# Use a matrix or equations to find the value of $\sin(\pi/3)$

I was asked to use $\sin(0)=0$, $\sin(\pi/2)=1$, and $\sin(\pi)=0$ to calculate the value of $\sin(\pi/3)$ using matrices or equations. I honestly have no idea how to solve this.

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Well, $\sin \frac{\pi}{2} = 1, \sin \pi = 0$, so I'm not sure how that will help you? – copper.hat Feb 28 '13 at 1:13
Perhaps you're supposed to use the triple-angle formula for sine? But I don't see how $\sin(\pi/2)$ or $\sin(0)$ are relevant. – Zev Chonoles Feb 28 '13 at 1:15
@copper.hat I'm sorry, those values are given, my bad. – ChairOTP Feb 28 '13 at 1:17
@Zev Chonoles. I honestly have no idea, and I'm really confused because this doesn't really have much to do with what we're learning. – ChairOTP Feb 28 '13 at 1:18
Draw the unit circle around the origin in the $xy$ plane and then inscribe a regular hexagon with one vertex at $(1,0).$ Think about it. – Will Jagy Feb 28 '13 at 1:41

So I looked at the answer and apparently they used curve fitting in order to find an approximation of $sin(\frac \pi3)$, and somehow they got to this:

$$P(x)=-\frac {4x^2}{\pi^2}+\frac {4x}{\pi}$$

So I set a table with the values of $x$ as the angle, and the answers as $sin(x)$, and then got a $2x3$ matrix

$$\begin{matrix} \frac {\pi^2}4&\frac \pi2&1 \\ \pi^2&\pi&0\\ \end{matrix}$$ And finally got to to the equation mentioned above, it's not the exact value of $sin(\frac \pi3)$ though, but it's close to it.

Thank you everyone anyway.

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Let denote $a=\cos(\frac{\pi}{3})$ and $b=\sin(\frac{\pi}{3})$ and note that $a$ and $b$ are positive. So $$(a+ib)^3=e^{i\pi}=-1.$$

Now, we expand $(a+ib)^3=a^3+3a^2ib-3ab^2-ib^3=-1$, then we take out the real and imaginary part and we find $$\left\{\begin{array}{llr} a^3-3ab^2&=&-1\\ 3a^2b-b^3&=&0 \end{array}\right.,$$ Hence, we find from the second equation $b^2=3a^2$ and then first equation give $8a^3=1$. Finally, we conclude that $a=\frac{1}{2}$ and $b=\frac{\sqrt{3}}{2}$.

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So basically this is using complex numbers and finding a non-trivial cube root of -1. I like it, but I wonder if the OP could quite appreciate it, seems to me complex numbers and the formula $e^{i\theta} = \cos \theta + i\sin \theta$ may be something the OP is unfamiliar with. +1 from me though :) – adam W Feb 28 '13 at 14:04
Actually I am familiar with the formula. Never used it before, and I doubt this is what my professor was asking for, but this is a really good way to find it! Thank you! – ChairOTP Feb 28 '13 at 15:40
I still have a question though, why is $3a^2b-b^3=0$ instead of $3a^2b-b^3=-1$? – ChairOTP Feb 28 '13 at 15:46
@ChairOTP $3a^2b-b^3$ is the imaginary part of $-1$ so it's $0$. – user63181 Feb 28 '13 at 16:00