# 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?

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.

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 $\frac{\pi}{12}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\pi}{3}$.

• I don't think you understand what I mean. 19π/12 how do I get two difference numbers? So that I can use the formula sin(s-t)=sin(s)cos(t) - cos(s)sin(t)? Jul 18 '14 at 19:44
• Yes, you're right. First you can calculate $\sin \frac{3\pi}{2}=\sin(\pi + \frac{\pi}{2})$ using this formula, next $\sin (\frac{3\pi}{2}+\frac{\pi}{12})$. $\sin \frac{\pi}{12}$ and $\cos \frac{\pi}{12}$ you calculate using for example $\sin 2x=2\sin x \cos x$ and $\cos 2x=\cos^2 x- \sin^2 x$.
– agha
Jul 18 '14 at 19:48

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.

• So it is a guessing game? Trial and error? Jul 18 '14 at 19:46
• Why π/4? and 4π/3? Jul 18 '14 at 19:50
• It doesn't have to be Pi/4 and 4*Pi/3. It could be any combination of angles that add up to 19*Pi/12. I just wanted ones whose denominators were 3 and 4, because of the unit circle denominators. You should have a unit circle in your book to look at while you're doing this. Jul 18 '14 at 19:52
• This is not correct. I need two numbers that when I subtract them they equal 19π/2 Jul 18 '14 at 19:57
• Then replace the (+) sign in my above solution with a (-) sign and solve again. So you would need (3x-4y)=19. I think x=9 and y=2 would work. Jul 18 '14 at 19:58

$2\pi$ 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 $\sin \left(\frac{19}{12}\pi\right) = -\frac14 (\sqrt 2 + \sqrt 6) \approx -0.9659$

The denominator is $3\cdot4$ and you (should) know the values for denominators $3$ and $4$, so just decompose in two fractions

$$\frac{4\pi}3+\frac\pi4.$$

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.

$$\frac{19\pi}{12}=\pi+\frac{3\pi}4-\frac{\pi}6$$

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}