Question #47049

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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!