1. ## inverse of sine?

Hey guys, I just have a quick question. Is there a true inverse of sine? now I know that an inverse of sine wouldn't be a function, but what if we restricted it. then wouldn't you be able to make a kind-of inverse function?

Ok I i'm just going to come out and say it, can you solve this? (for 'y' of course)

Code:
`x=(sqrt(y-2)^2)+(sqrt(-y+4)^2)+sin(y)-2`
this question has been eating at my soul for the past few days and if someone could provide some in-site that would be great

actually I need it for x=0 to try to find Pi so in that case you can just solve this...

Code:
`0=(sqrt(y-2)^2)+(sqrt(-y+4)^2)+sin(y)-2`
God... that's ugly.

-thanks again

btw: my logic has been all over the place lately so if something looks strange, that's why. like if i am using Pi to find Pi, that's what my friend told me I am doing... 2. You can find the inverse of sine modulo 2*Pi, it's called arcsine. 3. y = asin(x)

If x is zero, so is y.

Everything else cancels out. This equation gets you no closer to finding Pi. 4. If x is 0, y is not 0 (because of -2 term).
That equation cannot be solved arithmetically AFAIK. You'd have to do it numerically.
But solve an equation for finding Pi...? Why would you do that? 5. 6. that would be insine.

180 degrees out of phase of sin.

<joke> 7. Huh Elysia?

Here's how everything cancels out:

Original:
x = (sqrt(y-2)^2) + (sqrt(-y+4)^2) + sin(y) - 2

Taking square-root then immediately squaring means you haven't done anything.
This is of course assuming the argument doesn't go negative ... but one could use imaginary numbers and get away with it.

x = (y - 2) + (-y + 4) + sin(y) - 2

removing the parenthesis and moving the sin to the last

x = y - 2 - y + 4 - 2 + sin(y)

grouping variables and constants together

x = y - y - 2 + 4 - 2 + sin(y)

Things cancel out to give

x = sin(y)

Now if x is pinned at zero, y is zero. Or it could be any multiple of Pi since Sine is cyclical. 8. Ah, I didn't think of sqrt as actual square root. I think I just saw them as sines or something. My bad. 9. Originally Posted by nonoob Original:
x = (sqrt(y-2)^2) + (sqrt(-y+4)^2) + sin(y) - 2

Taking square-root then immediately squaring means you haven't done anything.
This is of course assuming the argument doesn't go negative ... but one could use imaginary numbers and get away with it.

x = (y - 2) + (-y + 4) + sin(y) - 2
You can't do that; -y+4 isn't the same as (sqrt(4-y))^2, which actually reduced to |4-y|, where || is the absolute value function. 10. Originally Posted by User Name: You can't do that; -y+4 isn't the same as (sqrt(4-y))^2, which actually reduced to |4-y|, where || is the absolute value function.
You are confusing sqrt(x^2) with (sqrt(x))^2 .
It gives |x| only for the former...but the later remains x. Popular pages Recent additions 