# A Trigonometric Identity And Restrictions on Domains

Show that: $$\arcsin y=\arccos \sqrt{1-y^2}$$

We know that the domain of $\arcsin$ is $[-1,1]$, and the domain of $\arccos$ is also $[-1,1]$. Since,$\sqrt{1-y^2} \geq 0$, therefore $0 \leq \sqrt{1-y^2} \leq 1$, and if the identity is to be true, then $0\leq y \leq 1$. For example, if $y=-\frac{1}{\sqrt{2}}$, then $\arcsin(-\frac{1}{\sqrt{2}})=-\frac{\pi}{4} \neq \arccos(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$. So I think it is necessary that one provide the condition $0 \leq y \leq 1$ in the statement of the problem, but there is no such condition given in my textbook, and I want to confirm whether what I am thinking is right, or condition should be taken as "understood". Please help!

• See this. – Git Gud Mar 8 '15 at 16:50

## 1 Answer

Let $T$ be a right triangle with angle $\theta$, opposite leg of length $y$ and hypotenuse length $1$. Then $\sin(\theta) = y$, so $\theta = \arcsin(y)$. However, we can solve for the length of the adjacent leg with the pythagorean theorem to get $\cos(\theta) = \sqrt{1-y^2}$ and $\theta = \arccos\sqrt{1-y^2}$. You have two equations for $\theta$, so the result you want follows.

As for the discrepancy you noted when $y$ is negative, I think this is the result of a difference in the ranges of arcsine and arccosine. Arcsine has a range of $[-\pi/2,\pi/2]$ while arccosine has a range of $[0,\pi]$. It's not inherently obvious, but it seems to be remedied easily enough by adding to the equation: $$\arcsin(y) = \text{sgn}(y)\cdot\arccos\left(\sqrt{1-y^2}\right)$$ where $\text{sgn}(y)=0$ if $y=0$, $\text{sgn}(y)=-1$ if $y<0$ and $\text{sgn}(y)=1$ if $y>0$.

So you are right to question negative $y$ values, and I think the text book is a bit lazy for not saying something.

• We must have $0 \leq y \leq 1$, right? – Samama Fahim Mar 8 '15 at 17:11
• @SamamaFahim ah I see what you are getting at. I'll edit my question. – graydad Mar 8 '15 at 17:18
• What I mean is this: We know that $\sin^2 \alpha + \cos^2 \alpha = 1$, where if $\alpha = \arccos x$ and $\alpha = \arcsin y$, then $\alpha \in [0, \pi] \cap [-\pi/2, \pi/2]$ – Samama Fahim Mar 8 '15 at 17:49