Question #47049

Question

3449 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-22T01:41:12

#y = +-sqrt(1/2 - 1/2x)" "{x|-1 ≤ x ≤ 1}#

#y = sin(t) -> t = arcsiny#

We know that #cos(2alpha) = 1 - 2sin^2alpha#:

#x = 1 - 2sin^2(t)#

#x = 1 - 2sin^2(arcsiny)#

#x = 1 - 2y^2#

Since it is often better regarded to have a function defined by #y#:

#x - 1 = -2y^2#

#-1/2x + 1/2= y^2#

#y = +-sqrt(1/2 - 1/2x)#

However, we must restrict the function.

The domain of #y = arcsinx# and #y = arccosx# is #-1 ≤ x ≤ 1#.

Hopefully this helps!