7
$\begingroup$

$\def\a#1{#1\unicode{xB0}}$ So I was looking through the homepage of Youtube to see if there were any math problems that I thought that I might be able to solve when I came across this video by Blackpenredpen which was about finding the exact value of $\sin(10\unicode{xB0})$ knowing that $\sin\left(\dfrac x2\unicode{xB0}\right)=\pm\sqrt{\dfrac12(1-\cos(x))}$ which I thought that I might be able to do. Here is my attempt at finding the exact value of $\sin(\a{10})$:$$\sin(\a{30})$$$$=\sin(\a{10}+\a{20})$$$$=\sin(\a{10})\cos(\a{20})+\cos(\a{10})\sin(\a{20})$$$$=\sin(\a{10})(1-2\sin(\a{20}))+2\sin(\a{10})\cos(\a{10})\cos(\a{10})$$$$=\sin(\a{10})-2\sin^3(\a{10})+2\sin(\a{10})-2\sin^3(\a{10})$$$$=3\sin(1\a0)-4\sin^3(1\a0)$$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,$$-4\left(\sin\left(\dfrac x3\right)\right)^3+3\left(\sin\left(\dfrac x3\right)\right)-\sin(x)=0$$$$4\left(\sin\left(\dfrac x3\right)\right)^3-3\left(\sin\left(\dfrac x3\right)\right)+\sin(x)=0$$$$\left(\sin\left(\dfrac x3\right)\right)^3-\dfrac34\left(\sin\left(\dfrac x3\right)\right)+\dfrac14\sin(x)=0$$Now, for any $y^3+\color{red}{p}y+\color{red}{q}=0$,$$y=\sqrt[3]{-\dfrac{\color{red}{q}}2+\sqrt{\dfrac{\color{red}{q}^2}4+\dfrac{\color{red}{p}^3}{27}}}+\sqrt[3]{-\dfrac{\color{red}{q}}2-\sqrt{\dfrac{\color{red}{q}^2}4+\dfrac{\color{red}{p}^3}{27}}}$$$$\therefore\sin\left(\dfrac x3\right)=\sqrt[3]{-\dfrac12\left(\dfrac14\sin(x)\right)+ \sqrt{\dfrac14\left(\dfrac14\sin(x)\right)^2+\dfrac1{27}\left(-\dfrac34^3\right)}}$$$$+\sqrt[3]{-\dfrac12\left(\dfrac14\sin(x)\right)^2-\sqrt{\dfrac14\left(\dfrac14\sin(x)\right)^2+\dfrac1{27}\left(-\dfrac34\right)^3}}$$$$=\dfrac12\sqrt[3]{-\sin(x)\pm\sqrt{-\cos^2(x)}}$$$$\dfrac12\sqrt[3]{-\sin(x)\pm i\cos(x)}$$Meaning that $\sin\left(\dfrac x3\right)$ is equal to$$\dfrac12\sqrt[3]{-\sin(x)\pm i\cos(x)}$$Now, when we plug in $\a{30}$ for $x$ (since $\dfrac{30}3=10$) we get$$\sin(\a{10})=\dfrac12\sqrt[3]{-\sin(\a{30})\pm i\cos(\a{30})}$$$$=\dfrac12\sqrt[3]{-\dfrac12\pm\dfrac{i\sqrt3}2}$$$$=\dfrac12\sqrt[3]{-\dfrac{1\pm i\sqrt3}2}$$$$\sin(1\a0)=\dfrac i2\sqrt[3]{\dfrac{1\pm i\sqrt3}2}$$


My question


Is my solution for the exact value of $\sin(1\a0)$ correct, or what could I do to attain the correct solution/attain it more easily?



Mistakes that I might have made


  1. Trigonometric identities
  2. Cubic formula
  3. Definition of the cubic formula for any $y+py+q=0$
  4. Incorrect that I put
$\endgroup$
7

1 Answer 1

4
$\begingroup$

Let $$\omega = \cos\left({10^o} \right) + i \sin\left(10^o\right)$$ Then $$\omega^3 = \cos(30^o) + i \sin(30^o) = \frac{\sqrt{3}+i}{2}$$ and $$ \omega = \left(\frac{\sqrt{3}+i}{2}\right)^{1/3} $$ You have to be careful with branches of the cube root, but this turns out to be correct if you use the principal branch. Then $\sin(10^o)$ is the imaginary part of this, i.e. $$ \sin(10^o) = \frac{\omega - \overline{\omega}}{2i} = -\frac{i}{2} \left( \left(\frac{\sqrt{3}+i}{2}\right)^{1/3} - \left(\frac{\sqrt{3}-i}{2}\right)^{1/3}\right)$$

$\endgroup$
2
  • $\begingroup$ Thank you very much for the good answer! However, there is something that I do not seem to understand. Would you please explain how you got $\omega^3=\cos(30\unicode{xB0})+i\sin(30\unicode{xB0})$ from $\omega=\cos(10\unicode{xB0}+i\sin(10\unicode{xB0})$? $\endgroup$
    – CrSb0001
    Jun 1, 2023 at 1:01
  • 4
    $\begingroup$ De Moivre's formula $\endgroup$ Jun 1, 2023 at 3:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .