<s>[INST] Proving uniqueness of solutions to $\sin^2A + \sin^2B = \sin (A+B)$ without using multivariable calculus In the course of solving a trigonometric problem (see $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle), in one approach the following equation needed to be solved:
$$\boxed{\sin^2A + \sin^2B = \sin (A+B)}$$
subject to $A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),\pi-(A+B) > \max(A,B)$, i.e. $A$ and $B$ are the two angles of a triangle not opposite the longest side.
Clearly, any $A,B\:|\:A+B=\frac{\pi}{2}$ is a family of solutions. Since multivariable calculus is presumably beyond the level of the original problem: 
How to prove that there are no other solutions $\underline{\text{without}}$ using multivariable calculus?
[I don't think a trig identity will suffice as there are other solutions if the restrictions on $A,B$ are relaxed.]

(For completeness - using multivariable calculus)
Part 1
Proof that $\sin^2A + \sin^2B < \sin (A+B)$ over region $R_1=\{0<A,B\land A+B<\frac{\pi}{2}\}$. 
Consider $$f(x,y)=\sin^2x + \sin^2y - \sin (x+y)$$ 
Then
$$\begin{align}
f_x &= \sin 2x - \cos(x+y) \\
f_y &= \sin 2y - \cos(x+y) \\
f_{xx} &= 2\cos 2x + \sin(x+y) \\
f_{yy} &= 2\cos 2y + \sin(x+y) \\
f_{xy} &= \sin(x+y) &= f_{yx} \\
\end{align}$$
For local extrema we require $f_x=f_y=0$. But
$$f_x=f_y=0 \implies \sin2x=\sin2y \implies y=x \lor y=\frac{\pi}{2}-x$$
Exclude $y=\frac{\pi}{2}-x$ as it violates the restriction that $x+y<\frac{\pi}{2}$.
If $y=x$, $f_x=0 \implies \sin2x=\sin2y \implies x=y=\frac{\pi}{8}$.
The determinant of the Hessian is 
$$D(x,y) = \begin{vmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{vmatrix} = f_{xx}f_{yy} - f_{xy}f_{yx} = f_{xx}f_{yy} - (f_{xy})^2$$
At $P(\frac{\pi}{8},\frac{\pi}{8})$, we have 
$$\begin{align}
f_{xx}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{yy}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{xy}&=\sin\frac{\pi}{4}=\frac{1}{\sqrt2} \\
D(x,y)&=\frac{9}{2}-\frac{1}{2}=4
\end{align}$$
Since both $f_{xx}$ and $D$ are positive at $P$, this is a local minimum (with $f(x,y)|_P=1-\sqrt2$).
On the boundaries:


*

*$f(0,y)=\sin^2y-\sin y<0$ on $x=0,y\in(0,\frac{\pi}{2})$

*$f(x,0)=\sin^2x-\sin x<0$ on $x\in(0,\frac{\pi}{2}),y=0$

*$f(0,0)=0$

*$f(x,y)=0$ for $x,y\geq0,x+y=\frac{\pi}{2}$


Since $(0,0)$ is not part of the domain, and there are no other local extrema, we have $f(x,y)<0$ over $R_1$.
Part 2
It will be sufficient to prove that $\sin^2A + \sin^2B > \sin (A+B)$ over region $R_2=\{0<A,B<\frac{\pi}{2} \land A+B>\frac{\pi}{2}   \}$.

 [/INST] Well, fix $B$ and show that there's a unique solution for $A$.  In other words, we want to show that for $0 < B < \pi/2$ the function
$$f_B(A) = \sin^2(A) + \sin^2(B) - \sin(A+B)$$
has a unique zero on the interval $(0, \pi/2)$.
First, note that
$$f_B(0) = \sin^2(B) - \sin(B) = \sin(B)[\sin(B) - 1] < 0$$
since $0 < \sin(B) < 1$.  Now consider
\begin{align*}
f'_B(A) &= 2\sin(A) \cos(A) - \cos(A+B) \\
  &= \sin(2A) - \cos(A+B)
\end{align*}
We have
$$f'(0) = - \cos(B) < 0$$
so $f_B$ starts out negative and decreasing.  Now, where can $f'_B$ vanish?  Well, that happens when
$$\sin(2A) = \cos(A+B).$$
Now $\sin(2A)$ increases from zero to one on $(0, \frac{\pi}{4})$ while $\cos(A+B)$ decreases from $\cos(B) > 0$ to $\cos(B + \frac{\pi}{4})$, which is between 0 and 1, on the same interval.  Consequently there is some solution to this equation with $A \in (0, \pi/4)$.  Moreover, there is only one solution on $(0, \pi/4)$, since to the right of this solution we have $\sin(2A)$ increasing and $\cos(A+B)$ decreasing.
What about solutions to 
$$\sin(2A) = \cos(A+B).$$
when $\frac{\pi}{4} < A < \frac{\pi}{2}$?  Well,
$$\cos(A+B) < \cos(A) < \sin(2A)$$
on this interval.
So we've shown that on the interval $(0, \pi/2)$ the function $f'_B(A)$ has a unique zero.  Together with the facts that $f_B(0) < 0$ and $f'_B(0) < 0$ this means that on $(0, \pi/2)$ the function $f_B$ decreases from a negative value at 0 to a negative minimum, and then increases for the rest of the interval.  Consequently, $f_B$ can have at most one zero on this interval.

I suspect if you also object to single-variable calculus you could eliminate it from the argument and do things entirely with trigonometry, although the argument would necessarily be longer.
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