The discussion revolves around evaluating the product of sines from 1 to 89 degrees, specifically the expression $P = \sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ$. Participants explore various methods to approach this problem without the use of a calculator.
Participants generally agree on the approach using the sine product formula, but there is no explicit consensus on the evaluation method as multiple hints and approaches are requested and discussed.
The discussion includes references to specific mathematical properties and formulas, but does not resolve the underlying assumptions or dependencies on definitions related to the sine function and its properties.
Theia said:Could you give a hint please? (Angel)
lfdahl said:Evaluate without the use of a calculator the product:$P = \sin 1\sin 2\sin 3…\sin 89$
(all angles in degrees)
[sp]Another way to do this is to use the formula $\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots\,\sin\frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}}$ (posed as a challenge question by greg1313 in https://mathhelpboards.com/challenge-questions-puzzles-28/trigonometric-product-challenge-21867.html, and neatly solved by kaliprasad using complex roots of unity). If you put $m= 180$ in that formula then it becomes $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 179^\circ = \frac{180}{2^{179}}.$$ But $\sin \theta = \sin(180^\circ - \theta)$, and $\sin 90^\circ = 1$, so we can write that as $$\bigl(\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ\bigr)^2= \frac{180}{2^{179}} = \frac{36\times 5}{2^{179}}.$$ Now take square roots to get $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ = \frac{3}{2^{88}}\sqrt{\frac 52}.$$lfdahl said:Evaluate without the use of a calculator the product:$P = \sin 1\sin 2\sin 3…\sin 89$
(all angles in degrees)
I like Serena said:Using the identities
$$\sin x\sin(60-x)\sin(60+x)=\frac 14\sin 3x \tag 1$$
$$\sin x\sin(90-x)=\frac 12\sin 2x \tag 2$$
$$\sin 36 = \sqrt{\frac 58 - \frac{\sqrt 5}8} \tag 3$$
$$\sin 72 = \sqrt{\frac 58 + \frac{\sqrt 5}8} \tag 4$$
we get:
$$
\overbrace{\sin 1\sin2 ...\sin 89}^{89\text{ factors}}
=\frac 1 {4^{29}} \overbrace{\sin 3\sin 6...\sin 87}^{29}\cdot \sin 30 \cdot\sin 60 \\
=\frac 1 {4^{29+9}} \overbrace{\sin 9\sin 18...\sin 81}^{9}\cdot \sin^2 30 \cdot\sin^2 60 \\
=\frac 1 {4^{38}} \frac 1{2^4}\overbrace{\sin 18\sin 36\sin 54\sin 72}^{4}\cdot \sin^2 30 \cdot\sin 45 \cdot \sin^2 60 \\
=\frac 1 {4^{38}} \frac 1{2^{4+2}}\overbrace{\sin 36\sin 72}^{2}\cdot \sin^2 30 \cdot\sin 45 \cdot \sin^2 60 \\
=\frac 1{2^{82}}\cdot\sqrt{\frac 58 - \frac{\sqrt 5}8}\sqrt{\frac 58 + \frac{\sqrt 5}8}\cdot \frac 14 \cdot\sqrt{\frac 12} \cdot \frac 34 \\
=\frac 3{2^{86}}\cdot \sqrt{\frac{5}{2^4}}\cdot\sqrt{\frac 12}\\
=\frac {3}{2^{88}}\sqrt{\frac 52}
$$
Opalg said:[sp]Another way to do this is to use the formula $\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots\,\sin\frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}}$ (posed as a challenge question by greg1313 in https://mathhelpboards.com/challenge-questions-puzzles-28/trigonometric-product-challenge-21867.html, and neatly solved by kaliprasad using complex roots of unity). If you put $m= 180$ in that formula then it becomes $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 179^\circ = \frac{180}{2^{179}}.$$ But $\sin \theta = \sin(180^\circ - \theta)$, and $\sin 90^\circ = 1$, so we can write that as $$\bigl(\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ\bigr)^2= \frac{180}{2^{179}} = \frac{36\times 5}{2^{179}}.$$ Now take square roots to get $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ = \frac{3}{2^{88}}\sqrt{\frac 52}.$$
[/sp]