Answer
Details of proof is shown below.
- Do direct calculation when $x=-1$, $x=0$ and $x=1$.
- To check on $(-1,0)$, replace $x=-a$ with $a\gt0$ then use the known identities to calculate.
Work Step by Step
$$\sin^{-1}x+\cos^{-1}x=\pi/2$$
We need to check this identity on $[-1,1]$
1) For $x=-1$:
$\sin^{-1}(-1)+\cos^{-1}(-1)=-\frac{\pi}{2}+\pi=\frac{\pi}{2}$
2) For $x=0$:
$\sin^{-1}0+\cos^{-1}0=0+\frac{\pi}{2}=\frac{\pi}{2}$
3) For $x=1$
$\sin^{-1}1+\cos^{-1}1=\frac{\pi}{2}+0=\frac{\pi}{2}$
So the identity has been proved at $x=-1$, $x=0$ and $x=1$.
4) Prove the identity on $(-1,0)$
The identity has already been proved on $x\in(0,1)$. So if we take $x=-a$, $a\gt0$, then:
- $\sin^{-1}(-a)=-\sin^{-1}a$ (shown in Eqs.2)
- $\cos^{-1}(-a)=\pi-\cos^{-1}a$ (shown in Eqs.4)
So, $\sin^{-1}(-a)+\cos^{-1}(-a)=-\sin^{-1}a+\pi-\cos^{-1}a=-(\sin^{-1}a+\cos^{-1}a)+\pi$
- Since $a\gt0$, $\sin^{-1}a+\cos^{-a}=\frac{\pi}{2}$ has already been proved. Therefore,
$$\sin^{-1}(-a)+\cos^{-1}(-a)=-\frac{\pi}{2+\pi}=\frac{\pi}{2}$$
So the identity is also true as $x\lt0$, or $x\in(-1,0)$ since $x\in[-1,1]$.
In conclusion, the identity $\sin^{-1}x+\cos^{-1}x=\pi/2$ has been shown on $[-1,1]$.
