Explain why $\arccos(t)=\arcsin(\sqrt{{1}-{t^2}})$ when $0Explain why $\arccos(t)=\arcsin(\sqrt{{1}-{t^2}})$ when $0<t≤1$. 
I tried researching online, couldn't find anything related to this question though. Know this equation is correct and make sense, just don't know how to explain it using algebra only, solving for the left side of the equation.
 A: Algebraic proof:
Let $\theta = \arccos(t)$, so $\cos(\theta) = t$. Note that since $0 < t \leq 1$, we have $0 \leq \theta < \frac{\pi}{2}$. Recall that $\sin^2(\theta) + \cos^2(\theta) = 1$, so $\sin(\theta) = \pm\sqrt{1 - \cos^2(\theta)}$. Since $0 \leq \theta < \frac{\pi}{2}$, we have $\sin(\theta) \geq 0$, so $\sin(\theta) = \sqrt{1 - \cos^2(\theta)} \Rightarrow \sin(\theta) = \sqrt{1 - t^2} \Rightarrow \theta = \arcsin(\sqrt{1 - t^2})$. Thus, $\arccos(t) = \arcsin(\sqrt{1 - t^2})$
Geometrical proof:

$$
\theta = \arccos(t) = \arcsin(\sqrt{1 - t^2})
$$
A: Here is a calculus proof: Given the integral definitions of $\arcsin$ and $\arcsin$, which respectively are 
$$\arcsin t=\int_0^t \frac{dz}{\sqrt{1-z^2}},\\ \arccos t=\int_t^1 \frac{dz}{\sqrt{1-z^2}}$$
for $-1\leq t\leq 1$, the substitution $w=\sqrt{1-z^2}$ yields (note that $w\,dw=-z\,dz$)
$$\arccos t=\int_t^1\frac{dz}{\sqrt{1-z^2}} = \int_{\sqrt{1-t^2}}^0 \frac{(-w/z)}{w}dw=\int_0^{\sqrt{1-t^2}}\frac{dw}{\sqrt{1-w^2}}=\arcsin(\sqrt{1-t^2})$$ as desired.
