Question

Prove that $sin(\frac{\pi}{5}) . sin(\frac{2\pi}{5}) . sin(\frac{3\pi}{5}) . sin(\frac{4\pi}{5}) = \frac{5}{16}$.

Answer 1

Hi there!
Here's the answer:

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Let A + B = Π
=> B = Π - A

=> Sin B = Sin (Π -A) = Sin A

{°•° Sin(Π-x) = sin x}


Now,

• $\frac{Π}{5} + \frac{4Π}{5} = Π$

=> $Sin(\frac{Π}{5}) = Sin(\frac{4Π}{5})$


• $\frac{2Π}{5} + \frac{3Π}{5} = Π$

=> $Sin(\frac{2Π}{5}) = Sin(\frac{3Π}{5})$


$LHS\: =$

= $sin(\frac{\pi}{5}) . sin(\frac{2\pi}{5}) . sin(\frac{3\pi}{5}) . sin(\frac{4\pi}{5})$

= $sin(\frac{\pi}{5}) . sin(\frac{2\pi}{5}) . sin(\frac{2\pi}{5}) . sin(\frac{\pi}{5})$

= $[sin(\frac{\pi}{5})sin(\frac{\pi}{5})]^{2}$

= (sin 36° sin 72°)²

= (sin 36° sin 18°)²

= $[\frac{\sqrt{10-2\sqrt{5}}}{4} × \frac{\sqrt{10+2\sqrt{5}}}{4}]^{2}$

= $[\frac{10-2\sqrt{5}}{16} × \frac{10+\sqrt{5}}{16}]$

= $\frac{100-20}{256}$

= $\frac{80}{256}$

= $\frac{5}{16}$

$=\: RHS$


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