How do you write an algebraic expression that is equivalent to #cos(arcsin((x-h)/r))#?

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Answer 1 · 2016-12-30T16:45:11

#cos(arcsin((x-h)/r)) = sqrt (1-((x-h)/r)^2)#

By definition:

#y=arcsin((x-h)/r)#

is a number such that:

#siny = (x-h)/r#

Then we have that:

#sqrt(1-sin^2y) = cosy#

and thus:

#cos(arcsin((x-h)/r)) = sqrt (1-((x-h)/r)^2)#

Answer 2 · 2016-12-30T16:56:04

Use the identity #cos(theta) = +-sqrt(1 - sin^2(theta))# and then simplify.

We substitute #sin^-1((x-h)/r)# for #theta#:

#cos(sin^-1((x-h)/r)) = +-sqrt(1 - sin^2(sin^-1((x - h)/r)))#

A property of a function and its inverse, #f(f^-1(x)) = x#, makes the sine and the inverse sine disappear:

#cos(sin^-1((x-h)/r)) = +-sqrt(1 - ((x - h)/r)^2)#

Make a common denominator:

#cos(sin^-1((x-h)/r)) = +-sqrt(r^2/r^2 - (x - h)^2/r^2)#

#cos(sin^-1((x-h)/r)) = +-sqrt((r^2 - (x - h)^2)/r^2)#

#cos(sin^-1((x-h)/r)) = +-sqrt(r^2 - (x - h)^2)/r#