Question

• Show that \(\arcsin y=\arccos x\) is the equation of a circle.
• Note that the equation of a circle is \(x^2+y^2=1\).

Solution

$$ \begin{align*} \arcsin y&=\arccos x\\ y&=\sin(\arccos x)\\ y&=\sqrt{1-\cos^2(\arccos x)}\\ y^2&=1-x^2\\ x^2+y^2&=1 \end{align*} $$

Alternative solution (the cooler way)

Let \(\theta=\arcsin y=\arccos x\)

$$ \begin{align*} \sin\theta&=y\\ \cos\theta&=x\\\\ \because\cos^2\theta+\sin^2\theta&=1\\ \therefore x^2+y^2&=1 \end{align*} $$