# Spivak, Ch. 15, Trigonometric Functions", Proof of $\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha \sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$.

The following problem is from Chapter 15 "Trigonometric Functions" from Spivak's Calculus

1. Prove that

$$\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha \sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$

indicating any restrictions on $$\alpha$$ and $$\beta$$.

My question is about the restrictions on $$\alpha$$ and $$\beta$$. I will show the solution from the solution manual first, and then specify my question.

Here is the solution manual solution

From the addition formula for $$\sin$$ we obtain, for $$|\alpha|\leq 1$$ and $$|\beta|\leq 1$$,

$$\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\sin{(\arcsin{\alpha})\cos{(\arcsin{\beta})}}+\cos{(\arcsin{\alpha})\sin{(\arcsin{\beta})}}$$

$$=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}$$

Note that though it is not mentioned a significant step is taken in showing that

$$\cos{(\arcsin{x})}=\sqrt{1-x^2}$$

This is done by computing the derivative of $$\sin(x)$$ as the reciprocal of the derivative of $$\arcsin$$ at $$\sin{x}$$.

Consequently

$$\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}\tag{1}$$

provided that $$-\pi/2\leq\arcsin{\alpha}+\arcsin{\beta}\leq\pi/2$$. [If $$\pi/2<\arcsin{\alpha}+\arcsin{\beta}\leq \pi$$, the right side must be replaced with $$\pi-\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$, and if $$-\pi\leq \arcsin{\alpha}+\arcsin{\beta}\leq -\pi/2$$, replaced with $$-\pi-\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$.]

My question is about the last paragraph.

Let's start at the point where we have

$$\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}$$

and we want to take the $$\arcsin$$ of each side to obtain $$(1)$$.

Now, $$\arcsin{\alpha}$$ and $$\arcsin{\beta}$$ are each in $$(-\pi/2, \pi/2)$$. Their sum is in $$(-\pi, \pi)$$, and the whole left expression is thus in $$[-1,1]$$.

$$\pm 1$$ occur when $$\arcsin{\alpha}+\arcsin{\beta}=\pm \frac{\pi}{2}$$.

If one of these two cases occurs, then $$\arcsin$$ isn't defined for $$\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\pm 1$$.

EDIT: the sentence above is is incorrect. I confused $$\arcsin$$ with its derivative $$\arcsin'$$.

So my first question is about the following snippet of the solution manual solution

provided that $$-\pi/2\leq\arcsin{\alpha}+\arcsin{\beta}\leq\pi/2$$

where the values $$\pi/2$$ and $$-\pi/2$$ are included in the possibilities. Are these values really allowed?

EDIT: given the first edit above, yes, these values are allowed.

My second question is about the discussion when $$\pi/2<\arcsin{\alpha}+\arcsin{\beta}\leq \pi$$ or $$-\pi\leq \arcsin{\alpha}+\arcsin{\beta}\leq -\pi/2$$. What is happening in these cases, I don't know what Spivak's solution manual did there.

• "If one of these two cases occurs, then arcsin isn't defined for sin(arcsinα+arcsinβ)=±1." I am not getting how you commented this.
– user1078285
Commented Jul 18, 2022 at 1:12
• You know what, I think I have been staring at these functions for too long today, and I have confused the derivative of $\arcsin$ with $\arcsin$ itself.
– xoux
Commented Jul 18, 2022 at 1:18
• This silly mistake solves the first question, but the second question remains.
– xoux
Commented Jul 18, 2022 at 1:21
• For the second question, while taking arcsin on both sides of $\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}$,we have to make cases as $\sin${($\arcsin${$\alpha$}=$\alpha$ only when $\pi/2 \leq$ $\alpha$ $\leq\ \pi/2$.
– user1078285
Commented Jul 18, 2022 at 1:35

I'd start from the conditions first.

When it exists, the right-hand side is an angle in the interval $$[-\pi/2,\pi/2]$$, so the left-hand side must as well.

The existence condition is (with $$a=\alpha,b=\beta$$ for ease of typing) $$\textstyle \lvert\, a\sqrt{1-b^2}+b\sqrt{1-a^2}\,\rvert\le 1\tag{*}$$ In particular, $$|a|\le1$$ and $$|b|\le1$$, but this was already known from examining the left-hand side.

