How to refactor $\pi / 12$ into $\pi /4 - \pi /3$ (standard calculus angles)? How do you know that $\pi / 12$ converts exactly into $\pi /4 - \pi /3$? Do I need to memorize that or is there a simple procedure to follow in order to refactor any angle into combination of standard calculus angles ($\pi /4, \pi /3, \pi /6$) for which sin/cos values are well known?
Similar for $7\pi / 12 = \pi /4 + \pi /3$
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$\begingroup$I'm assuming you are able to go from $\pi/4-\pi/6$ to $\pi/12$, but are asking about how to go the other way.
Maybe you will find it easier if you think in degrees: $\pi$ is $180^{\circ}$, so $\pi/12$ is $15^{\circ}$, which is clearly representable in terms of "standard" angles as $\color{blue}{45^\circ} - \color{red}{30^\circ}$, i.e. $\color{blue}{\pi/4}-\color{red}{\pi/6}$. Similarly, $7\pi/12$ is $105^\circ$, which is representable in terms of standard angles as $45^\circ+60^\circ$, i.e. $\pi/4+\pi/3$.
By thinking in degrees like this, it may help you more easily spot how to split up angles into a sum of standard angles.
(If using this method, it's good to keep in mind standard angles' conversions between degrees and radians, like $\pi/6\leftrightarrow 30^\circ,\pi/4\leftrightarrow 45^\circ,\pi/3\leftrightarrow 60^\circ$, etc.)
$\endgroup$ $\begingroup$I can write the following system: $$\left\{\begin{matrix} a\cdot b=12 \\ a+b=1 \end{matrix}\right.$$With $a,b \in Z$. This have solutions, for: $b\cdot (-b+1)=12$; so $b^2-b+12=0$ and $(b-4)\cdot (b+3)=0$; $b=+4 \lor b=-3$. Also: $a=-3 \lor a=4$. From this, I have:$$\frac{\pi}{12}=\frac{\pi}{a}+\frac{\pi}{b}=\frac{\pi}{4}-\frac{\pi}{3}$$The result is the same for others values of $a$ and $b$. For $7\pi / 12 = \pi /4 + \pi /3$, the system is:$$\left\{\begin{matrix} a\cdot b=12 \\ a+b=7 \end{matrix}\right.$$With the same solutions of your post.
$\endgroup$ $\begingroup$A good place to start is by expressing each of the standard angles as multiples of $\frac{\pi}{12}$.\begin{align*} \frac{\pi}{6} & = \frac{2\pi}{12}\\ \frac{\pi}{4} & = \frac{3\pi}{12}\\ \frac{\pi}{3} & = \frac{4\pi}{12} \end{align*}From there, you can immediately see that \begin{align*} \frac{\pi}{12} & = \frac{\pi}{4} - \frac{\pi}{6}\\ & = \frac{\pi}{3} - \frac{\pi}{4} \end{align*}and that \begin{align*} \frac{5\pi}{12} & = \frac{\pi}{4} + \frac{\pi}{6}\\ \frac{7\pi}{12} & = \frac{\pi}{3} + \frac{\pi}{4} \end{align*}For a larger angle such as $17\pi/12$, look for linear combinations of the special angles. For instance, \begin{align*} \frac{17\pi}{12} & = \frac{9\pi}{12} + \frac{8\pi}{12}\\ & = \frac{3\pi}{4} + \frac{2\pi}{3} \end{align*}It is less useful to express $17$ as $10 + 7$ since $7$ is not a multiple of $2$, $3$, or $4$. Similarly, it is not helpful to express $17$ as $11 + 6$, $12 + 5$, $13 + 4$, $14 + 3$, or $16 + 1$ since one of the summands is not a multiple of $2$, $3$, or $4$. On the other hand, we can express $17$ as $15 + 2$, which yields\begin{align*} \frac{17\pi}{12} & = \frac{15\pi}{12} + \frac{2\pi}{12}\\ & = \frac{5\pi}{4} + \frac{\pi}{6} \end{align*}
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