Inv Trig (1 Viewer)

Average Boreduser · Nov 21, 2022

How would I find the domain for this?

Average Boreduser · Nov 21, 2022

I also got an issue w/

Drongoski · Nov 21, 2022

Is last one: $sin^{-1} \frac {\sqrt {15} + \sqrt 8}{12}$ ??

Average Boreduser · Nov 21, 2022

yes that's the answer, how'd you do that?

Drongoski · Nov 21, 2022

The other two; are these answers correct?:

$$1) $ y = tan^{-1}\sqrt{x^2 - 1} \\ \\ \text { Domain: } x \leq -1 $ or $ x \geq 1 \text { and Range: } 0 \leq y < \frac {\pi}{2} \\ \\ \\ $2) $ y = cos^{-1}(e^x) \\ \\ \text{ Domain: } -\infty < x \leq 0 \text { and Range: } 0 \leq y < \frac {\pi}{2}$

Drongoski · Nov 21, 2022

Pethmin said:
yes that's the answer, how'd you do that?

I'll explain it later. Please bear with me.

Drongoski · Nov 21, 2022

For last one:

$\text {Let: }sin^{-1} \frac {1}{3} = A \text { and } sin^{-1} \frac {1}{4} = B\\ \\ \text {Now draw 2 right-angled triangle, one with angle A, sides 1, 3 and } \sqrt 8 \text { and the other with angle B and sides 1, 4 and } \sqrt {15}\\ \\ \text {From these triangles, you can read off cos A and cos B} \\ \\ \text {Now: } sin \left (sin^{-1} \frac {1}{3} + sin^{-1}\frac {1}{4} \right) = sin(A + B)\\ \\ = sinAcosB + cosAsinB = \frac {1}{3} \times \frac {\sqrt{15}}{4} + \frac {\sqrt 8}{3} \times \frac {1}{4} = \frac {\sqrt {15} + \sqrt 8}{12} \text {( = 0.5584..., within the required range between -1 and +1)}\\ \\ \therefore sin^{-1}\frac {1}{3} + sin^{-1}\frac {1}{4} = sin^{-1}\frac {\sqrt {15} + \sqrt 8}{12}$

QED

hornsbystudylad · Nov 22, 2022

Drongoski said:
For last one:

$\text {Let: }sin^{-1} \frac {1}{3} = A \text { and } sin^{-1} \frac {1}{4} = B\\ \\ \text {Now draw 2 right-angled triangle, one with angle A, sides 1, 3 and } \sqrt 8 \text { and the other with angle B and sides 1, 4 and } \sqrt {15}\\ \\ \text {From these triangles, you can read off cos A and cos B} \\ \\ \text {Now: } sin \left (sin^{-1} \frac {1}{3} + sin^{-1}\frac {1}{4} \right) = sin(A + B)\\ \\ = sinAcosB + cosAsinB = \frac {1}{3} \times \frac {\sqrt{15}}{4} + \frac {\sqrt 8}{3} \times \frac {1}{4} = \frac {\sqrt {15} + \sqrt 8}{12} \text {( = 0.5584..., within the required range between -1 and +1)}\\ \\ \therefore sin^{-1}\frac {1}{3} + sin^{-1}\frac {1}{4} = sin^{-1}\frac {\sqrt {15} + \sqrt 8}{12}$

QED

random question - how do you type in maths notation? Is there a website or software to switch from maths and natural input?

Etho_x · Nov 22, 2022

hornsbystudylad said:
random question - how do you type in maths notation? Is there a website or software to switch from maths and natural input?

It's Latex. Personally I don't know how to use it yet but there is an option on the toolbar in the text box, denoted by "f(x)".

Etho_x · Nov 22, 2022

Pethmin said:
How would I find the domain for this?

Recall that the domain of arctan x is all real x. However, for this question you have a restriction, that is the domain of sqrt(x^2 - 1).

