Inv Trig (1 Viewer)

Pethmin · Nov 21, 2022

How would I find the domain for this?

Pethmin · Nov 21, 2022

I also got an issue w/

Drongoski · Nov 21, 2022

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

Pethmin · 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

Pethmin · 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?

Aeonium · Nov 23, 2022

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

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

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.

Pethmin · Nov 23, 2022

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

they vague asf

Pethmin · Nov 23, 2022

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

Aeonium said:

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

Click to expand...

If I'm being honest, no one really uses it and the homework help zooms are packed every day with at least 6 qns each from each student which correlates to hours upon hours of wait times. So usually what happens is I resort to BOS

Aeonium · Nov 23, 2022

damn

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

Pethmin · 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

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