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The problem is to prove that

$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$

All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, all these attempts were futile.

Any hints are welcomed.

Blue
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4 Answers4

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Denote $Q = \sin 20 \sin 40 \sin 60 \sin 80$. Observe first that

\begin{align} \sin 60 & = \sin (20+40) \\ & = \sin 20 \cos 40 + \cos 20 \sin 40 \\ & = \sin 20 (2\cos^2 20 - 1) + \cos 20 (2 \sin 20 \cos 20) \\ & = (4\cos^2 20 - 1) \sin 20 \end{align}

We can now write

\begin{align} Q & = \sin 20 \sin (60-20) \sin 60 \sin (60+20) \\ & = \sin 20 (\sin 60 \cos 20 - \cos 60 \sin 20) \sin 60 (\sin 60 \cos 20 + \cos 60 \sin 20) \\ & = \sin 20 \sin 60 (\sin^2 60 \cos^2 20 - \cos^2 60 \sin^2 20) \\ & = \sin 20 \sin 60 \frac{3\cos^2 20 - \sin^2 20}{4} \\ & = \sin 20 \sin 60 \frac{4\cos^2 20 - 1}{4} \\ & = \sin 60 \frac{(4\cos^2 20 - 1)\sin 20}{4} \\ & = \sin 60 \frac{\sin 60}{4} \\ & = \frac{\sin^2 60}{4} = \frac{3}{16} \end{align}

Brian Tung
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HINT:

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin x[\sin^2120^\circ-\sin^2x]=\dfrac{3\sin x-4\sin^3x}4$$

$$\implies4\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin3x$$

Set $x=20^\circ$

Community
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lab bhattacharjee
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Hint:

Use the so-called Morrie's law :

$\cos(20°)\cos(40°)\cos(80°)=1/8$

and

$\sin(20°)\sin(40°)\sin(80°)=\dfrac{\sqrt{3}}{8}$.

Emilio Novati
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  • Please consider using (10^{\circ}) to get $(10^{\circ})$ instead of (10°) which, in LaTeX gives the rather oddly spaced $(10°)$. Although, a constant use of the degree sign is not necessary. – Fly by Night Jul 11 '15 at 23:11
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We have:

$\sin20^{\circ} = \sin(30^\circ - 10^\circ)= \frac{1}{2}\cos10^\circ - \frac{\sqrt{3}}{2}\sin10^\circ$

$\sin40^{\circ} = \sin(30^\circ + 10^\circ)= \frac{1}{2}\cos10^\circ + \frac{\sqrt{3}}{2}\sin10^\circ$

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\sin80^\circ =\cos 10^\circ$

Therefore,

$A = \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \frac{\sqrt{3}}{2} \left((\cos10^\circ )^2 - \frac{3}{4} \right)\cos10^\circ$

Moreover :

$ \cos30^\circ = 4(\cos10^\circ )^3 - 3\cos 10^\circ$, from this cubic equation we find out the value of $cos10^\circ$, and then plug it to the expression of $A$ we will get the answer.

Hoping_Blessing
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