I am trying to find an expression for $\sin(40)$ in the following way: Let
$\theta$ = 40
3$\theta$=120
$\sin(3\theta)$=sin(120)
$-4\sin^3(\theta)+3\sin(\theta)=\frac {\sqrt 3} 2 $
Put $\sin(\theta)=x$
$-4x^3+3x=\frac {\sqrt 3}2$
$x^3-\frac 34x=\frac {\sqrt 3}{-8}$
We have to solve this cubic equation. I will brought it in the the form $x^3+q=px$. Now I apply the Cardan Formula.The original formula shows roots for $x^3+px=q$. I am putting $-p$ for $p$ and $-q$ for $q$ to get the formula:
$x=\sqrt[3] {\frac {-q}2+ \sqrt{{\frac{q^2}4}-{\frac{p^3}{27}}}}+\sqrt[3] {\frac {-q}2- \sqrt{{\frac{q^2}4}-{\frac{p^3}{27}}}}$
The quantity under square root reduces to $\frac {-1}{256}$.Upon Simplyfying I found the that
$x=\sqrt[3] \frac {-\sqrt 3 +\sqrt{-1}} {16}+\sqrt[3] \frac {-\sqrt 3 -\sqrt{-1}} {16}$
$x=\frac 12{\sqrt[3] \frac {-\sqrt 3 +\sqrt{-1}} {2}+\sqrt[3] \frac {-\sqrt 3 -\sqrt{-1}} {2}}$
This is one complex root of the cubic. Now I simplified this cube roots:
$-\sqrt 3 +i=\text{(polar form)}\:2,150$
$\sqrt[3] {-\sqrt 3 +i}=2^{1/3},50=2^{1/3}(\cos(50)+i \sin(50))$
$\sqrt[3] {-\sqrt 3 -i}=2^{1/3},10=2^{1/3}(\cos(10)+i \sin(10))$
Am I right here? Otherwise How should I continue to find an expression for sin(40).
Please help.
1 Answer
You'll get a cleaner answer using this identity: $$\sin x=\frac{1}{2} i e^{-i x}-\frac{1}{2} i e^{i x}$$ Thus for $40^\circ=\frac{2\pi}9$, you get: $$\begin{align}\sin 40^\circ&=\frac{1}{2} i e^{-\frac{1}{9} (2 i \pi )}-\frac{1}{2} i e^{\frac{2 i \pi }{9}}\\ &=-\frac{1}{2} (-1)^{5/18} \left((-1)^{4/9}-1\right) \end{align}$$
From $f(x)=-4x^3+3x-\frac{\sqrt3}{2}=0$, we get the roots using Cardano's for $P(x)=ax^3+bx^2+cx+d$ formula using: $$x_1=-\frac{b}{3 a}+S+T\\ x_2=-\frac{b}{3 a}+\frac{1}{2} \left(i \sqrt{3}\right) (S-T)-\frac{S+T}{2}\\ x_3=-\frac{b}{3 a}-\frac{1}{2} \left(i \sqrt{3}\right) (S-T)-\frac{S+T}{2}$$ Where: $$S=\sqrt[3]{\sqrt{Q^3+R^2}+R}\\ T=\sqrt[3]{R-\sqrt{Q^3+R^2}}$$ Where: $$Q=\frac{3 a c-b^2}{9 a^2}\\ R=\frac{-27 a^2 d+9 a b c-2 b^3}{54 a^3}$$ For $f(x)$, we get: $$Q=-\frac{1}{4}\\ R=-\frac{\sqrt{3}}{16}$$ For $S$ and $T$, we get: $$S=\sqrt[3]{-\frac{\sqrt{3}}{16}+\frac{i}{16}}\\ T=\sqrt[3]{-\frac{\sqrt{3}}{16}-\frac{i}{16}}$$ And thus we get roots $x_1,x_2,x_3$ as: $$x_1=\frac{\sqrt[3]{-\sqrt{3}-i}+\sqrt[3]{-\sqrt{3}+i}}{2 \sqrt[3]{2}}\\ x_2=\frac{\left(-1-i \sqrt{3}\right) \sqrt[3]{-\sqrt{3}-i}+i \sqrt[3]{-\sqrt{3}+i} \left(\sqrt{3}+i\right)}{4 \sqrt[3]{2}}\\ x_3=\frac{i \left(\left(-\sqrt{3}+i\right)^{4/3}-\left(-\sqrt{3}-i\right)^{4/3}\right)}{4 \sqrt[3]{2}}$$ Evaluating these numerically, we get that $x_1$ is the only correct answer. Take not too that all these roots are real because $D<0$
