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- Jan 26, 2012

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Hint:

\(\displaystyle \sin(5\theta)=5\sin(\theta)-20\sin^3(\theta)+16\sin^5(\theta)\)

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- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,043

Hint:

\(\displaystyle \sin(5\theta)=5\sin(\theta)-20\sin^3(\theta)+16\sin^5(\theta)\)

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- Jan 26, 2012

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1) Sudharaka

Solution:

This is a pretty well known problem and the solution can be found many places online. Here is the solution Sudharaka submitted which is the standard one.

\(\mbox{Substituting }\theta=\dfrac{\pi}{5}\mbox{ we get,}\)

\[\displaystyle \sin\left(\frac{\pi}{5}\right)\left\{5-20\sin^2\left(\frac{\pi}{5}\right)+16\sin^4\left( \frac{\pi}{5}\right) \right\}=0\]

\(\mbox{Since, }\sin\left(\frac{\pi}{5}\right)\neq 0\mbox{ ;}\)

\[5-20\sin^2\left(\frac{\pi}{5}\right)+16\sin^4\left( \frac{\pi}{5}\right)=0\]

\[\Rightarrow \sin^{2}\left(\frac{\pi}{5}\right)=\frac{5\pm\sqrt{5}}{8}\]

\[\Rightarrow \sin\left(\frac{\pi}{5}\right)=\pm\sqrt{\frac{5\pm \sqrt{5}}{8}}\]

The Sine function is positive in the first quadrant. Therefore,

\[\sin\left(\frac{\pi}{5}\right)=\sqrt{\frac{5\pm \sqrt{5}}{8}}\]

The Sine function increases in the first quadrant. Hence we have, \(\sin\left(\frac{\pi}{5}\right)<\sin\left(\frac{ \pi}{4}\right)=\frac{1}{\sqrt{2}}\). But,

\[5>4\Rightarrow \frac{5}{8}>\frac{1}{2}\Rightarrow \frac{5+\sqrt{5}}{8}>\frac{1}{2}\Rightarrow \sqrt{\frac{5+\sqrt{5}}{8}}>\frac{1}{\sqrt{2}}\]

Therefore the only possible solution is,

\[\sin\left(\frac{\pi}{5}\right)=\sqrt{\frac{5-\sqrt{5}}{8}}\]

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