Finding the exact value of cos(arcsin())

[opus]

Homework Statement

Find the exact value of $y=cos\left(arcsin \frac {5} {11} \right)$

Homework Equations

The Attempt at a Solution

I'm stuck at square one. I know that I must evaluate arcsin first, which will give me an angle θ. From that angle, I can take the cosine of it and that much makes sense.
However, in this "guided homework problem" it is stating information that doesn't seem so obvious to me. I've posted a picture.
In the guided problem, it states that θ is in the range of the inverse function. I don't see how this is the case. The range of the inverse function is in the inclusive range of $\frac {-π}{2}$ and $\frac {π}{2}$. In looking at the problem, I can tell that sine is positive, but that doesn't imply that it's in range as it could be in quadrant II.
What should I be looking for?

 

[opus]

Attached image

 

[ehild]

Homework Statement

Find the exact value of $y=cos\left(arcsin \frac {5} {11} \right)$

Homework Equations

The Attempt at a Solution

I'm stuck at square one. I know that I must evaluate arcsin first, which will give me an angle θ. From that angle, I can take the cosine of it and that much makes sense.
However, in this "guided homework problem" it is stating information that doesn't seem so obvious to me. I've posted a picture.
In the guided problem, it states that θ is in the range of the inverse function. I don't see how this is the case. The range of the inverse function is in the inclusive range of $\frac {-π}{2}$ and $\frac {π}{2}$. In looking at the problem, I can tell that sine is positive, but that doesn't imply that it's in range as it could be in quadrant II.
What should I be looking for?

No, quadrant II is π/2<θ<π. . The range of the inverse sine function is -π/2 < θ < π/2 . Here cos(θ) is positive.

 

[Ray Vickson]

Homework Statement

Find the exact value of $y=cos\left(arcsin \frac {5} {11} \right)$

Homework Equations

The Attempt at a Solution

I'm stuck at square one. I know that I must evaluate arcsin first, which will give me an angle θ. From that angle, I can take the cosine of it and that much makes sense.
However, in this "guided homework problem" it is stating information that doesn't seem so obvious to me. I've posted a picture.
In the guided problem, it states that θ is in the range of the inverse function. I don't see how this is the case. The range of the inverse function is in the inclusive range of $\frac {-π}{2}$ and $\frac {π}{2}$. In looking at the problem, I can tell that sine is positive, but that doesn't imply that it's in range as it could be in quadrant II.
What should I be looking for?

Never mind trying to find the angle $\theta$; it won't really help you solve the problem. All you need to figure out is whether or not $\cos \theta$ will be $> 0$ or $< 0$.

 

[Mark44]

Since 5/11 is positive, that means that $\theta$ lies between 0 and $\pi/2$; i.e., is in the first quadrant.

 

[vela]

In the guided problem, it states that θ is in the range of the inverse function. I don't see how this is the case. The range of the inverse function is in the inclusive range of $\frac {-π}{2}$ and $\frac {π}{2}$. In looking at the problem, I can tell that sine is positive, but that doesn't imply that it's in range as it could be in quadrant II. What should I be looking for?

You're thinking of the solutions to $\sin\theta = \frac {5}{11}$, which has solutions in quadrant I and II. That's not the same as $\theta = \arcsin \frac{5}{11}$ because the range of the arcsin function corresponds to angles in only the first and fourth quadrants.

To make arcsin a function, it can only map 5/11 to one of the two possible values (between 0 and $2\pi$) for $\theta$ for which $\sin\theta = \frac {5}{11}$. The arcsin function is defined so that the value it returns is the one in the first quadrant and not the second.

 

[robphy]

Draw a right triangle. What is the adjacent side?

 

[Mark44]

The arcsin function is defined so that the value it returns is the one in the first quadrant and not the second.

... because the range of arcsin() is $[-\frac \pi 2, \frac \pi 2]$.

 

[vela]

I think you mean the range, right? The domain is [-1,1].

 

[Mark44]

Yes, that's what I meant. I've edited my earlier response.

