The product of sines from 1 to 89 degrees can be evaluated using the formula $\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots\,\sin\frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}}$. By substituting $m=180$, the equation simplifies to $\bigl(\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ\bigr)^2=\frac{180}{2^{179}}$. This leads to the final result of $\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ = \frac{3}{2^{88}}\sqrt{\frac{5}{2}}$. The solution was confirmed through the use of complex roots of unity.
PREREQUISITESMathematicians, students studying trigonometry, educators teaching sine functions, and anyone interested in advanced trigonometric identities.
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 $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ = \frac{3}{2^{88}}\sqrt{\frac 52}.$$
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