what is the exact value of sin (pi/5) ?
Formula:
\(\begin{array}{|rcll|} \hline \sin(5\varphi) = 16 \sin^5(\varphi) - 20 \sin^3(\varphi) + 5\sin(\varphi) \\ \hline \end{array}\)
Ansatz:
\(\small{ \begin{array}{|lrcl|} \hline (1): & \sin(360^{\circ}) = 0 = \sin(5\cdot 72^{\circ}) = 16 \sin^5(72^{\circ}) - 20 \sin^3(72^{\circ}) + 5\sin(72^{\circ})\\ & 16 \sin^5(72^{\circ}) - 20 \sin^3(72^{\circ}) + 5\sin(72^{\circ}) &=& 0 \quad & | \quad : \sin(72^{\circ}) \\ & 16 \sin^4(72^{\circ}) - 20 \sin^2(72^{\circ}) + 5 &=& 0 \quad &| \quad x = \sin^2(72^{\circ}) \\ & 16 x^2 - 20 x + 5 &=& 0 \\\\ (2): & \sin(180^{\circ}) = 0 = \sin(5\cdot 36^{\circ}) = 16 \sin^5(36^{\circ}) - 20 \sin^3(36^{\circ}) + 5\sin(36^{\circ}) \\ & 16 \sin^5(36^{\circ}) - 20 \sin^3(36^{\circ}) + 5\sin(36^{\circ}) &=& 0 \quad & | \quad : \sin(36^{\circ}) \\ & 16 \sin^4(36^{\circ}) - 20 \sin^2(36^{\circ}) + 5 &=& 0 \quad &| \quad x = \sin^2(36^{\circ}) \\ & 16 x^2 - 20 x + 5 &=& 0 \\ \hline \end{array} }\)
The quadratic polynomial \(16 x^2 - 20 x + 5\) has following roots: \(<~ \sin^2(72^{\circ}),\ \sin^2(36^{\circ}) ~>\)
The general quadratic equation is: \( {\displaystyle ax^{2}+bx+c=0.}\)
The quadratic formula is: \({\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}\)
\(\begin{array}{|rcll|} \hline x &=& \displaystyle { \frac {20\pm {\sqrt {20^{2}-4\cdot 16 \cdot 5}}}{2\cdot 16}} \\\\ &=& \frac {20\pm {\sqrt { 400 - 320 }}} {32} \\\\ &=& \frac {20\pm {\sqrt { 80 }}} {32} \\\\ &=& \frac {20\pm {\sqrt { 16\cdot 5 }}} {32} \\\\ &=& \frac {20\pm 4 {\sqrt { 5 } }} {32} \\\\ &=& \frac {5\pm {\sqrt { 5 } }} {8} \\\\ \sqrt{x} &=& \sqrt{ \frac {5\pm {\sqrt { 5 } }} {8} } \\ \hline \end{array} \)
Because \(\sin(72^{\circ}) > \sin(36^{\circ})\)
we have:
\(\begin{array}{|rcll|} \hline \sin(72^{\circ}) = \sqrt{ \frac {5 + {\sqrt { 5 } }} {8} } \\ \sin(36^{\circ}) = \sqrt{ \frac {5 - {\sqrt { 5 } }} {8} } \\ \hline \end{array}\)
