Explain 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.

  • $\begingroup$ $\cos^2\theta+\sin^2\theta=1$ $\endgroup$ Dec 2, 2019 at 2:55

2 Answers 2


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:

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$$ \theta = \arccos(t) = \arcsin(\sqrt{1 - t^2}) $$

  • $\begingroup$ Is it possible to prove with algebra only? $\endgroup$ Dec 2, 2019 at 1:58
  • 2
    $\begingroup$ I've added the algebraic proof. It's worth mentioning that the idea of the algebraic proof is built around the geometrical proof. $\endgroup$ Dec 2, 2019 at 2:05

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.


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