Need help with a question

[zool3]

Can anyone do a step by step explanation of how to get from cos (arcsin x) to √(1- x^2)

Thanks

[HarveyD]

let u = arcsin x and use chain rule
i.e
dy/dx = dy/du x du/dx

Edit: assumed you were looking for the derivative...
whoops

[Special At Specialist]

cos(arcsin(x))
Let u = arcsin(x)
sin(u)
= sin(arcsin(x))
= x
We know that sin^2(x) + cos^2(x) = 1, so:
cos^2(x) = 1 - sin^2(x)
cos(x) = sqrt(1 - sin^2(x))
We worked out that sin(u) = x, so:
cos(u) = sqrt(1 - x^2)
cos(arcsin(x)) = sqrt(1 - x^2)

[kamil9876]

Poster didn't mention anything about derivative, I'm assuming it's to actually show they are equal. Hint: use . Also you should justify why you take the positive square root.

[zool3]

cos(arcsin(x))
Let u = arcsin(x)
sin(u)
= sin(arcsin(x))
= x

why did you do this step?

[Special At Specialist]

cos(arcsin(x))
Let u = arcsin(x)
sin(u)
= sin(arcsin(x))
= x

why did you do this step?

To put it in the form:
sin^2(u) + cos^2(u) = 1
Since I knew what sin(arcsin(x)) was, it was only a matter of transposition to solve for cos(arcsin(x)).

[zool3]

ohh okay! thanks for that!