Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees

One can deduce this identity from the conjunction of the law of sines and the law of cosines.

The law of sines says that for the three angles $A$, $B$, $C$ of a triangle, with opposite sides $a$, $b$, $c$, we have $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d. $$ The last equality merely defines $d$, and one can omit it and still have a statement of the law of sines. The common value $d$ is actually the diameter of the circumscribed circle.

If things are scaled so that $d=1$, then we have

$$a=\sin A,\quad b=\sin B,\quad c=\sin C\tag{1}$$

The law of cosines says $$ a^2+b^2-2ab\cos C = c^2.\tag{2} $$ Substitute the expressions in $(1)$ into the appropriate places in $(2)$ and you get $$ \sin^2 A+\sin^2 B - 2\sin A\sin B\cos C = \sin^2 C $$ and there's your identity.

This still leaves the problem of how to prove the law of sines and the law of cosines. And if you want to use your identity to prove the law of cosines, then this reasoning would be circular. But if you can take the two laws to be established already, then this does it.

Let's take the Right Hand Side. $$2 \sin B \cos C = \sin(B+C) + \sin(B-C) = \sin A + \sin(B-C)$$ Therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin A \sin(B-C)$ Now, $$2 \sin A \sin (B-C) = \cos (A - B +C) - \cos (A + B -C) = \cos 2C - \cos 2B = 2 \sin^2 B - 2 \sin ^2 C$$ Cancelling the $2$ gives us therefore, that $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$.

To finish, we already established $2 \sin A \sin B \cos C = \sin^2 A = \sin A \sin (B-C)$. Now, since $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$, therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin^2 B - \sin^2 C$.

$\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin(B+C)\sin(B-C)$ either using the identity $\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$

or $\sin^2B-\sin^2C=\frac{1}{2}(2\sin^2B-2\sin^2C)=\frac{1}{2}(1-\cos2B-(1-\cos2C))=\frac{1}{2}(\cos2C-\cos2B)=-\frac{1}{2}2\sin(B+C)\sin(C-B)=\sin(B+C)\sin(B-C)$
as $\sin(-x)=-\sin(x)$

Now, $\sin(B+C)=\sin(180^\circ-A)=\sin{A}$

So, $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin{A} \sin(B-C)$

$=\sin{A}(\sin{A}+\sin(B-C))$

$=\sin{A}(\sin(B+C)+\sin(B-C))$ replacing $\sin{A}$ with $\sin(B+C)$

$=\sin{A}(2\sin{B}\cos{C})$

$=2\sin{A}\sin{B}\cos{C}$

This may not be the shortest way, but is pretty systematic and doesn't involve any real trickery. First, we eliminate the $C$ angles from the equation, using that $C = 180 - A - B$. We write $\sin(C) = \sin(180 - A - B) = \sin(A + B)$, and $\cos(C) = \cos(180 - A - B) = -\cos(A + B)$ and what you need to prove is $$\sin^2A + \sin^2B - \sin^2(A + B) = -2\sin A\sin B \cos(A+B)$$ Inserting the sine and cosine addition formulas in this, what you need to prove becomes $$\sin^2A + \sin^2B - (\sin A\cos B + \cos A \sin B)^2 = -2\sin A \sin B(\cos A \cos B - \sin A \sin B)$$ Writing this out, it becomes $$\sin^2A + \sin^2B - \sin^2 A\cos^2 B - 2\sin A \cos A \sin B \cos B -\cos^2 A \sin^2B$$ $$ = -2\sin A\sin B \cos A \cos B + 2\sin^2 A \sin^2 B$$ Cancelling terms, your goal is to prove $$\sin^2A + \sin^2B - \sin^2 A\cos^2 B-\cos^2 A \sin^2B = 2\sin^2 A \sin^2 B$$ Equivalently, $$\sin^2A(1 - \cos^2B) + \sin^2B(1 - \cos^2A) = 2\sin^2 A \sin^2 B$$ This last equation is true since $1 - \cos^2 = \sin^2$. So going backwards in the above steps, one obtains the original equation $\sin^2A + \sin^2B - \sin^2C = 2\sin A \sin B \sin C$ as desired.