Homework Help: Two Inverse Trig Graphs

1. Jan 31, 2012

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 cant find it.

All im 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 im 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 !

2. Feb 1, 2012

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.

3. Feb 1, 2012

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?

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?

4. Feb 1, 2012

Char. Limit

... have I got that right?

It is not clear what your strategy entails besides plotting the line y=x.

restricting your domain as you did, 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?[/QUOTE]

I think he meant arsinx, or arcsin(x). But I'm not entirely certain myself.

5. Feb 1, 2012

Jet1045

oh holy eff, i clearly cant 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 :(

6. Feb 1, 2012

SammyS

Staff Emeritus
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.

7. Feb 1, 2012

Simon Bridge

Oh gotcha... that's easier.

8. Feb 1, 2012

Jet1045

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 im 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 wouldnt 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 :/

Last edited: Feb 1, 2012
9. Feb 1, 2012

SammyS

Staff Emeritus
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  ?  .

10. Feb 1, 2012

Jet1045

Ugh i dont know why im 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, im just having the hardest time grasping this :/

11. Feb 1, 2012

Jet1045

oh wait wait.
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?

12. Feb 1, 2012

SammyS

Staff Emeritus
Yes to this
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.