Inv Trig (1 Viewer)

Pethmin

Well-Known Member
How would I find the domain for this?

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Pethmin

Well-Known Member
I also got an issue w/

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Drongoski

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

Pethmin

Well-Known Member
yes that's the answer, how'd you do that?

Drongoski

Well-Known Member
The other two; are these answers correct?:

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

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Drongoski

Well-Known Member
yes that's the answer, how'd you do that?
I'll explain it later. Please bear with me.

Drongoski

Well-Known Member
For last one:

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

QED

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New Member
For last one:

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

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

Well-Known Member
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

Well-Known Member
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

Well-Known Member
random question - how do you type in maths notation? Is there a website or software to switch from maths and natural input?
$\bg_white \text{LaTeX}$

Anaya R

Well-Known Member
$\bg_white \text{LaTeX}$
Actually that guide is a bit outdated.
I'm currently learning how to do LaTeX, so I'd recommend their website:

Pethmin

Well-Known Member
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

Active Member
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

Well-Known Member
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

Well-Known Member
really got the du qns - does the homework hinting stuff or whatever suck that much
they vague asf

Pethmin

Well-Known Member
really got the du qns - does the homework hinting stuff or whatever suck that much
[/QUOTE
really got the du qns - does the homework hinting stuff or whatever suck that much
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

damn

Deem_Skills

Member
The other two; are these answers correct?:

$\bg_white 1) y = tan^{-1}\sqrt{x^2 - 1} \\ \\ \text { Domain: } x \leq -1 or x \geq 1 \text { and Range: } 0 \leq y < \frac {\pi}{2} \\ \\ \\ 2) y = cos^{-1}(e^x) \\ \\ \text{ Domain: } -\infty < x \leq 0 \text { and Range: } 0 \leq y < \frac {\pi}{2}$
I'm late but how did u find the domain and range for the second question? And the range for the first

Pethmin

Well-Known Member
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