 

[opus]

So in looking at arcsin $\left(\frac {5} {11}\right)$, I know that y is a positive value, which limits my options to either quadrant I or quadrant II. But since we are taking the arcsin, who's range is [$\frac {-π}{2} , \frac{π}{2}$], I can conclude that it has to be in quadrant I, because quadrant II would be out of the range of arcsin. So does this hold true for any value given? $\frac {11}{2}$? $\frac {1}{50}$? $\frac {777}{800}$? If I take the arcsin of any value, it will produce an angle in the allowable range? Or is this problem cherry picked to be in the range of arcsin?

 

[tnich]

So in looking at arcsin $\left(\frac {5} {11}\right)$, I know that y is a positive value, which limits my options to either quadrant I or quadrant II. But since we are taking the arcsin, who's range is [$\frac {-π}{2} , \frac{π}{2}$], I can conclude that it has to be in quadrant I, because quadrant II would be out of the range of arcsin. So does this hold true for any value given? $\frac {11}{2}$? $\frac {1}{50}$? $\frac {777}{800}$? If I take the arcsin of any value, it will produce an angle in the allowable range? Or is this problem cherry picked to be in the range of arcsin?

Try it out. First, think about what the domain of arcsin(x) is. Then try the values of x at the endpoints of the domain and some in the middle as inputs to the function. What do you get for arcsin(x)?

 

[vela]

So in looking at arcsin $\left(\frac {5} {11}\right)$, I know that y is a positive value, which limits my options to either quadrant I or quadrant II. But since we are taking the arcsin, who's range is [$\frac {-π}{2} , \frac{π}{2}$], I can conclude that it has to be in quadrant I, because quadrant II would be out of the range of arcsin. So does this hold true for any value given? $\frac {11}{2}$? $\frac {1}{50}$? $\frac {777}{800}$? If I take the arcsin of any value, it will produce an angle in the allowable range? Or is this problem cherry picked to be in the range of arcsin?

What's the definition of the range of a function?

 

[haruspex]

If I take the arcsin of any value

If working with real functions, you can only take arcsin of a value in [-1,1]. Each will return a unique value in [-π/2,π/2].

 

[Let'sthink]

Because the range for arc sin (5/11) is already given as (-pi/2, pi/2) theta is just first quadrant. Just find cos theta for that theta for which is sin theta = 5/11. I do not think there is any problem

 

[haruspex]

I must evaluate arcsin first, which will give me an angle θ

Going that route, I do not see how you will satisfy:

Find the exact value

Try @robphy's hint in post #7 (but whether that leads to an exact value that's any more a solution than what you started with is a matter of opinion).

 

[mattt]

Do you know that $$1=cos^2(A) + sin^2(A)$$ for any real number "A" ?

So $$1=cos^2(arcsin(\frac{5}{11}) + sin^2(arcsin(\frac{5}{11}))$$, from here it is quite straight forward, isn't it?

 

[Ray Vickson]

Do you know that $$1=cos^2(A) + sin^2(A)$$ for any real number "A" ?

So $$1=cos^2(arcsin(\frac{5}{11}) + sin^2(arcsin(\frac{5}{11}))$$, from here it is quite straight forward, isn't it?

This is what I was hoping would be the response of the OP to my message #4, but I did not want to give away the whole solution.

 

[OmCheeto]

Do you know that $$1=cos^2(A) + sin^2(A)$$ for any real number "A" ?

So $$1=cos^2(arcsin(\frac{5}{11}) + sin^2(arcsin(\frac{5}{11}))$$, from here it is quite straight forward, isn't it?

Not for me.
This just proves that 1 = 1, when I plug the numbers into my calculator.

@opus , did you ever figure this out?

ps. It took me a couple of days to figure out the solution. I think I used a different method than everyone else so far.

 

[mattt]

This is what I was hoping would be the response of the OP to my message #4, but I did not want to give away the whole solution.

Yes, one does never know where to put the line between "just a hint" and "almost solving the whole thing". :-)

 

[robphy]

Do you know that $$1=cos^2(A) + sin^2(A)$$ for any real number "A" ?

