Prove that cos( arcsin( x ) ) = sqrt( 1 - x ** 2 )
A push in the correct direction would be appreciated.

not sure how you're supposed to prove it.
cos( arcsin(x) ) = sin( arccos(x) ) = sqrt(1 - x[sup]2[/sup]) is a trigonometric identity. Maybe look at some other trig identities and see if you can derive it.

Where is the calculus part?

[quote author=dxoigmn link=board=36;threadid=8705;start=0#msg80478 date=1095390988]
Where is the calculus part?
[/quote]It's a question from my calculus book, which I use in my calculus course.

[quote author=Eibro[yL] link=board=36;threadid=8705;start=0#msg80485 date=1095392385]
It's a question from my calculus book, which I use in my calculus course.
[/quote]

Looks more like trig, but I guess they review trig in the beginning calc courses because you need it a lot later on. Anyway.

[code]
sin**2(x) + cos**2(x) = 1
cos**2(x) = 1 - sin**2(x)
cos(x) = sqrt(1 - sin**2[x])

Let x = arcsin(x)

cos(arcsin[x]) = sqrt(1 - sin**2[arcsin(x)])
cos(arcsin[x]) = sqrt(1 - x**2)
[/code]

I did it backwards ;) I'm sure you can work backwards.

[quote author=dxoigmn link=board=36;threadid=8705;start=0#msg80497 date=1095393499]
[quote author=Eibro[yL] link=board=36;threadid=8705;start=0#msg80485 date=1095392385]
It's a question from my calculus book, which I use in my calculus course.
[/quote]

Looks more like trig, but I guess they review trig in the beginning calc courses because you need it a lot later on. Anyway.

[code]
sin**2(x) + cos**2(x) = 1
cos**2(x) = 1 - sin**2(x)
cos(x) = sqrt(1 - sin**2[x])

Let x = arcsin(x)

cos(arcsin[x]) = sqrt(1 - sin**2[arcsin(x)])
cos(arcsin[x]) = sqrt(1 - x**2)
[/code]

I did it backwards ;) I'm sure you can work backwards.
[/quote]Ahoy, thanks mate.
Yes, I believe we are still reviewing.

I am stuck again
sin( arctan( x ) ) = x / sqrt( 1 + x ** 2 )

I figure it's something like...
sin( arcsin( x ) / arccos( x ) ) =
x / arccos( x ) =
Looks similar, we just need arccos( x ) -> sqrt( 1 + x ** 2 )

[quote author=Eibro[yL] link=board=36;threadid=8705;start=0#msg80895 date=1095623343]
I am stuck again
sin( arctan( x ) ) = x / sqrt( 1 + x ** 2 )

I figure it's something like...
sin( arcsin( x ) / arccos( x ) ) =
x / arccos( x ) =
Looks similar, we just need arccos( x ) -> sqrt( 1 + x ** 2 )
[/quote]

Try this:

let x = tan(y) = sin(y) / cos(y)

sin(y) = sin(y) / (cos(y) * sqrt(1 + sin2(y) / cos2(y))
cos(y) * sqrt(1 + sin2(y) / cos2(y)) = 1
sqrt(cos2(y) + sin2(y)) = 1
cos2(y) + sin2(y) = 1 -- a basic identity



Your "sin( arcsin( x ) / arccos( x ) ) = x / arccos( x )" seems invalid