Changed codomain of inverse trigonometric functions

Question

If codomain of $\arcsin(x)$ is $(\pi/2 , 3\pi/2)$ and codomain of $\arccos(x)$ is $(\pi , 2\pi)$ then what should be $\arcsin + \arccos$ equal to ?

I thought of putting $x = \sin \theta$

But then stuck , what to do next ?

Answer 1

Let $\theta=\sin^{-1}x$. Then, $$x=\sin\theta$$ $$x=\cos\left(\frac{\pi}2-\theta\right)$$ $$\cos^{-1}x=\cos^{-1}\left(\cos\left(\frac{\pi}2-\theta\right)\right)$$

Since $\theta=\sin^{-1}x$, we have $\theta\in\left(\frac{\pi}2,\frac{3\pi}2\right)$. $$\frac{\pi}2-\theta\in(-\pi,0)$$ $$\cos\left(\frac{\pi}2-\theta\right)=\cos\left(\frac{\pi}2-\theta+2\pi\right)$$ where $\frac{\pi}2-\theta+2\pi\in(\pi,2\pi)$.

Thus, $$\cos^{-1}x=\cos^{-1}\left(\cos\left(\frac{\pi}2-\theta+2\pi\right)\right)=\frac{5\pi}2-\theta=\frac{5\pi}2-\sin^{-1}x$$

$$\sin^{-1}x+\cos^{-1}x=\frac{5\pi}2$$

Answer 2

I'll denote by $\arcsin_s$ and $\arccos_s$ these two “shifted” functions.

Since $\arcsin_s x+\arccos_s x$ is easily seen to be constant, just compute it for $x=0$.

Why is it constant? Because the derivative of $\arcsin_s x$ is $1/\sqrt{1-x^2}$ and the derivative of $\arccos_s x$ is $-1/\sqrt{1-x^2}$ (for $x\in(-1,1)$).