For the function y = sqrt(x^2 - 1), the domain is that x >= 1, or x <= -1. Since the domain of arctan x is all real x, the domain of sqrt(x^2 - 1) is the domain of arctan(sqrt(x^2 - 1)), that is x >= 1, or x <= -1.

carrotsss · Nov 22, 2022

hornsbystudylad said:
random question - how do you type in maths notation? Is there a website or software to switch from maths and natural input?

$\text{LaTeX}$

A Short Guide to LaTeX

Hi everyone, With the use of LaTeX becoming more widespread, I've decided to write a short guide to LaTeX. If you have any problems, post in this thread or message me somehow. jetblack2007. The graphical editor supplied to us is excellent, though I think at least a small amount of knowledge of...

boredofstudies.org

Anaya R · Nov 22, 2022

carrotsss said:
$\text{LaTeX}$

A Short Guide to LaTeX

Hi everyone, With the use of LaTeX becoming more widespread, I've decided to write a short guide to LaTeX. If you have any problems, post in this thread or message me somehow. jetblack2007. The graphical editor supplied to us is excellent, though I think at least a small amount of knowledge of...

boredofstudies.org

Actually that guide is a bit outdated.
I'm currently learning how to do LaTeX, so I'd recommend their website:

Learn LaTeX in 30 minutes - Overleaf, Online LaTeX Editor

An online LaTeX editor that’s easy to use. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more.

www.overleaf.com

Average Boreduser · Nov 23, 2022

Etho_x said:
Recall that the domain of arctan x is all real x. However, for this question you have a restriction, that is the domain of sqrt(x^2 - 1).

For the function y = sqrt(x^2 - 1), the domain is that x >= 1, or x <= -1. Since the domain of arctan x is all real x, the domain of sqrt(x^2 - 1) is the domain of arctan(sqrt(x^2 - 1)), that is x >= 1, or x <= -1.

So essentially is the domain of the composite function inside brackets transferred into the whole function?

Etho_x · Nov 23, 2022

Pethmin said:
So essentially is the domain of the composite function inside brackets transferred into the whole function?

Yeah, but that’s only because the parent function (arctan x) has a domain of all real x. If it was a function like arccos x with a domain between -1 and 1 inclusive, I think there would be those restrictions which you’d have to consider.

Average Boreduser · Nov 23, 2022

Aeonium said:
really got the du qns - does the homework hinting stuff or whatever suck that much

they vague asf

Deem_Skills · Nov 24, 2022

Drongoski said:
The other two; are these answers correct?:

$$1) $ y = tan^{-1}\sqrt{x^2 - 1} \\ \\ \text { Domain: } x \leq -1 $ or $ x \geq 1 \text { and Range: } 0 \leq y < \frac {\pi}{2} \\ \\ \\ $2) $ y = cos^{-1}(e^x) \\ \\ \text{ Domain: } -\infty < x \leq 0 \text { and Range: } 0 \leq y < \frac {\pi}{2}$

I'm late but how did u find the domain and range for the second question? And the range for the first

Average Boreduser · Nov 24, 2022

Deem_Skills said:
I'm late but how did u find the domain and range for the second question? And the range for the first

let u=e^x, therefore y=arccosu, find the common restriction of 'u' and find D: and R: using both graphs

Drongoski · Nov 24, 2022

Deem_Skills said:
I'm late but how did u find the domain and range for the second question? And the range for the first

But I still don't know if my answers were correct?

Average Boreduser · Nov 24, 2022

Drongoski said:
But I still don't know if my answers were correct?

Sorry forgot to confirm, yes they (all answers you provided) are correct

Average Boreduser · Nov 24, 2022

Deem_Skills said:
I'm late but how did u find the domain and range for the second question? And the range for the first

Range for 1st: using composite function again (let u=sqrtx^2-1) therefore, y=arctanu- find common restriction for 'u' and use both graphs to using the restriction to determine range and Domain

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