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This question :

https://math.stackexchange.com/questions/1411700/whats-the-size-of-the-x-angle

has the answer $10°$. This follows from the equation

$$2\sin(80°)=\frac{\sin(60°)}{\sin(100°)}\times \frac{\sin(50°)}{\sin(20°)}$$

which is indeed true , which I checked with Wolfram.

  • How can this equation be proven ?
Community
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Peter
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  • There should be an easier way to show that the angle is $10°$, but I failed to find one. – Peter Aug 27 '15 at 22:54
  • Perhaps by noting that $80=60+20$, $50=30+20$, and $100 = 120-20$. – John Joy Aug 27 '15 at 22:57
  • I already used several manipulations to arrive at the given equation, and certainly, I tried the addition-theorems. Perhaps, someone can show $$2cos^2(10°)sin(20°)=cos(40°)sin(60°)$$, which is equivalent to the given equation. – Peter Aug 27 '15 at 22:59

3 Answers3

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$$\begin{align}\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}&=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}\\&=2\sin 80^\circ\sin 20^\circ \cos 50^\circ\\&=2\left(-\frac 12(\cos 100^\circ-\cos 60^\circ)\right)\cos 50^\circ\\&=(\cos 60^\circ-\cos 100^\circ)\cos 50^\circ\\&=\frac 12\cos 50^\circ-\cos 100^\circ\cos 50^\circ\\&=\frac 12\cos 50^\circ-\frac 12(\cos 150^\circ+\cos 50^\circ)\\&=-\frac 12\times\left(-\frac{\sqrt 3}{2}\right)\\&=\frac{\sqrt 3}{4}\\&=\frac 12\sin 60^\circ\end{align}$$

mathlove
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2

$$ 2\cdot\sin80^\circ\cdot \underbrace{\sin100^\circ}_{\sin80^\circ}\cdot\sin20^\circ=2\cdot \sin80^\circ\cdot2\sin40^\circ\underbrace{\cos40^\circ}_{\sin50^\circ}\cdot\sin20^\circ=\\ =\underbrace{4\cdot\sin80^\circ\cdot\sin40^\circ\cdot\sin20^\circ}_{\sqrt{3}/2}\cdot\sin50^\circ=\sin60^\circ\cdot\sin50^\circ. $$ The only non-obvious identity is Morrie's law for sine.

A.Γ.
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$$\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}=2\sin 80^\circ\sin 20^\circ\cos 50^\circ=2\sin 80^\circ\sin 20^\circ\sin40^\circ$$

as $\cos 50^\circ=\sin40^\circ$

Now by Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, $$\sin\left(60^\circ-x\right)\sin\left(60^\circ+x\right)=\sin^260^\circ-\sin^2x$$

$$\implies\sin x\sin\left(60^\circ-x\right)\sin\left(60^\circ+x\right)=\dfrac{\sin3x}4$$

Here $x=20^\circ$

Community
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lab bhattacharjee
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