Inverse Trig Graphs: Domain and Range

[Jet1045]

Hey everyone. Sorry to post another topic, but i thought this would be easy to find on google, to help me with , but i can't find it.

All I am really lookin for is someone to let me know if my answers are right, and if not, then how i can fix them :)

Anyways, i was asked to graph y=sin(arsinc)
So i graphed the line y=x with the doman as -1 to 1 and the range as -pi/2 to pi/2

Then i neded to graph y=arsin(sinx)
So i graphed the line y=x with the domain as -pi/2 to pi/2 and the range as -1 to 1

This is just wat i came up with based on my profs notes, but I am not sure if they are right, and are just kinda guesses i came up with on what he wrote. I just tried a few things and thought this made sense. haha

if someone could let me know if this is right, i'd greatly appreciate it !

 

[Char. Limit]

I think you might need to switch them. I know that sin(f(x)) doesn't have a range from -pi/2 to pi/2 no matter WHAT f(x) is. But I think your domain and range in 1 is correct for 2, and your domain and range in 2 is correct for 1.

 

[Simon Bridge]

Hang on, you have to graph:y=\sin ( \text{sinc}^{-1}x ) \qquad : \qquad \text{sinc}(x)=\frac{\sin(x)}{x}... have I got that right?

https://www.physicsforums.com/showthread.php?t=282941
It is not clear what your strategy entails besides plotting the line y=x.

restricting your domain etc, wouldn't you want to plot y=sinc(x) then reflect it in y=x to get the inverse function - then take the sine of that? You are doing this numerically I hope?

 

[Char. Limit]

Hang on, you have to graph:y=\sin ( \text{sinc}^{-1}x ) \qquad : \qquad \text{sinc}(x)=\frac{\sin(x)}{x}

... have I got that right?<br /> <br /> <a href="https://www.physicsforums.com/showthread.php?t=282941" class="link link--internal">https://www.physicsforums.com/showthread.php?t=282941</a><br /> It is not clear what your strategy entails besides plotting the line y=x.<br /> <br /> restricting your domain as you did, wouldn&#039;t you want to plot y=sinc(x) then reflect it in y=x to get the inverse function - then take the sine of that? You are doing this numerically I hope?[/QUOTE]<br /> <br /> I think he meant arsinx, or arcsin(x). But I&#039;m not entirely certain myself.

 

[Jet1045]

oh holy eff, i clearly can't spell. sorry about that!
yeah i was asked to graph

sin(arcsinx)
and
arcsin(sinx)

I just have no ideas what to label the x and y-axis for each with :(

 

[SammyS]

Hey everyone. Sorry to post another topic, but i thought this would be easy to find on google, to help me with , but i can't find it.

All I'm really lookin for is someone to let me know if my answers are right, and if not, then how i can fix them :)

Anyways, i was asked to graph y=sin(arsinc)
So i graphed the line y=x with the domain as -1 to 1 and the range as -pi/2 to pi/2

Then i neded to graph y=arsin(sinx)
So i graphed the line y=x with the domain as -pi/2 to pi/2 and the range as -1 to 1

This is just what i came up with based on my profs notes, but I am not sure if they are right, and are just kinda guesses i came up with on what he wrote. I just tried a few things and thought this made sense. haha

if someone could let me know if this is right, I'd greatly appreciate it !

I think you have a typo in your first function.

Should it be y = sin(arcsin(x)) ?

The domain & range you give for this are actually only for the arcsine function, although this is also the correct domain for this composite function. As for the range: What is the result of putting all possible values from -π/2 to π/2 into the sine function?

As for the second function, arcsin(sin(x)) :

What is the domain of sin(x) ?

Are any of the values from the range of sin(x) which are not in the domain of arcsin(x)?​

Answering those questions should get you started.

 

[Simon Bridge]

Oh gotcha... that's easier.

 

[Jet1045]

I think you have a typo in your first function.

Should it be y = sin(arcsin(x)) ?

The domain & range you give for this are actually only for the arcsine function, although this is also the correct domain for this composite function. As for the range: What is the result of putting all possible values from -π/2 to π/2 into the sine function?

