# Problem of the week #8 - May 21st, 2012

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#### Jameson

Staff member
Without using a calculator find the exact value of $$\displaystyle \sin \left( \frac{\pi}{5} \right)$$.

Hint:

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

#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

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.

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

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

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

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

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

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

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

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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