Expert Answer to "Geometric proof of sin⁡60∘sin⁡40∘=4sin⁡20∘sin⁡80∘"

amonitas3zeb

Answered question

2022-03-23

Geometric proof of

\frac{{\sin 60}^{\circ}}{{\sin 40}^{\circ}} = 4 {\sin 20}^{\circ} {\sin 80}^{\circ}

Nunnaxf18

Beginner2022-03-24Added 18 answers

By using Briggs formulas,

\sin A \sin B = \frac{1}{2} (\cos (A - B) - \cos (A + B))

\sin C \cos D = \frac{1}{2} (\sin (C + D) + \sin (C - D))

we have:

\sin A \sin B \sin C = \frac{1}{2} (\cos (A - B) \sin C - \cos (A + B) \sin C)

= \frac{1}{4} (\sin (C + A - B) + \sin (C - A + B) + \sin (A + B - C) - \sin (A + B + C))

and we just need to prove:

{\sin 20}^{\circ} + {\sin 100}^{\circ} + {\sin 60}^{\circ} - {\sin 140}^{\circ} = {\sin 60}^{\circ}

or:

{\sin 40}^{\circ} = {\sin 80}^{\circ} - {\sin 20}^{\circ} = 2 {\sin 30}^{\circ} {\cos 50}^{\circ}

That is trivial since

2 {\sin 30}^{\circ} = 1

and

{\sin 40}^{\circ} = {\cos 50}^{\circ}

Abdullah Avery

Beginner2022-03-25Added 19 answers

Another proof. The given identity is equivalent to:

\sin \frac{π}{9} \sin \frac{2 π}{9} \sin \frac{3 π}{9} \sin \frac{4 π}{9} = \frac{3}{16}

ot to:

\prod_{k = 1}^{8} \sin \frac{k π}{9} = \frac{9}{256}

that follows from the well-known identity:

\prod_{k = 1}^{n - 1} \sin \frac{π k}{n} = \frac{2 n}{2^{n}}

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