Solving the Square Root of a Difference: Integrating √(9-x^2)

[sponsoredwalk]

Ah, that explains everything.

:uhh:

 

[mrigmaiden]

As to the sin(arcsin) and cos(arcsin), I know sin(arcsin(x/3))=x/3 and I cheated and looked up cos(arcsin(x))=√(1-x^2) from one of my formula memorization sheets from class. So cos(arcsin(x/3)=√(9-x^2)/3 but this is one of those I don't know why.

Consider the triangles that the other poster with the videos showed. That gives you all the information you need. Sometimes it is easy to get confused with the trigonometric inverse functions so here is an example to lubricate your brain a bit:

arcsin(1)=90 degrees=pi/2 radians That is to say that the angle that produces a sin equal to 1 is pi/2 radians. so arctrigfunction(number)=ANGLE (in degrees or radians)

Now to our problem, let's examine the expression cos(arcsin(x/3))=√(9-x^2)/3 by visualizing its associated right triangle. The triangle has theta as the angle of interest, the hypotenuse is of length 3, the leg adjacent to theta has length sqrt(9-x^2) and the leg opposite to theta has length x. Use the Pythagorean Theorem to prove that to yourself.

Now that we know what the triangle looks like, we can figure out how to evaluate the expression cos(arcsin(x/3)). Consider rephrasing the problem in your mind in the following way:

1.arcsin(x/3) gives me the ANGLE theta where sin(theta)=x/3

2.cos(arcsin(x/3)) means What is the cosine of the angle theta?

3.This verbal re-framing should help to parse the expression a bit more.

4.Recall that for right triangles:
a)cos(theta)=adjacent side length/hypotenuse length
b)sin(theta)=opposite side length/hypotenuse length
c)tan(theta)=opposite side length/adjacent side length

5.Now that we have the triangle and we have called arcsin(x/3)=theta, then cos(arcsin(x/3))=cos(theta)
=adjacent side length to theta/hypotenuse length
=(sqrt(9-x^2))/3

6. Sometimes it is better to learn how to derive expressions from first principles instead of looking them up all the time. The you can memorize the formulas after you understand how they came about. I'll explain briefly why cos(arcsin(x))=√(1-x^2) according to your class notes.

7. First draw the triangle arcsin(x)=theta=> sin(theta)=x/1=opp/hyp thus the opposite side length to theta has length equal to x and the hypotenuse has length equal to 1. Now, use the Pythagorean theorem to show that the side length adjacent to theta is of length sqrt(1-x^2). So the triangle has theta as the angle of interest, the adjacent side to theta is of length sqrt(1-x^2), the opposite side to theta is of length x and the hypotenuse is of length 1. So cos(arcsin(x))=cos(theta)=adjacent/hypotenuse=sqrt(1-x^2)/1=sqrt(1-x^2)

I hope this helps.

 

[mrigmaiden]

:uhh:

hehehe He's just trying to get a rise out of someone. Don't let it bother you. Take what he has to say regarding the math and physics and ignore the rest lol!