Question

Could someone clarify what is going on when you multiply things like sin(cos^-1y)?

Answer 1

You aren't multiplying those are you? Isn't it:
\[\sin(\arccos(y))?\]

Answer 2

yeah it's that

Answer 3

You're not multiplying them. The arccos(y) is the argument of sine. Arc trig functions are the "inverses" of trig functions. Similar to a square root and a^2 (though, technically, sqrt(x) isn't a TRUE inverse, I digress). So if you had:
\[\sin(\arcsin(3x))\]
That would just be equal to 3x. Because arcsin(3x) would give you an ANGLE which is what goes into trig functions, then the trig function would give you back out a number given the angle.
So if you have:
\[\sin(\arccos(y))=\sin(\arccos(\frac{y}{1}))\]
Cosine=adj/hyp
So you have y as adjacent and 1 as hypotenuse.
That means that the opposite side would be y^2+o^2=1^2
o=sqrt(1-y^2)
so that means the sin(arccos(y)) would be:
sqrt(1-y^2)
That is, of course, assuming that arccos(y) has an implied one and wasn't originally like y^2/y. And they just simplified it. But I hope that gives you a general idea.

Answer 4

I just don't understand where the y^2+0^2=1^2 comes in

Answer 5

nvm, I drew the triangle haha get it now

Answer 6

Okay, good :)

Answer 7

I didn't have a way to draw a triangle for you on here xDDD