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Help Eibroaaar with calculus

Started by Eibro, September 16, 2004, 08:50 PM

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Eibro

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

K

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.

dxoigmn


Eibro

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.
Eibro of Yeti Lovers.

dxoigmn

#4
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.

Eibro

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.
Eibro of Yeti Lovers.

Eibro

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 )
Eibro of Yeti Lovers.

Adron

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