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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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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