What is the value of the product $\sin(10) \sin(20) \sin(30) \sin(40) \sin(50) \sin(60) \sin(70) \sin80$, where all the angles are in degrees? Solve using complex numbers.
I found this in a book of mine. I know how to solve by pairing the terms one by one and using sum of cosines and sines formulas, but how do you do it using complex numbers? Any help would be appreciated!
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In this solution I use complex numbers to fast track the sum of angles identities, but I still use a significant amount of trigonometry, so I'm not sure if this is what you were looking for.
Instead of finding $p=\sin 10\sin 20\sin 30\sin 40\sin 50\sin 60\sin 70\sin 80$ directly, instead let's try to find $p^2$ $$\begin{array}{lll} p^2&=&p\cdot p\\ &=&p\cdot (\sin 10\sin 20\sin 30\sin 40\sin 50\sin 60\sin 70\sin 80)\\ &=&p\cdot (\sin 170\sin 160\sin 150\sin 140\sin 130\sin 120\sin 110\sin 100)\\ &=&p\cdot 1\cdot (\sin 170\sin 160\sin 150\sin 140\sin 130\sin 120\sin 110\sin 100)\\ &=&p\cdot \sin 90\cdot (\sin 170\sin 160\sin 150\sin 140\sin 130\sin 120\sin 110\sin 100)\\ &=&\sin 10\sin 20\sin 30\dots\sin 90\dots\sin150\sin160\sin170 \end{array}$$
Next, consider the following polynomial $x(x-1)P(x)$ $$\begin{array}{lll} x(x-1)P(x)&=&(x-\sin 0)(x-\sin 10)\dots(x-\sin 90)\dots(x-\sin 160)(x-\sin 170)\\ x(x-1)P(x)&=&x(x-1)(x-\sin 10)\dots(x-\sin 80)(x-\sin 100)\dots(x-\sin 170)\\ P(x)&=&(x-\sin 10)\dots(x-\sin 80)(x-\sin 100)\dots(x-\sin 170)\\ P(x)&=&x^{16}+x(\dots)-\sin10\sin20\dots\sin160\sin170\\ P(x)&=&x^{16}+x(\dots)-p^2 \end{array}$$
Additionally, consider $$e^{(18t)i}=(e^{ti})^{18}$$ $$\begin{array}{lll} \cos18t+i\sin18t &=& (\cos t+i\sin t)^{18}\\ &=& \sum_{k=0}^{17}\binom{18}{k}\cos^{18-k}t(i\sin t)^k\\ \text{Im}(\cos18t+i\sin18t)&=&\text{Im}\bigg( \sum_{k=0}^{17}\binom{18}{k}\cos^{18-k}t(i\sin t)^k\bigg)\\ \sin18t&=& \sum_{\substack{0\le k\le 17\\k\text{ odd}}}\binom{18}{k}(-1)^{\frac{k-1}{2}}\cos^{18-k}t\sin^k t\\ \sin18t&=& \cos t\sum_{\substack{0\le k\le 17\\k\text{ odd}}}\binom{18}{k}(-1)^{\frac{k-1}{2}}\cos^{17-k}t\sin^{k} t\\ \sin18t&=& \cos t\sum_{\substack{0\le k\le 17\\k\text{ odd}}}\binom{18}{k}(-1)^{\frac{k-1}{2}}(\cos^2t)^\frac{17-k}{2}\sin^{k} t\\ \sin18t&=& \cos t\sum_{\substack{0\le k\le 17\\k\text{ odd}}}\binom{18}{k}(-1)^{\frac{k-1}{2}}(1-\sin^2t)^\frac{17-k}{2}\sin^{k} t\\ \sin18t&=& \cos t\sin t\sum_{\substack{0\le k\le 17\\k\text{ odd}}}\binom{18}{k}(-1)^{\frac{k-1}{2}}(1-\sin^2t)^\frac{17-k}{2}\sin^{k-1} t\\ \end{array}$$ Next, suppose that $\sin 18t = 0,0^\circ\le t\lt 180^\circ$. One possibility is that $\cos t = 0$ (i.e. $t=90^\circ$); another is that $\sin t=0$ (i.e. $t=0^\circ$). Of particular interest are the roots of $$f(t) = Q(\sin t)=\sum_{\substack{0\le k\le 17\\k\text{ odd}}}\binom{18}{k}(-1)^{\frac{k-1}{2}}(1-\sin^2t)^\frac{17-k}{2}\sin^{k-1} t$$ Because we have already found $2$ of the $18$ roots of $\sin 18t$ ($0^\circ$ and $90^\circ$), the remaining $16$ roots are the zeros of $f$, namely $\{t\mid t=10j, j \in N^{+},j\le 17,j\ne 9\}$. But these are the same roots as those of $P(\sin t)$, so we can infer that $P$ and $Q$ have the same roots and (by the Polynomial Remainder Theorem), because $P$ and $Q$ are both of degree $16$, are linear combinations of one another.
Just to clarify, $$Q(x)=\sum_{\substack{0\le k\le 17\\k\text{ odd}}}\binom{18}{k}(-1)^{\frac{k-1}{2}}(1-x^2)^\frac{17-k}{2}x^{k-1}$$ and $$Q(x) = aP(x)$$ We know that the coefficient associated with $P(x)$'s $x^{16}$ term is $1$, so $a$ is identical to the coefficient associated with $Q(x)$'s $x^{16}$ term, namely, $$a = \binom{18}{1}+\binom{18}{3}+\binom{18}{5}+\binom{18}{7}+\binom{18}{9}+\binom{18}{11}+\binom{18}{13}+\binom{18}{15}+\binom{18}{17}$$ but this is just the binomial expansion of $$a = \frac{(1+1)^{18} - (1-1)^{18}}{2}=\frac{2^{18}}{2}=2^{17}$$ Thus $$P(x)\equiv\frac{Q(x)}{2^{17}}$$ Now just considering the constant term of $P(x)$ we have $$p^2 = \frac{\binom{18}{1}}{2^{17}}=\frac{18}{2^{17}} =\frac{9}{2^{16}}=\frac{3^2}{({2^{8}})^2}=\bigg(\frac{3}{256}\bigg)^2$$ taking the principle square root we have $$p = \frac{3}{256}$$
Using Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$,
$$\prod_{k=0}^{18-1}\sin(10^\circ k)=\frac{18}{2^{18-1}}$$
Now, $\sin(180^\circ-10^\circ n)=\sin10(18-n)^\circ $
$$\implies\sin(90^\circ)\prod_{k=0}^8\sin^2(10^\circ k)=\frac{18}{2^{18-1}}$$
Finally, $\sin(10^\circ k)>0$ for $0<k\le8$
More generally we can prove, $$\prod_{k=0}^{\left\lfloor\dfrac n2\right\rfloor}\sin\left(\dfrac{180^\circ}n k\right)=+\sqrt{\dfrac n{2^{n-1}}}$$