Finding arcsin, arccos, and arctan

[mollyk]

This is the work I’ve been able to complete so far. I was able to find sin, cos, and tan directly off the unit circle, but now I’m stuck with arcsin, arccos, and arctan. In class we have only learned how to find them from values (like 1/2, 0, 1,etc) and not the other way around with angles. I tried to do the opposite by doing arcsin of the values of sin (like arcsin(0) = 0, arcsin(1/2) = pi/6) but i’m not sure if that’s the correct process. Am I heading in the right direction or is there a different way to find arc based on angles? I hope that makes sense, Im a little lost.

 

[mollyk]

I placed what I think would be right? In the table but in this case would arc just the be same as the normal trig function? Or am I completely off?

 

[mollyk]

I was able to find sin, cos, and tan directly from the unit circle however I am now stuck on how to find arcsin, arccos, and arctan. I started to try it, but I’m not sure if I’m going in the right direction. I basically wrote down the same values as revalue sin and cos for the arc version but I’m not sure if that’s right. Here I have to find arc from the given theta angles on the unit circle but I’ve own found arc from given values like 1/2, 0, and 1 before. Am I heading in the right direction or is there a method Im completely missing? I hope that makes sense.

 

[topsquark]

You appear to have done the table for [math]sin( \theta )[/math], [math]cos( \theta )[/math], and [math]tan( \theta )[/math] correctly. Well done.

However, even though [math]asn \left ( \dfrac{ \pi }{2} \right )[/math], for example, does exist it won't look pretty. I think what the problem wanted was something like [math]asn \left ( \dfrac{ \sqrt{2}}{2} \right ) = \dfrac{ \pi }{2}[/math]. The sine of an angle is a number.... the inverse sine of a number is an angle.

-Dan

 

[pka]

The statement of this question is deeply flawed. It is only half correct.
The sine, cosine, and tangent functions can take those numbers as arguments.
BUT the functions arcsine, arccosine, and arctangent can not. In fact, they have different domains.
Please see this link. Please study the domains of the several inverse trigonometry functions.

 

[Steven G]

The problem in my opinion is not clear. Having said that, your answers are not reasonable answers.

Arcsin x = y, means that sin(y) =x AND -pi/2 < y <pi/2. That is, you take arcsin of numbers and get angles. That implies, to me, that you should have angles in the inverse columns, not numbers.

 

[jonah2.0]

Beer induced opinion follows.

View attachment 27997View attachment 27998
This is the work I’ve been able to complete so far. I was able to find sin, cos, and tan directly off the unit circle, but now I’m stuck with arcsin, arccos, and arctan. In class we have only learned how to find them from values (like 1/2, 0, 1,etc) and not the other way around with angles. I tried to do the opposite by doing arcsin of the values of sin (like arcsin(0) = 0, arcsin(1/2) = pi/6) but i’m not sure if that’s the correct process. Am I heading in the right direction or is there a different way to find arc based on angles? I hope that makes sense, Im a little lost.

It is quite possible that the screenshot from your book is one monumental typo by the publisher. It could have been been sin, cos, tan, csc, sec, and cot instead as that is the usual exercise given in most books of Trigonometry.

 

[Harry_the_cat]

Beer induced opinion follows.

It is quite possible that the screenshot from your book is one monumental typo by the publisher. It could have been been sin, cos, tan, csc, sec, and cot instead as that is the usual exercise given in most books of Trigonometry.

Yes that's what I thought too.