# How do you find sin(pi/12) and cos(pi/12)?

Feb 4, 2015

I would use the expansion in series of the two functions, as

Where a function (in a point) is given by an infinite sum of values.
The n! is called "factorial" and $x$ is in radians.

We choose few values only, depending upon the accuracy we want (basically, decimal digits you want).
For your case (3 decimals only):

$\sin \left(\frac{\pi}{12}\right) = \frac{\pi}{12} - {\left(\frac{\pi}{12}\right)}^{3} / \left(3 \cdot 2 \cdot 1\right) + {\left(\frac{\pi}{12}\right)}^{5} / \left(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\right) - \ldots .$
$= \frac{\pi}{12} - \frac{1}{6} {\left(\pi\right)}^{3} / \left({12}^{3}\right) + \frac{1}{120} \frac{{\pi}^{5}}{{12}^{5}} - \ldots = 0.261 - 0.003 + 0.000 \ldots =$

$= 0.258$

Now you can try to do the same by yourself with $\cos$ (which starts at $1$).

hope it helps

Feb 8, 2015

Here is another way to solve this problem.

It's known that ${\sin}^{2} \left(\phi\right) + {\cos}^{2} \left(\phi\right) = 1$ for any angle $\phi$.
Therefore,
${\sin}^{2} \left(\frac{\pi}{12}\right) + {\cos}^{2} \left(\frac{\pi}{12}\right) = 1$

Let's use a formula for a sine of a double angle:
$\sin \left(2 \phi\right) = 2 \cdot \sin \left(\phi\right) \cdot \cos \left(\phi\right)$

Using this formula,
$2 \cdot \sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{12}\right) = \sin \left(2 \cdot \frac{\pi}{12}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

Substitute for simplicity:
$x = \cos \left(\frac{\pi}{12}\right)$ and $y = \sin \left(\frac{\pi}{12}\right)$
Both are positive (since an angle $\frac{\pi}{12}$ is in the first quadrant).
Also $x > y$ since, as angle $\phi$ increases from 0 to $\frac{\pi}{12}$, $\cos \left(\phi\right)$ decreases from $1$ and sin(phi) increases from $0$. They meet only at $\frac{\pi}{4}$, so at $\frac{\pi}{12}$ function $\cos$ is greater than function $\sin$.

We have a system of two equations with two unknowns:
${x}^{2} + {y}^{2} = 1$
$2 x y = \frac{1}{2}$

Adding the second equation to the first, we get
${x}^{2} + 2 x y + {y}^{2} = \frac{3}{2}$ or
${\left(x + y\right)}^{2} = \frac{3}{2}$
Since both $x$ and $y$ are positive
$x + y = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$

Subtracting the second equation from the first, we get
${x}^{2} - 2 x y + {y}^{2} = \frac{1}{2}$ or
${\left(x - y\right)}^{2} = \frac{1}{2}$
Since $x > y$
$x - y = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$

So, we have a very simple system of two equations with two unknowns:
$x + y = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$
$x - y = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$

Adding and subtracting this equations, we find solutions:
$2 x = \frac{\sqrt{6} + \sqrt{2}}{2}$
$2 y = \frac{\sqrt{6} - \sqrt{2}}{2}$

Solutions are
$\cos \left(\frac{\pi}{12}\right) = x = \frac{\sqrt{6} + \sqrt{2}}{4}$
$\sin \left(\frac{\pi}{12}\right) = y = \frac{\sqrt{6} - \sqrt{2}}{4}$

Jun 22, 2016

Use the cosine and sine half angle formulas:

• $\sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(\theta\right)}{2}}$
• $\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \left(\theta\right)}{2}}$

First, let's solve for $\sin \left(\frac{\pi}{12}\right)$. If we let $\theta = \frac{\pi}{6}$, then we see that:

$\sin \left(\frac{\frac{\pi}{6}}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(\frac{\pi}{6}\right)}{2}}$

We will take the positive root since the angle is $\frac{\frac{\pi}{6}}{2} = \frac{\pi}{12}$, which is in the first quadrant, where sine is positive.

color(blue)(sin(pi/12))=sqrt((1-sqrt3/2)/2)=sqrt((2-sqrt3)/4)color(blue)(=sqrt(2-sqrt3)/2

The cosine method is almost identical. The positive root will again be taken because cosine is also positive in the first quadrant.

$\cos \left(\frac{\frac{\pi}{6}}{2}\right) = \sqrt{\frac{1 + \cos \left(\frac{\pi}{6}\right)}{2}}$

Hence:

$\textcolor{red}{\cos \left(\frac{\pi}{12}\right)} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} \textcolor{red}{= \frac{\sqrt{2 + \sqrt{3}}}{2}}$