Question 47049

Nov 22, 2016

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

Explanation:

$y = \sin \left(t\right) \to t = \arcsin y$

We know that $\cos \left(2 \alpha\right) = 1 - 2 {\sin}^{2} \alpha$:

$x = 1 - 2 {\sin}^{2} \left(t\right)$

$x = 1 - 2 {\sin}^{2} \left(\arcsin y\right)$

$x = 1 - 2 {y}^{2}$

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

$x - 1 = - 2 {y}^{2}$

$- \frac{1}{2} x + \frac{1}{2} = {y}^{2}$

$y = \pm \sqrt{\frac{1}{2} - \frac{1}{2} x}$

However, we must restrict the function.

The domain of $y = \arcsin x$ and $y = \arccos x$ is -1 ≤ x ≤ 1#.

Hopefully this helps!