How do I reduce radian fractions? For example, I need to know $\sin (19π/12)$. 
I need to use the subtraction formula. How do I get $(\text{what}) - (\text{what}) = 19π/12$? I am stuck at what are the radians
Do I divde it by something? What is the process? 
 A: After perusing the answers, I think the challenge you're facing has to do mostly with being comfortable manipulating rational numbers (i.e. fractions). There is no special method or trick here, you need to rearrange a rational number into the sum or difference of other rationals.
However, there is ONE guiding principle. Since you're looking to solve these problems by hand, and since you expect a somewhat clean answer, your goal should be to decompose whatever number $x\pi$ radians into a sum that involves targeted radian quantities that lead to simple trigonometric values. That is, you want to wind up with combinations of $\frac{1}6\pi, \frac{1}3\pi, \frac{1}{4}\pi, \frac12\pi, 1\pi$ and other common angles that you've seen on diagrams of the unit circle.

Let's talk through the example you give. To make it nicer, just keep the $\pi$ multiplied on the right of the fraction. You need to focus on the rational coefficient.
$$\sin\left(\frac{19}{12}\pi\right)$$
Your question specifically wants a difference of two numbers. In this case, we're inspired by $\frac13=\frac{4}{12}$, $\frac14=\frac{3}{12}$, and $\frac{1}{6}=\frac{2}{12}$ since the denominator is a $12$. Adding the second one to the original gives an even numerator, which might reduce nicely. We find that
$$\frac{3}{12} + \frac{19}{12} = \frac{22}{12} = \frac{11}{6}$$
which is another target unit circle angle. Rearranging this result, we have
$$\frac{19}{12} = \frac{11}{6}-\frac{1}{4}$$
so
$$\sin\left(\frac{19}{12}\pi\right) = \sin\left(\frac{11}{6}\pi-\frac{1}{4}\pi\right) = \sin\frac{11}{6}\pi\cos\frac{\pi}{4} - \cos\frac{11}{6}\pi\sin\frac{\pi}{4} $$
all of which can be related to special right triangles as found on the unit circle.

I wrote this last part earlier. This is another approach, which isn't as quick, but hopefully is instructive of the thinking that goes into these problems. This solution is similar to agha's.
Observe that the denominator is $12$. That's related to $3$ and $4$, which are involved in some of our target values mentioned in paragraph two. Also observe that the number $19/12$ is larger than $1$, so we can pull out this $1$ as follows:
$$\frac{19}{12} = \frac{12}{12} + \frac{7}{12} = 1+ \frac{7}{12}$$
Okay cool. We have split our original fraction into two nicer ones, so we can use a sum formula as follows 
$$\sin\left(\frac{19}{12}\pi\right) = \sin\left(\pi + \frac{7}{12}\pi\right)=\sin\pi\cos\frac{7}{12}\pi + \cos\pi\sin\frac{7}{12}\pi$$
The argument $\pi$ is a nice one, but $\frac{7}{12}\pi$ isn't. That's alright, since $7=3+4$. We can separately use an angle addition formula with
$$\frac{7}{12} = \frac{3}{12} + \frac{4}{12} = \frac14+ \frac13$$
Here we decomposed the ugly fraction into two target ones. You can use another sum formula for $\frac{7}{12}\pi$ and then plug that answer back into the sum formula from the previous step.
A: Note that $\frac{19\pi}{12}=\pi+\frac{7\pi}{12}=\pi+\frac{\pi}{12}+\frac{\pi}{2}$.
You know $\sin$ and $\cos$ of $\pi$, $\frac{\pi}{2}$ and $\frac{\pi}{12}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\pi}{3}$.
A: Well, $4 \cdot 3=12$, so you could write $$\frac{x}{4}+\frac{y}{3}=\frac{3x+4y}{12}=\frac{19}{12}.$$  Then try to think of an $x$ and $y$ that work.  Well $y=4$ and $x=1$ works.  So you could write that $$\frac{19\pi}{12}=\frac{\pi}{4}+\frac{4\pi}{3}.$$  I think you should be able to use sum and difference formulas from there!
By the way, I picked $4$ and $3$ as denominators because they are commonly used denominators on unit circles.
A: $$\frac{19\pi}{12}=\pi+\frac{3\pi}4-\frac{\pi}6$$
Since $\sin(\pi+x)=-\sin x$, $\sin\left(\frac{19\pi}{12}\right)=-\sin\left(\frac{3\pi}4-\frac{\pi}6\right)$.
And since $\sin(A-B)=\sin A\cos B-\cos A\sin B$:
$$\begin{align}
-\sin\left(\frac{3\pi}4-\frac{\pi}6\right)&=-\left(\sin\left(\frac{3\pi}4\right)\cos\left(\frac{\pi}6\right)-\cos\left(\frac{3\pi}4\right)\sin\left(\frac{\pi}6\right)\right) \\
&=-\left(\frac{\sqrt{2}}2*\frac{\sqrt{3}}2-\left(-\frac{\sqrt{2}}2 * \frac12\right)\right) \\
&=-\frac{\sqrt{2}+\sqrt{6}}4
\end{align}$$
A: $2\pi$ radians represents a full circle turn so 


*

*$\sin(2\pi+\theta)=\sin(\theta)$ and 

*$\sin(2\pi-\theta)=\sin(-\theta)=-\sin(\theta)$ 


So $\sin \left(\frac{19}{12}\pi\right) = \sin \left(2\pi -\frac{5}{12}\pi\right) = \sin \left(-\frac{5}{12}\pi\right) = -\sin \left(\frac{5}{12}\pi\right)$ 
which since $\frac{5}{12}\pi=\frac{1}{6}\pi+\frac{1}{4}\pi$  makes $\sin \left(\frac{5}{12}\pi\right) = \sin \left(\frac{\pi}{6}+\frac{\pi}{4}\right) =\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}\right) =\frac12 \sqrt{\frac12}+\frac{\sqrt{3}}{2} \sqrt{\frac12}$ and so gives $\sin \left(\frac{19}{12}\pi\right) = -\frac14 (\sqrt 2 + \sqrt 6) \approx -0.9659$
A: The denominator is $3\cdot4$ and you (should) know the values for denominators $3$ and $4$, so just decompose in two fractions
$$\frac{4\pi}3+\frac\pi4.$$
Now apply the addition formula.

This can be obtained by solving
$$\frac a3+\frac b4=\frac{19}{12}$$ or
$$4a+3b=19.$$
You can work this out by trial and error.
