this post was submitted on 03 Mar 2024
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It's homework help, but I'm not asking for the solution. The problem only asks for cos, sin, tan, cot, csc given sec. I found those pretty quickly on my own, and confirmed solutions with the back of the book.

Where I run into confusion is when I try to find angle theta on my own. Arccos of found cos gives 2.06, arcsin of found sin gives 1.08, and arctan of found tan gives -1.08. Problem givens exclude possibility of the negative angle found by arctan(-15/8), but the other two are possible and conflicting. And why wouldn't they all be the same? I reattempted because there were so many erase marks from trying to figure this out that it was almost illegible.

Am I wrong? Did the book give me a point not on the unit circle or something, assuming I wouldn't try to find theta on my own? Have I used arcfuncs wrong- I checked the domains against the function definitions? Have I found a hole in math?

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[–] [email protected] 5 points 5 months ago* (last edited 5 months ago) (2 children)

The correct answer it the 2.06 from the cosine, but with the sine you're just getting the first quadrant answer, since it's simmetrical.

[–] [email protected] 1 points 5 months ago* (last edited 5 months ago)

The correct answer is the 2.06 from the cosine, but with the sine you're just getting the first quadrant answer, since it's simmetrical.

[–] [email protected] 1 points 5 months ago (1 children)

That's definitely insightful- I found pi-arccos=arcsin. So is this a case where the range of arcsin is restricted to q1 and q4, while arccos is restricted to q1 and q2 so can give answers in q2? And why would arctan give the negative of arcsin?

I can tell I'm not quite understanding some fundamental concept, but I can't tell where to focus my review or if I just straight missed something.

[–] [email protected] 2 points 5 months ago (1 children)

Well basically yes you got it, since sin and cos repeat, you can't actually calculate the inverse for any angle, so the domain of the inverse functions is restricted to a range where they are bijective. But simply put, think it on the circle, can the height (sin) tell you if you are left (q2) or right (q1) ? No, but the x coordinate (cos) can :)

[–] [email protected] 1 points 5 months ago* (last edited 5 months ago)

Okay thanks. One more question: Unless quadrant is known beforehand (as in this case it was given), is there any way to tell which inverse function is "correct?"

Edit2: Oh, of course- you already said and I already read but forgot: It's based on cos and sine, whether they're negative.

[–] [email protected] 1 points 5 months ago

not sure if it may help, but maybe check out the graphs at the bottom of the basic concepts section here:

https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions