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 - x2) is a trigonometric identity. Maybe look at some other trig identities and see if you can derive it.
Where is the calculus part?
Quote from: dxoigmn on September 16, 2004, 10:16 PM
Where is the calculus part?
It's a question from my calculus book, which I use in my calculus course.
Quote from: Eibro[yL] on September 16, 2004, 10:39 PM
It's a question from my calculus book, which I use in my calculus course.
Looks more like trig, but I guess they review trig in the beginning calc courses because you need it a lot later on. Anyway.
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)
I did it backwards ;) I'm sure you can work backwards.
Quote from: dxoigmn on September 16, 2004, 10:58 PM
Quote from: Eibro[yL] on September 16, 2004, 10:39 PM
It's a question from my calculus book, which I use in my calculus course.
Looks more like trig, but I guess they review trig in the beginning calc courses because you need it a lot later on. Anyway.
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)
I did it backwards ;) I'm sure you can work backwards.
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 from: Eibro[yL] on September 19, 2004, 02:49 PM
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 )
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