Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$

I'm trying to prove that $$\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}.$$ Let $\alpha = 2 \arcsin x$ and $\beta = \arccos x$; meaning $\sin\alpha = \frac{x}{2}, \cos\beta = x$. We know that: $$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.$$ Finding $\cos\alpha$ and $\sin\beta$: \begin{align} \cos\alpha & = \frac{\sqrt{4-x^2}}{2}, \\[0.1in] \sin\beta & = \sqrt{1-x^2}. \end{align} So plugging everything in: \begin{align} \sin(2\arcsin x + \arccos x) & = \frac{x}{2} \cdot x + \frac{\sqrt{4-x^2}}{2} \cdot \sqrt{1-x^2}\\[0.1in] & = \frac{x^2 + \sqrt{4-x^2}\sqrt{1-x^2}}{2}. \end{align} But this doesn't seem to lead to the right side. Is my method incorrect?

• $\alpha = 2\arcsin x$ means that $x = \sin\frac{\alpha}{2}$, not that $\frac{x}{2} = \sin\alpha$. – rogerl Oct 10 '14 at 18:51

$$\arcsin x+\arccos x=\frac\pi2$$

So, we have $$\sin\left(\frac\pi2+\arcsin x\right)=\cos(\arcsin x)$$

• Why does $\sin(\frac{\pi}{2}+\arcsin x)=\cos(\arcsin x)$? – hohner Oct 10 '14 at 18:58
• @hohner, $\sin(\pi/2+y)=\cos y$ as sine ratio is positive in the second quadrant – lab bhattacharjee Oct 10 '14 at 18:59
• Thanks. Is there any other way of proving this? I'm studying $R\cos(x\pm\alpha)$ and inverse trig functions at the moment :) – hohner Oct 10 '14 at 19:06
• Also, why does $\arcsin x=\arccos\sqrt{1-x^2}$? – hohner Oct 10 '14 at 19:10
• @hohner, Have not you noticed the proof? – lab bhattacharjee Oct 10 '14 at 19:11

$f(x)=\sin(2\arcsin x + \arccos x) - \sqrt{(1-x^2)}=\sin(\arcsin x + \frac\pi 2) - \sqrt{(1-x^2)}=\cos(\arcsin x)-\sqrt{(1-x^2)}$.

$f'(x)=0$, hence $f$ is constant.

In particular $f(x)=f(0)=0$

$\alpha=2\arcsin x$ means that $x=\sin(\frac\alpha2)$, not that $\frac x2=\sin\alpha$

$\frac\alpha2=\arcsin x$

$x=\sin(\frac\alpha2)$

an alternate and faster way is to use

$\arcsin x+\arccos x=\frac\pi2$