inverse trig becos my head hurts (1 Viewer)

Sylfiphy · Mar 9, 2024

im back and I have two questions ..

1) how do I find the domain and range for y=ln(sin^-1x)
2) how do I find domain and range for y=cos^-1(cosx) and how do I graph it

Average Boreduser · Mar 9, 2024

Sylfiphy said:
im back and I have two questions ..

1) how do I find the domain and range for y=ln(sin^-1x)
2) how do I find domain and range for y=cos^-1(cosx) and how do I graph it

1. composite graphs (find common R)
2. Range is same as normal cosinv graph and domain let it lie between -1 and 1

Sylfiphy · Mar 9, 2024

Average Boreduser said:
1. composite graphs (find common R)
2. Range is same as normal cosinv graph and domain let it lie between -1 and 1

I don't get the common range bit tho because one of them has a restricted domain

Average Boreduser · Mar 9, 2024

Sylfiphy said:
I don't get the common range bit tho because one of them has a restricted domain

let u be sin^-1x. y=lnu. find your common R of variable u, then determine d and r

Yeah u have two restrictions on the domains, which is why you balance them by finding the restriciton which works for both domains if that makes sense

Sylfiphy · Mar 9, 2024

Average Boreduser said:
let u be sin^-1x. y=lnu. find your common R of variable u, then determine d and r

Yeah u have two restrictions on the domains, which is why you balance them by finding the restriciton which works for both domains if that makes sense

I get that but for one of the questions the common domain was -1≤u≤1 and then the range was all real values which makes no sense

Average Boreduser · Mar 9, 2024

Sylfiphy said:
I get that but for one of the questions the common domain was -1≤u≤1 and then the range was all real values which makes no sense

yeah because it repeats the period. it resembles a zigzag pattern

Sylfiphy · Mar 9, 2024

Average Boreduser said:
yeah because it repeats the period. it resembles a zigzag pattern

I got that now

thanks bro ur always answering my questions

Luukas.2 · Mar 9, 2024

Sylfiphy said:
im back and I have two questions ..

1) how do I find the domain and range for y=ln(sin^-1x)
2) how do I find domain and range for y=cos^-1(cosx) and how do I graph it

Question 1:
$\text{$y = \sin^{-1}{x}$ takes ratios between $-1 \le x \le 1$ and outputs angles between $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.}$

$\text{However, you are using this output as an input into a log function, which can only accept positive values. Hence, you need $\sin^{-1}{x} > 0$.}$

$\text{You are left with the domain of $y = \ln{\left(\sin^{-1}{x}\right)}$ as $0 < x \le 1$ and the range as $y \le \ln{\frac{\pi}{2}}$.$

$\text{Or, in interval notation, $x \in (0,\ 1]$ and $y \in \left(-\infty,\ \ln{\frac{\pi}{2}}\right]$.$

Question 2:
$\text{$y = \cos{x}$ takes any angle and outputs a ratio between $-1 \le y \le 1$, which will serve as an input for $\cos^{-1}{x}$.}$

$\text{This will, in turn, output an angle between $0$ and $\pi$ radians.}$

$\text{You are left with the domain of $y = \cos^{-1}{\left(\cos{x}\right)}$ as $x \in \mathbb{R}$ and the range as $0 \le y \le \pi$.$

As has been noted, it is a saw-tooth function, which can be recognised by plotting points:

$\begin{align*} \text{When $x = 0$:} \qquad y &= \cos^{-1}{\left(\cos{x}\right)} = \cos^{-1}{\left(\cos{0}\right)} = \cos^{-1}{\left(1\right)} = 0 \\ \text{When $x = \pm\frac{\pi}{2}$:} \qquad y &= \cos^{-1}{\left(\cos{x}\right)} = \cos^{-1}{\left[\cos{\left(\pm\frac{\pi}{2}\right)}\right]} = \cos^{-1}{\left(0\right)} = \frac{\pi}{2} \\ \text{When $x = \pm\pi$:} \qquad y &= \cos^{-1}{\left(\cos{x}\right)} = \cos^{-1}{\left[\cos{\left(\pm\pi\right)}\right]} = \cos^{-1}{\left(-1\right)} = \pi \\ \text{When $x = \pm\frac{3\pi}{2}$:} \qquad y &= \cos^{-1}{\left(\cos{x}\right)} = \cos^{-1}{\left[cos{\left(\pm\frac{3\pi}{2}\right)}\right]} = \cos^{-1}{\left(0\right)} = \frac{\pi}{2} \\ \text{When $x = \pm2\pi$:} \qquad y &= \cos^{-1}{\left(\cos{x}\right)} = \cos^{-1}{\left[\cos{\left(\pm2\pi\right)}\right]} = \cos^{-1}{\left(1\right)} = 0 \end{align*}$

inverse trig becos my head hurts (1 Viewer)

Sylfiphy

Member

Average Boreduser

Rising Renewal

Sylfiphy

Member

Average Boreduser

Rising Renewal

Sylfiphy

Member

Average Boreduser

Rising Renewal

Sylfiphy

Member

Luukas.2

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)