So some time ago, I asked this question on finding the exact value of the sine of ten degrees. While I did get the wrong answer, I was wondering if I would be able to get the correct value of the sine of ten degrees using the method that I had used. Here is my attempt at doing so:$\def\a#1{#1\unicode{xB0}}$
$$\sin(30\unicode{xB0})$$$$=\sin(\a{10}+\a{20})$$$$=\sin(\a{10})\cos(\a{20})+\cos(\a{10})\sin(\a{20})$$$$=\sin(\a{10})(1-2\sin(\a{20}))+2\sin(\a{10})\cos(\a{10})\cos(\a{10})$$$$=\sin(\a{10})-2\sin^3(\a{10})+2\sin(\a{10})-2\sin^3(\a{10})$$$$=3\sin(1\a0)-4\sin^3(1\a0)$$Now, letting$$1\a0=\dfrac x3$$$$3\sin(1\a0)-4\sin^3(1\a0)\gets3\sin\left(\dfrac x3\right)-4\sin^3\left(\dfrac x3\right)$$Now, we set what we have gotten equal to $\sin(x)$:$$\sin(x)=3\sin\left(\dfrac x3\right)-4\sin^3\left(\dfrac x3\right)$$
Then, we put this into a cubic formula situation:$$4\sin^3\left(\dfrac x3\right)-3\sin\left(\dfrac x3\right)+\sin(x)=0$$$$\dfrac{\sin^3\left(\dfrac x3\right)}{\dfrac14}-\dfrac{\sin\left(\dfrac x3\right)}{\dfrac13}+\sin(x)=0$$$$4\sin^3\left(a\right)-3\sin(a)+\sin(3a)=0$$Now here's something that I then found out when writing this out:$$4\sin^3(x)-3\sin(x)\equiv-\sin(3x)$$$$\therefore-\sin(3x)+\sin(x)=0$$$$\implies-\sin(x)+\sin\left(\dfrac x3\right)=0$$But I realized that this might be wrong, so I went back to the equation and did this:$$4\sin^3\left(\dfrac x3\right)-3\sin\left(\dfrac x3\right)+\sin(x)=0$$Now, according to De Moivre's Formula:$$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$$Now you may be asking, how does this relate to the exact value of $\sin(10\unicode{xB0})$ when you're already doing the route of the cubic formula solution? Here's how:
We don't need to use the cubic formula. Instead, we expand what we have using De Moivre's Formula so then we can plug in $30$ for $x$ and then simplify to get the exact value of $\sin(10\unicode{xB0})$, and here's how we do that:$$4\sin^3\left(\dfrac x3\right)-3\sin\left(\dfrac x3\right)+\sin(x)\equiv-4i\left(i\sin^3\left(\dfrac x3\right)\right)$$$$-\left(-3i\left(i\sin\left(\dfrac x3\right)\right)\right)+\sin(x)$$$$\equiv-4i\left((\cos(x)+i\sin(x))^2(\cos(x)-i\sin(x))\right)^\frac13+4i\cos^3\left(\dfrac x3\right)$$$$-\left(-3i\left(\cos(x)+i\sin(x)\right)^\frac13\right)-3i\cos\left(\dfrac x3\right)+\sin(x)$$$$\text{Which can be rearranged as }-4i\left((\cos(x)+i\sin(x))^2(\cos(x)-i\sin(x))\right)^\frac13$$$$+4i\cos^3\left(\dfrac x3\right)+3i(\cos(x)+i\sin(x))^\frac13-3i\cos(x)+\sin(x)$$Now plugging in thirty for $x$ and simplifying, we get$$-4i\left(\dfrac{\sqrt3+i}2\right)^\frac13-2i\sqrt3+3i\left(\dfrac{\sqrt3+i}2\right)^\frac13-\dfrac{3i\sqrt3}2+2.5\equiv\sin(10\unicode{xB0})$$
My question
Is my solution for the value of $\sin(10\unicode{xB0})$ correct using the method that I have used here, or what could I do to achieve the solution/achieve it more quickly?
Mistakes I might have made
- The expansion using De Moivre's Formula
- Simplifying the expression after expanding using De Moivre's Formula