So $$1=cos^2(arcsin(\frac{5}{11}) + sin^2(arcsin(\frac{5}{11}))$$, from here it is quite straight forward, isn't it?

Not for me.
This just proves that 1 = 1, when I plug the numbers into my calculator.

Don't use a calculator for the $\cos^2$ term... think of it as $C^2$, the square of the unknown.
Don't use a calculator for the $\sin^2$ term... can you evaluate it without a calculator?

 

[mattt]

Not for me.
This just proves that 1 = 1, when I plug the numbers into my calculator.

@opus , did you ever figure this out?

ps. It took me a couple of days to figure out the solution. I think I used a different method than everyone else so far.

You don't need to use any calculator. They ask you to calculate, say, "A" and you already know that $$A>0$$ and that $$1 = A^2 + \left(\frac{5}{11}\right)^2$$, I think it is quite straight forward from here...don't you think? :-)

 

[OmCheeto]

Don't use a calculator for the cos2\cos^2 term... think of it as C2C^2, the square of the unknown.
Don't use a calculator for the sin2\sin^2 term... can you evaluate it without a calculator?

You don't need to use any calculator. They ask you to calculate, say, "A" and you already know that

A>0​

A>0 and that

1=A2+(511)2​

1 = A^2 + \left(\frac{5}{11}\right)^2, I think it is quite straight forward from here...don't you think? :-)

As I stated, I figured out the solution, in two days, meaning Feb 24th, meaning 8 days ago.
At first I cheated, and googled.
This gave me a solution, but I didn't understand "why" it worked.
The numerous hints didn't help either.
Although, in hindsight, I understand where everyone was coming from.

Being not the OP, and having had my "epiphany", I'll wait for opus to respond.

 

[OmCheeto]

Draw a right triangle. What is the adjacent side?

I think opus did that, in post #2

In retrospect, yours was the best hint, IMHO.
Equations might yield the answer, but a picture is worth a thousand "Ah ha!"s.

 

[opus]

Shoot! That's a lot of great posts! Apologies for the very late response. I've had the worst flu of my life over the past two weeks.
I think I had a faulty idea of the inverse Trig functions, but while I had the flu I watched about 20 different videos on them I think I've got a much better understanding of them now. What I didn't fully grasp was that we take the sine of angles and arcsin of values which made things quite confusing. In thinking about the domain to keep the function one-to-one, I thought about a similar idea in Algebra but with quadratic graphs and how we need to restrict the domain to make them one-to-one.
I really appreciate everyone's help. I've got a great teacher but this book is really terrible. There are missing words and typos throughout and I tried to fill in the blanks with my own intuition and I filled in the blanks with my own faulty information. I downloaded an OpenStax book and have been going off of that and now things are running smoothly.

 

[Let'sthink]

Homework Statement

Find the exact value of $y=cos\left(arcsin \frac {5} {11} \right)$

Homework Equations

The Attempt at a Solution

I'm stuck at square one. I know that I must evaluate arcsin first, which will give me an angle θ. From that angle, I can take the cosine of it and that much makes sense.
However, in this "guided homework problem" it is stating information that doesn't seem so obvious to me. I've posted a picture.
In the guided problem, it states that θ is in the range of the inverse function. I don't see how this is the case. The range of the inverse function is in the inclusive range of $\frac {-π}{2}$ and $\frac {π}{2}$. In looking at the problem, I can tell that sine is positive, but that doesn't imply that it's in range as it could be in quadrant II.
What should I be looking for?

I think Op has already got the answer. But has he really understood why the range has been specified for the inverse function? Someone tried to tell you that but have you understood. Hint for that you need to know and understand the definition of "function"

 

[opus]

So the domain for the sine function is restricted to [-π/2, π/2] so that it will be one-to-one and can have an inverse. If it wasn't restricted, we could have an input value for the inverse map to more than one output value. The range is remains the same at [-1,1].
For the arcsin function, what was the range from the sine function is now the domain for the arcsine, in this case [-1,1] is the domain of arcsine. The range for the arcsin is going to be [-π/2, π/2] because this is what we restricted our sine function domain to, and the range for the arcsin is the domain of the sine function.
Did I explain this correctly?

