Simplifying Trigonometric Identities for Derivatives

[miglo]

http://www.wolframalpha.com/input/?i=sqrt%281-%28%28sqrt%281-x^2%29%29^2%29

how is it that the sqrt(1-(sqrt(1-x^2))^2)=sqrt(x^2)?
shouldnt it be sqrt(-x^2)?, because (sqrt(1-x^2))^2=1-x^2, so then you have sqrt(1-1-x^2)
which would give you sqrt(-x^2)

im trying to find the derivative of arcsin(sqrt(1-x^2)) ,0<x<1
and well it might be that I've forgotten about some algebraic trick i can do here its been bugging me for a while now that i can't solve this
can someone please explain this to me, its really confusing me
also not asking for the answer to my derivative, I am pretty sure i can finish it off once i know why the above statement is true

thanks in advance

 

[tiny-tim]

hi miglo!

(have a square-root: √ and try using the X² icon just above the Reply box )

… so then you have sqrt(1-1-x^2)
which would give you sqrt(-x^2)

nooo …

1 - (1 - x²) = 1 - 1 + x² = x²

(but anyway isn't it obvious that arcsin√(1 - x²) = arccos(x)? )

 

[miglo]

ohh wow, i totally didnt see that -1 while trying to work out this problem, thanks tiny-tim!

and no i don't see how that's obvious, after using sqrt(x^2) instead of sqrt(-x^2) i get as my answer to be -1/(sqrt(1-x^2)), which is the derivative of arccos, but i don't see how arcsin(sqrt(1-x^2))=arccos, care to explain?

 

[tiny-tim]

"arcsin√(1 - x²) = arccos(x)"

is another way of saying that the angle (in [0,π/2) anyway) whose cos is x is the angle whose sin is √(1 - x²) …

isn't it? ​

 

[miglo]

oh man i was never really good with trigonometry, could this also be explained using the pythagorean theorem? but anyways thanks again, ill keep this in mind, hopefully itll make sense to me soon

 

[tiny-tim]

oh man i was never really good with trigonometry, could this also be explained using the pythagorean theorem?

yes, but better would be to learn all the standard trigonometric identities …

in this case, cos² + sin² = 1 ​