As for the second function, arcsin(sin(x)) :

What is the domain of sin(x) ?

Are any of the values from the range of sin(x) which are not in the domain of arcsin(x)?​

Answering those questions should get you started.

Thanks Sammy! So for the first function y=sin(arcsin(x)) by putting any value from -pi/2 to pi/2 into sinx, you get values from -1 to 1. So does this mean that the range of this function is also -1 to 1, just like the domain?

For the second one I am a little omfused. So the domain of sinx when you restrict it so that you can graph just arcsinx, is from -pi/2 to pi/2 correct?
And for your second tip, you mean the range of sin being -1 to 1, are any of those values not in the domain of arcsin? So if the domain of just arcsin is -1 to 1, then wouldn't all of those values be in it??

AGGHHH i hattteee trig... hahah and I've never understood how to work with graphs of composite functions :/

 

[SammyS]

Thanks Sammy! So for the first function y=sin(arcsin(x)) by putting any value from -pi/2 to pi/2 into sinx, you get values from -1 to 1. So does this mean that the range of this function is also -1 to 1, just like the domain?

For the second one I am a little omfused. So the domain of sinx when you restrict it so that you can graph just arcsinx, is from -pi/2 to pi/2 correct?
And for your second tip, you mean the range of sin being -1 to 1, are any of those values not in the domain of arcsin? So if the domain of just arcsin is -1 to 1, then wouldn't all of those values be in it??

AGGHHH i hattteee trig... hahah and I've never understood how to work with graphs of composite functions :/

But even if you don't restrict the domain of sin(x), (and in this case there is no reason to restrict it) the range of sin(x) is [-1, 1].

Think about x going from -π/2 to π/2. Then sin(x) goes from -1 to 1. So arcsin(sin(x)) goes from -π/2 to π/2. Correct?

Now the strange part begins...

Let x go from π/2 to 3π/2. What does sin(x) do? It goes from 1 to -1. But, all that the arcsin function "knows" is that it's being fed values from 1 to -1. It doesn't know where these values came from to start with ... So arcsin(sin(x)) goes from  ?  to  ?  .

 

[Jet1045]

But even if you don't restrict the domain of sin(x), (and in this case there is no reason to restrict it) the range of sin(x) is [-1, 1].

Think about x going from -π/2 to π/2. Then sin(x) goes from -1 to 1. So arcsin(sin(x)) goes from -π/2 to π/2. Correct?

Now the strange part begins...

Let x go from π/2 to 3π/2. What does sin(x) do? It goes from 1 to -1. But, all that the arcsin function "knows" is that it's being fed values from 1 to -1. It doesn't know where these values came from to start with ... So arcsin(sin(x)) goes from  ?  to  ?  .

Ugh i don't know why I am finding this so hard.. :/

so for y=sin(arcsinx) the domain and range are both -1 to -1.
is there a general rule to work with when dealing with composite trigonometric functions and their Domains and Ranges?

K so for y=arcsin(sinx) where you gave me the fill in the blanks. arcsin-1 = -pi/2 and arcsin 1 = pi/2 so therefore the range of this function would go from -pi/2 to pi/2? and the domain would then be -1 to 1? Sorry if i completely misread what you were trying to explain, I am just having the hardest time grasping this :/

 

[Jet1045]

oh wait wait.
i think i got it. after reading your posts again>
so the D and R of sin(arcsinx) are both -1 to 1
and the D and R of arcsin(sinx) are both -pi/2 to pi/2
is this correct?

 

[SammyS]

oh wait wait.
i think i got it. after reading your posts again>
so the D and R of sin(arcsinx) are both -1 to 1

Yes to this

and the D and R of arcsin(sinx) are both -pi/2 to pi/2
is this correct?

No to the domain of arcsin(sin(x)), but yes to the range.

What is sin(11π/4), for instance? This is defined, correct?

It's (√2)/2.

What is arcsin((√2)/2) ? It's π/4 .

sin(x) is defined for all real numbers.