{-1}(x) = \sin^{-1}(x)$ , as $\sin^{-1}(x)$ is already a function defined to have a different range. But other than that it's fine - you can be specific about the range, unlike the answers.