Trigonometry: Value of constants I have this equation where $A\sin b=B\sin(a-b)$
I have to find the value of A and B. Is there any other way except for assuming $A=\sin(a-b)$ (if it's even correct in the first place) even though it is giving me the answer I want?
Note the $\sin$ values are not zero neither the constants. The a and b are specific points, nos. radians,...
edit:$sin(a)$ and $sin(a-b)$ is a constant value except zero.
 A: If you have specific values of $a$ and $b$, plug them into
$$A\sin(b) = B\sin(a-b) \tag{1}\label{eq1}$$
to get something like
$$Ac_1 = Bc_2 \tag{2}\label{eq2}$$
where $c_1 = \sin(b)$ and $c_2 = \sin(a-b)$. If $c_1$ and $c_2$ are not $0$, then there is no unique value for $A$ and $B$. Instead, you can choose one of the values and get the other one in terms of the first. In particular, you can assume a value for $B$ and get $A = \frac{Bc_2}{c_1}$, or assume a value for $A$ and get $B = \frac{Ac_1}{c_2}$.
A: $$A\sin b=B\sin(a-b)\iff \frac {A}{B}= \frac {\sin (a-b)}{\sin b}$$
$$A=\lambda  \sin (a-b)$$ and$$ B=\lambda \sin b$$ where $\lambda $ is an arbitrary real number.
A: Assuming that $a, b, A,$ and $B$ are parts of a triangle (which imposes some restrictions on their values) by Law of Sines we have
$$\frac{\sin a}{A} = \frac{\sin b}{B}.$$
If we un-cross-multiply your equation, we have
$$\frac{\sin (a-b)}{A} = \frac{\sin b}{B}.$$
We conclude that $\sin a = \sin(a-b).$
Under our restrictions, this forces $b=0$.
