simplify radicals

Question

In the book $A=B$ (link) page 11 example 1.5.1 it is written

By setting $\sin a = x$ and $\sin b = y$, we see that the identity $\sin(a + b) = \sin a \cos b + \sin b \cos a$ is equivalent to $$ \arcsin x + \arcsin y = \arcsin(x \sqrt{1-y^2} + y \sqrt{1-x^2}) $$ When $x = 0$ this is tautologous, so it suffices to prove that the derivatives of both sides with respect to $x$ are the same. This is a routinely verifiable algebraic identity.

By differentiating w.r.t. $x$ I obtained: \begin{gather}\frac 1 {\sqrt{1-x^2}}+ \frac {\frac{dy}{dx}} {\sqrt{1\!-\!y^2}} =\frac{\frac d {dx}\left (x\sqrt{1\!-\!y^2}+y\sqrt{1\!-\!x^2}\right)} {\sqrt{1-\left (x\sqrt{1\!-\!y^2}+y\sqrt{1\!-\!x^2}\right)^2}}=\frac{\sqrt{1\!-\!y^2}\!-\!\frac {xy}{\sqrt{1\!-\!x^2}}+\left(\sqrt{1\!-\!x^2}\!-\!\frac {xy}{\sqrt{1\!-\!y^2}}\right) \!\frac{dy}{dx}} {\sqrt{1-\left (x\sqrt{1\!-\!y^2}+y\sqrt{1-x^2}\right)^2}} \end{gather}

where $y$ is somewhat arbitrary function of $x$. But I cannot reduce this (especially the radical in the denominator of the RHS) to show that the equality holds. Seems to me there is no way they can be equal so do I just assume it is a mistake in the text or is there someone who can show equality?

He seems to go about it by implying the squares of both sides are equal using Maple code but I can't show that either (and I don't have Maple so i can't test it that way).

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Petkovšek, Marko; Wilf, Herbert S.; Zeilberger, Doron, $A=B$. With foreword by Donald E. Knuth, Wellesley, MA: A. K. Peters. xii, 212 p. (1996). ZBL0848.05002.

Answer 1

The identity is false. The left-hand side can take any value between $-\pi$ and $\pi$, whereas the right-hand side can only take on values in the interval $[-\pi/2,\pi/2]$.

For instance, if $y=1$, you'd like to prove that $$ \arcsin x+\frac{\pi}{2}=\arcsin\sqrt{1-x^2} $$ Is this valid? Not for all $x\in[-1,1]$. The derivatives are not equal. Indeed, the derivative of the right-hand side is $$ \frac{1}{\sqrt{1-(1-x^2)}}\frac{-x}{\sqrt{1-x^2}}=-\frac{x}{|x|}\sqrt{1-x^2} $$ which is the same as $1/\sqrt{1-x^2}$ only for $x\in(-1,0)$.

Check at Desmos: https://www.desmos.com/calculator/kavz3yliyt

Answer 2

Note: Check also egreg's answer about the validity of this equation.

How to proof that $$\arcsin(x)+\arcsin(y)=\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})\tag{1}$$ holds? From the book A=B by Marko Petkovsek, Herbert Wilf, and Doron Zeilberger (Example 1.5.1 on page 11):

When $x = 0$ this is tautologous, so it suffices to prove that the derivatives of both sides with respect to $x$ are the same.

Yes, because if $x=0$ the left side is $\arcsin(y)$ and the right side is $\arcsin(y)$.

So let's check if the derivates of both sides with respect to $x$ are the same. Note that $y$ is not a function of $x$ as the OP assumes, but $x$ and $y$ are independent variables, so $\frac d{dx}y=0$, because $\frac d{dx}\text{const}=0.$ The OP already calculated the derivates and if we consider that $\frac{dy}{dx}=0$ we get:

$$\frac{d}{dx } \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})$$ $$=\frac{\sqrt{1\!-\!y^2}\!-\!\frac {xy}{\sqrt{1\!-\!x^2}}}{\sqrt{1-\left (x\sqrt{1\!-\!y^2}+y\sqrt{1-x^2}\right)^2}}$$ $$=\frac {\sqrt{(1-y^2)(1-x^2)}-xy} {\sqrt{1-x^2)}{\sqrt{1-\left (x\sqrt{1\!-\!y^2}+y\sqrt{1-x^2}\right)^2}}}$$ $$=\frac1{\sqrt{1-x^2}}{\frac {\sqrt{(1-y^2)(1-x^2)}-xy}{\sqrt{1-\left (x\sqrt{1\!-\!y^2}+y\sqrt{1-x^2}\right)^2}}}$$ $$=\pm \frac1{\sqrt{1-x^2}}$$

because $$\Bigg({\frac {\sqrt{(1-y^2)(1-x^2)}-xy}{\sqrt{1-\left (x\sqrt{1\!-\!y^2}+y\sqrt{1-x^2}\right)^2}}}\Bigg)^2=1$$ can be easily shown. So $$\frac {\sqrt{(1-y^2)(1-x^2)}-xy}{\sqrt{1-\left (x\sqrt{1\!-\!y^2}+y\sqrt{1-x^2}\right)^2}}$$ is either $+1$ or $-1$.

But egreg's answer shows us that this must not be the case that the sign is always $+1$.

But note that we have shown that the square of the derivate of the left hand side with respect to $x$ is equal to the square of the derivate of the right hand side with respect to $x$. The book authors leave this proof to Maple.

The book can bee downloaded here.

The book shows the following Maple program

f:=arcsin(x) + arcsin(y) :
g:= arcsin(x*(1-y**2)**( 1/2) + y*(1-x**2)**(1/2));
f1:=diff(f,x): g1:=diff(g,x):
normal(simplify(expand(g1**2))-f1**2);

So the first two lines define the functions that represent the LHS and the RHS of the equations. The third line calculates the derivates of these functions with respect to $x$ and the final line calculates f the difference of squares of these derivates. This should be $0$. The authors assist Maple to do these calculation by calculating the difference of the squares of the derivates instead of the difference of the derivates and by using the commands expand , simplify and normal, which transform expressions to equivalent expression. I assume without these assistance Maple will return an expression containing a lot of square roots, similar to the expressions in our calculation and it is unclear if this expression is equals to $0$ or not. Note again that the authors request from Maple only to show that the squares of the derivattives are equal.

A free computer algebra system is Maxima. Here you can do the calculation with the following commands:

f:asin(x)+asin(y);
g:asin(x*sqrt(1-y^2)+y*sqrt(1-x^2));
f1:diff(f,x);
g1:diff(g,x);
ratsimp(g1^2-f1^2);

The last command does some simplifications and returns $0$.

You can try this out (at your own risk) here