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The original question was to find $$\prod_{r=1}^{45} \sin\left(\left(2r-1\right)^\circ\right)$$ Using $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=\frac{e^{-ix}}{2i}\cdot(e^{2ix}-1)$ and simplifying a little I ultimately got $$\frac{1}{2^{45}i}\cdot e^{-i\pi/4}\cdot\prod_{r=1}^{45} \left(-1+e^{\frac{\pi i}{90}\cdot(2r-1)} \right)$$

I cannot proceed any further. Seeing $2r-1$ immediately made by think of complex roots of $-1$. However even after multiple tries I could not find any correlation with the given product. I would be very grateful if one could give an approach to find this.

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  • Do you accept the q-pochhammer symbol or hypergeometric functions in your simplification? – Kevin Dietrich Feb 16 '23 at 05:40
  • I am sorry but that is quite way beyond high school level. I had hoped I could solve this only using roots of unity. – Popular Power Feb 16 '23 at 08:44
  • Hint: Let $\alpha_r = -1 + e^{\frac{\pi i}{90}(2r-1)}$. Notice $i\alpha_r \alpha_{46-r} \in \mathbb{R}{+}$ for $r=1,\ldots,22$, so $\prod{r=1}^{45} \alpha_r = (1-i)s$ for some $s \in \mathbb{R}{+}$. What is $\left|\prod{r=1}^{45} \alpha_r\right|^2$? – achille hui Feb 16 '23 at 14:51

1 Answers1

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Without complex analysis

This question is best targeted by manipulating trig identities. Nonetheless, there is a suggestion to solving the product using roots of unity in the method below this one.

Observe a similar product: $$\prod_{r = 1}^{89}{\sin{r^\circ}} = \sin{1^\circ} \cdot \sin{2^\circ} \cdot \sin{3^\circ} \cdot \sin{4^\circ} \cdot \dots \cdot \sin{89^\circ} \\ \quad \\ =\sin{1^\circ} \cdot \sin{89^\circ} \cdot \sin{2^\circ} \cdot \sin{88^\circ} \cdot \dots \cdot \sin{45^\circ} \\ \quad \\ =\sin{1^\circ} \cdot \sin{(90 - 1)^\circ} \cdot \sin{2^\circ} \cdot \sin{(90 - 2)^\circ} \cdot \dots \cdot \sin{45^\circ} \\ \quad \\ =\sin{1^\circ} \cdot \cos{1^\circ} \cdot \sin{2^\circ} \cdot \cos{2^\circ} \cdot \dots \cdot \sin{45^\circ} \\ \quad \\ = \frac{1}{2}\sin{2^\circ} \cdot \frac{1}{2}\sin{4^\circ} \cdot \dots \cdot \sin{45^\circ}$$

Now, $$\prod_{r = 1}^{45}{\sin{(2r - 1)^\circ}} = \frac{\prod_{r = 1}^{89}{\sin{r^\circ}}}{\prod_{r = 1}^{45}{\sin{2r^\circ}}}$$

With complex analysis

Due to a very nice Q&A, there is no actual working provided, rather a visualisation of how "unity of roots" maybe employed here.

Observe the following circle with equation $|z| = 1$.

Unit Circle

With $z_n = \cos{2n} + i\sin{2n}$, ($180$ total)

we see $|z_0 - z_n| = |1 - \cos{2n} - i\sin{2n}| = \sqrt{\left(1 - \cos{2n}\right)^2 + \sin^2{2n}} = \sqrt{2 - 2\cos{2n}} \\ = \sqrt{2 - 2\left(2\cos^2{2n} - 1\right)} = 2\sin{n}$

Hence,

$$\prod_{r = 1}^{179}{\sin{r^\circ}} = 2^{-89} \cdot |z_0 - z_1| \cdot |z_0 - z_2| \cdot |z_0 - z_3| \cdot |z_0 - z_4| \cdot \dots \cdot |z_0 - z_{179}| \\= \frac{z_0^{180} - 1}{z_0 - 1} = z_0^{179} + z_0^{178} + \dots + 1$$

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