Inverse Trigonometric Equation problem

[acen_gr]

EDIT: 1.Homework Statement
The problem asks to simplify this: cos(arcsin x)

Homework Equations

let arc sin x = θ

The Attempt at a Solution

Hoping for some explanation. I really don't have any idea. Thank you so much in advance!

EDIT: I have to edit because I received a warning for not posting for #3 and #2. I originally posted "cos(arcsin x)" for #2 because that was what I know that is relevant to the problem. And I apologize for I didn't know it wasn't enough (I'm new here and sorry if I didn't understand the rules well, I thought it was okay to remove the format instructions and just put what is being asked instead. Quoted above is my previous post #3. As I said there, I really don't have any idea and therefore asks for some (can't put any attempt) When I got the hint from the mentors, I was able to work on this:

The problem asks to simplify this: cos(arcsin x)

Let arcsin x = θ

if arcsin x = θ, then x = sinθ

get the cos of θ from given x/1 and should get for cosθ = √1-x² (by pythagorean theorem)

the answer is cos(θ) = √1-x²

 

[Mentallic]

Let $\theta = \arcsin(x)$ so then $x=\sin(\theta)$. This expression is just simple trig where you first learn that the sine of an angle is OPP/HYP in a right triangle right? So do just that. Create the right-triangle with an angle of $\theta$ such that $\sin(\theta)=x$ (which is equivalent to x/1).

 

[acen_gr]

Let $\theta = \arcsin(x)$ so then $x=\sin(\theta)$. This expression is just simple trig where you first learn that the sine of an angle is OPP/HYP in a right triangle right? So do just that. Create the right-triangle with an angle of $\theta$ such that $\sin(\theta)=x$ (which is equivalent to x/1).

So what should be my final answer? Is it Cos θ? Sorry if I ask too many. :(

EDIT: Ok sir. Nevermind this. I already got the answer. Thank you so much! :)

 

[HallsofIvy]

Good! I am glad you were able to get the answer- I hope you were able to work it out for yourself from the hint acen gr gave you.

For others who might wonder: Yes, "the answer is $cos(\theta)$" but that is not the end of it. As acen gr suggested, "Create the right-triangle with an angle of θ such that sin(θ)=x (which is equivalent to x/1)." Then we have a right triangle with "opposite side" of length x and "hypotenuse" of length 1. By the Pythagorean theorem, the "near side" has length $\sqrt{1- x^2}$. Since cosine is "near side over hypotenuse", $cos(\theta)= cos(cos^{-1}(x))= \frac{(\sqrt{1- x^2}}{1}= \sqrt{1- x^2}$

You could also do it this way: for any angle $\theta$, $sin^2(\theta)+ cos^2(\theta)= 1$ so that $cos(\theta)= \pm\sqrt{1- sin^2(\theta)}$. That is, $cos(sin^{-1}(x))= \pm\sqrt{1- sin^2(sin^{-1}(x))}= \pm\sqrt{1- x^2}$. Because the standard domain of $sin^{-1}$ is $-\pi/2$ to $\pi/2$, the sign is the same as the sign of x.

(The sign did not come up in the previous paragraph because being able to draw a triangle with side "x" assumes x is positive.)

 

[Mentallic]

HallsofIvy, I think you're getting a little mixed up with who is helping who :tongue:

 

[acen_gr]

Good! I am glad you were able to get the answer- I hope you were able to work it out for yourself from the hint acen gr gave you.

For others who might wonder: Yes, "the answer is $cos(\theta)$" but that is not the end of it. As acen gr suggested, "Create the right-triangle with an angle of θ such that sin(θ)=x (which is equivalent to x/1)." Then we have a right triangle with "opposite side" of length x and "hypotenuse" of length 1. By the Pythagorean theorem, the "near side" has length $\sqrt{1- x^2}$. Since cosine is "near side over hypotenuse", $cos(\theta)= cos(cos^{-1}(x))= \frac{(\sqrt{1- x^2}}{1}= \sqrt{1- x^2}$

You could also do it this way: for any angle $\theta$, $sin^2(\theta)+ cos^2(\theta)= 1$ so that $cos(\theta)= \pm\sqrt{1- sin^2(\theta)}$. That is, $cos(sin^{-1}(x))= \pm\sqrt{1- sin^2(sin^{-1}(x))}= \pm\sqrt{1- x^2}$. Because the standard domain of $sin^{-1}$ is $-\pi/2$ to $\pi/2$, the sign is the same as the sign of x.

(The sign did not come up in the previous paragraph because being able to draw a triangle with side "x" assumes x is positive.)

Sir, so the final answer should be Cosθ = $\sqrt{1- x^2}$? Am I right?

EDIT: Anyway sir, I am the one who was helped by Mentallic :)
EDIT: How do I make square root symbol here?

 

[HallsofIvy]

HallsofIvy, I think you're getting a little mixed up with who is helping who :tongue:

One of these days, I really need to learn to read!

 

[HallsofIvy]

Sir, so the final answer should be Cosθ = $\sqrt{1- x^2}$? Am I right?

Assuming that x is positive, yes. Otherwise $cos(sin(\theta))= \pm\sqrt{1- x^2}$.

EDIT: Anyway sir, I am the one who was helped by Mentallic :)
EDIT: How do I make square root symbol here?

There is a LaTeX tutorial at https://www.physicsforums.com/showthread.php?p=3977517&posted=1#post3977517.

For just a single square root symbol, if you click on the "Go Advanced", you will see a table of "Quick Symbols" and there is a "√" there- notice that you had better use ( ) with that!

 

[acen_gr]

Assuming that x is positive, yes. Otherwise $cos(sin(\theta))= \pm\sqrt{1- x^2}$.

Wee! Thanks mentor! :) I love this site!

 

[Mentallic]

Sir, so the final answer should be Cosθ = $\sqrt{1- x^2}$? Am I right?

Yes that's correct.

How do I make square root symbol here?

You already have! Using tex tags as you've done.

One of these days, I really need to learn to read!

Create a homework thread and I'm sure some of us would be delighted to help you

 

[Mentallic]

Assuming that x is positive, yes. Otherwise $cos(sin(\theta))= \pm\sqrt{1- x^2}$.

Just to clarify, we're talking about $$\cos(\sin^{-1}\theta)$$ and not $$\cos(\sin\theta)$$

Also, the range of $\sin^{-1}\theta$ is $[-\pi/2, \pi/2]$ and thus the range of $\cos(\sin^{-1}\theta)$ is $[0,1]$ so the answer is only $\sqrt{1-x^2}$

 

[acen_gr]

Just to clarify, we're talking about $$\cos(\sin^{-1}\theta)$$ and not $$\cos(\sin\theta)$$

Also, the range of $\sin^{-1}\theta$ is $[-\pi/2, \pi/2]$ and thus the range of $\cos(\sin^{-1}\theta)$ is $[0,1]$ so the answer is only $\sqrt{1-x^2}$

Oh I see. No wonder why the actual question here in my notes asks for an exact answer;) Thanks Metallic! :D