 

[SammyS]

So the domain for the sine function is restricted to [-π/2, π/2] so that it will be one-to-one and can have an inverse. If it wasn't restricted, we could have an input value for the inverse map to more than one output value. The range is remains the same at [-1,1].
For the arcsin function, what was the range from the restricted sine function is now the domain for the arcsine, in this case [-1,1] is the domain of arcsine. The range for the arcsin is going to be [-π/2, π/2] because this is what we restricted our sine function domain to, and the range for the arcsin is the domain of the restricted sine function.
Did I explain this correctly?

That is essentially correct.

I would modify that just a bit, as I have suggested with the additional red text above.

The result of restricting the domain of a function is a new, different function.

 

[Let'sthink]

So the range for the arcsine function is restricted to [-π/2, π/2] so that it will be one-to-one and sine function can have an inverse. If it wasn't restricted, we could have an input value for the inverse map to more than one output value. The domain remains the same at [-1,1].
For the arcsin function, what was range for the sine function is now the domain for the arcsine, in this case [-1,1] is the domain of arcsine. The range for the arcsin is going to be [-π/2, π/2] y?

It appears you have understood but your statements needs to be correctly worded you have made a slight jumble. I tried to correct it in your quote.

 

[opus]

Ok thank you for responding. That is more clear and precise. I felt myself rambling and talking in circles trying to explain it, but your edits are much better. Thank you again!

 

[Let'sthink]

Ok thank you for responding. That is more clear and precise. I felt myself rambling and talking in circles trying to explain it, but your edits are much better. Thank you again!

Just to test your understanding please tell me the unrestricted domain of the sine function. When the domain is not restricted is it one-one or many one function? Also state the definition of a function using all terms such as "domain" correctly.

 

[opus]

Just to test your understanding please tell me the unrestricted domain of the sine function. When the domain is not restricted is it one-one or many one function? Also state the definition of a function using all terms such as "domain" correctly.

The graph of the sine function goes on forever in both the positive and negative directions and we can make as many positive or negative unit circle rotations as we’d like. For these reasons the unrestricted domain of the sine function is[-∞,∞]. This would be a one to many function where an infinite amount of inputs would map to a single output. For example, (π/2) and (-3π/2) would both map to 1.
For the last part, do you mean something like f is a function that maps an input,x, to one and only one output, y, for all values of x in the domain of f? I didnt look this one up because I wanted to attempt to define it on my own to test my understanding. It doesn't seem wholly sufficient.

 

[Let'sthink]

The graph of the sine function goes on forever in both the positive and negative directions and we can make as many positive or negative unit circle rotations as we’d like. For these reasons the unrestricted domain of the sine function is[-∞,∞]. This would be a many to one function where an infinite number of inputs would map to a single output in a limited range. For example, (π/2) and (-3π/2) would both map to 1.
For the last part, do you mean something like f is a function that maps an input,x, to one and only one output, y, for all values of x in the domain of f? I didnt look this one up because I wanted to attempt to define it on my own to test my understanding. It doesn't seem wholly sufficient.

Again you are almost right and like earlier have made some terminological mistakes! I correct them by writing in red. Your definition is almost correct. But just I write it again in so called my words. Function is a rule which associates with EVERY element in the domain set, with a UNIQUE element in the Range set. That is why a function can be one one or many one (but never one many, because then it will not be function at all) So the inverse of a many one function cannot be a function unless we restrict the range set and make it one one. This problem illustrates these ideas.

 

[opus]

Again you are almost right and like earlier have made some terminological mistakes! I correct them by writing in red. Your definition is almost correct. But just I write it again in so called my words. Function is a rule which associates with EVERY element in the domain set, with a UNIQUE element in the Range set. That is why a function can be one one or many one (but never one many, because then it will not be function at all) So the inverse of a many one function cannot be a function unless we restrict the range set and make it one one. This problem illustrates these ideas.

Ah that is much more precise. I definitely feel much better about my understanding of functions after going over this. Thank you so much.