Condition (*) is satisfied for $$|a|\le 1$$ and $$|b|\le 1$$ by using Cauchy-Schwarz with the vectors $$\textstyle v=(a,\sqrt{1-a^2}\,)\qquad w=(\sqrt{1-b^2},b)$$ and the standard CS inequality $$|v\cdot w|\le\|v\|\,\|w\|$$ becomes $$\textstyle \lvert\, a\sqrt{1-b^2}+b\sqrt{1-a^2}\,\rvert\le\sqrt{a^2+1-a^2}\sqrt{1-b^2+b^2}=1$$

OK, this leaves only the condition that $$-\frac{\pi}{2}\le \arcsin a+\arcsin b\le\frac{\pi}{2} \tag{**}$$ Under this condition, we can take the sine of both sides. From the left-hand side we get $$\textstyle \sin(\arcsin a)\cos(\arcsin b)+\cos(\arcsin a)\sin(\arcsin b) =a\sqrt{1-b^2}+b\sqrt{1-a^2}$$ because, for $$|x|\le 1$$, we have $$\textstyle\cos(\arcsin x)=\sqrt{1-\sin^2(\arcsin x)}=\sqrt{1-x^2}$$ because $$-\pi/2\le\arcsin x\le\pi/2$$ and so $$\cos\arcsin x\ge0$$.

Thus we conclude that, for $$|a|\le1$$ and $$|b|\le 1$$,

$$\arcsin a+\arcsin b=\arcsin\bigl(a\sqrt{1-b^2}+b\sqrt{1-a^2}\,\bigr)$$ if and only if $$\lvert\arcsin a+\arcsin b\rvert\le \pi/2$$

Let's start at the point where we have

$$\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\tag{1}$$

and we want to take the $$\arcsin$$ of each side to obtain $$(1)$$.

First, let's just note that $$\alpha$$ and $$\beta$$ are each in $$[-1,1]$$, and thus $$\arcsin{\alpha}$$ and $$\arcsin{\beta}$$ are each in $$[-\pi/2,\pi/2]$$. The sum $$(\arcsin{\alpha}+\arcsin{\beta})$$ is thus in $$[-\pi,\pi]$$.

We need to consider what happens for all such possible values of $$(\arcsin{\alpha}+\arcsin{\beta})$$.

Case 1: $$-\pi/2\leq \arcsin{\alpha}+\arcsin{\beta}\leq\pi/2$$

The equation

$$\arcsin{(\sin{(\arcsin{\alpha}+\arcsin{\beta})})}=\arcsin{\alpha}+\arcsin{\beta}\tag{2}$$

is true.

We can thus take the $$\arcsin$$ of both sides of $$(1)$$ and end up with

$$\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$

Case 2: $$\pi/2\leq \arcsin{\alpha}+\arcsin{\beta}\leq\pi$$

Equation $$(2)$$ is no longer true (indeed, it doesn't even make sense because of the domain of $$\arcsin$$).

However, it is true that $$\sin{x}=\sin{\pi-x}$$. Thus,

$$(\pi-\arcsin{\alpha}-\arcsin{\beta}) \in (-\pi/2,0]$$

and

$$\sin{(\pi-\arcsin{\alpha}-\arcsin{\beta})}=\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\tag{3}$$

Now it is true that

$$\arcsin{(\sin{(\pi-\arcsin{\alpha}-\arcsin{\beta})})}=\pi-\arcsin{\alpha}-\arcsin{\beta}$$

So we can take the $$\arcsin$$ of the first and last expressions in $$(3)$$ to obtain

$$\pi-\arcsin{\alpha}-\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$

and the desired

$$\arcsin{\alpha}+\arcsin{\beta}=\pi-\arcsin{(\alpha\sqrt{1-\beta^2}-\beta\sqrt{1-\alpha^2})}$$

Case 3: $$-\pi\leq \arcsin{\alpha}+\arcsin{\beta}<-\pi/2$$

This case is similar to case 2. Since

$$\sin{(-\pi-x)}=\sin{x}$$

we have that

$$(-\pi-\arcsin{\alpha}-\arcsin{\beta}) \in (-\pi/2,0]$$

$$\sin{(-\pi-\arcsin{\alpha}-\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\tag{3}$$

and taking the $$\arcsin$$ of each side we obtain

$$-\pi-\arcsin{\alpha}-\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}-\beta\sqrt{1-\alpha^2})}$$

$$\arcsin{\alpha}+\arcsin{\beta}=-\pi-\arcsin{(\alpha\sqrt{1-\beta^2}-\beta\sqrt{1-\alpha^2})}$$

$$\blacksquare$$

Spivak's solution manual sure does skip a lotta steps.

HINT

• $$\arcsin(\sin\alpha)=\alpha$$ only when $$\alpha\in[-\frac\pi2,\frac\pi2]$$

• $$\arcsin(\sin\alpha)=\pi-\alpha$$ when $$\alpha\in[\frac\pi2,\pi]$$

• $$\arcsin(\sin\alpha)=-\pi-\alpha$$ when $$\alpha\in[-\pi,-\frac\pi2]$$