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I have the following equality and I wonder if there is an exact way to prove that it is true, without using the calculator of course:

$$\frac{\sqrt{\frac{5}{2}-\frac{\sqrt{5}}{2}}}{\sin \frac{2 \pi}{5}} = \frac{1}{\sin \frac{3 \pi}{10}}.$$

Thank you.

Honestly I don't know where even to start...

Maybe I'd move $\sin\frac{2\pi}{5}$ to the right-hand side but I don't see how this could be helpful.

I see $\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}},$ but I don't know how to continue.

Also I don't see what sense adding those numbers up make...

Michael Rozenberg
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zest16
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    I solved your problem. If you want to see my solution, show please your attempts. – Michael Rozenberg Jan 26 '20 at 21:28
  • Honestly I don't know where even to start... Maybe I'd move $sin \frac{2}{5} \pi$ to the right-hand side but I don't see how this could be helpful. – zest16 Jan 26 '20 at 21:34
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    $3\pi/10$ is half $3\pi/5$, and $(3\pi/5)+(2\pi/5)=\pi$. – Gerry Myerson Jan 26 '20 at 21:35
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    @zest16 Try more. I wait your another attempts. – Michael Rozenberg Jan 26 '20 at 21:35
  • Well based on GerryMyerson's comment I see $sin \frac{\alpha}{2} = \pm \sqrt{\frac{1-cos \alpha}{2}}$, but I don't know how to continue. Also I don't see what sense adding those numbers up make... – zest16 Jan 26 '20 at 21:48
  • Gerry's comment is suggesting that you can express $\frac{\sin a}{\sin b}$ in a simpler form if you know that $b$ relates to $a$ via $\pi$. Recall that $\sin(A+B)=\sin A \cos B + \cos A \sin B$. – Jam Jan 26 '20 at 21:55

2 Answers2

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It suffices to remark that:

$$2 \sin\left(\frac{\pi}{5}\right)=\sqrt{\frac{5}{2}-\frac{\sqrt{5}}{2}}.$$ Then using $\sin(2x)=2\sin(x)\cos(x)$, the equality is equivalent to: $$2 \sin\left(\frac{\pi}{5}\right) \sin\left(\frac{3\pi}{10}\right)=2 \sin\left(\frac{\pi}{5}\right) \cos\left(\frac{\pi}{5}\right),$$ which is equivalent to:

$$\sin\left(\frac{3\pi}{10}\right)=\cos\left(\frac{\pi}{5}\right).$$ The latter is true since $\dfrac{3\pi}{10}+\dfrac{\pi}{5}=\dfrac{\pi}{2}$.

S. Maths
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    As for why $\sin\frac{\pi}{5}$ has that value, it helps to chase similar triangles in a regular pentagon with its diagonals inscribed. – J.G. Jan 26 '20 at 21:51
  • Well my equality actually comes from an exercise I'm doing to prove the method to draw a regular pentagon is correct, so I guess it would be kind of a circular definition, wouldn't it? – zest16 Jan 26 '20 at 22:07
  • @zest16 You can derive the side length of the radius $1$, regular pentagon using elementary geometry without trigonometry. You can then relate this to your trigonometric problem without problems of circularity. – Jam Jan 27 '20 at 11:05
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We need to prove that: $$\sqrt{\frac{5-\sqrt5}{2}}=\frac{\sin72^{\circ}}{\cos36^{\circ}}$$ or $$\frac{5-\sqrt5}{2}=4\sin^236^{\circ}$$ or $$\frac{5-\sqrt5}{2}=2(1-\cos72^{\circ})$$ or $$\sqrt5=4\sin18^{\circ}+1$$ or $$4\sin^218^{\circ}+2\sin18^{\circ}-1=0$$ or $$2(1-\cos36^{\circ})+2\sin18^{\circ}-1=0$$ or $$2\cos36^{\circ}-2\sin18^{\circ}=1,$$ which is true because $$2\cos36^{\circ}-2\sin18^{\circ}=\frac{2\sin36^{\circ}\cos36^{\circ}-2\sin36^{\circ}\sin18^{\circ}}{\sin36^{\circ}}=$$ $$=\frac{\sin72^{\circ}-\sin108^{\circ}+\sin36^{\circ}}{\sin36^{\circ}}=1.$$

Michael Rozenberg
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