How can we find the value of sin 3 , sin 6 , sin 9 \sin 3, \sin 6, \sin 9 sin 3 , sin 6 , sin 9
**Solution:**
Using that
sin 18 = 5 − 1 4 cos 18 = 10 + 2 5 4 \sin 18 = \frac{\sqrt{5} - 1}{4} \quad \cos 18 = \frac{\sqrt{10 + 2\sqrt{5}}}{4} sin 18 = 4 5 − 1 cos 18 = 4 10 + 2 5 sin 15 = 6 − 2 4 cos 15 = 6 + 2 4 \sin 15 = \frac{\sqrt{6} - \sqrt{2}}{4} \quad \cos 15 = \frac{\sqrt{6} + \sqrt{2}}{4} sin 15 = 4 6 − 2 cos 15 = 4 6 + 2 sin 3 = sin ( 18 − 15 ) = sin 18 ∗ cos 15 − cos 18 ∗ sin 15 \sin 3 = \sin(18 - 15) = \sin 18 * \cos 15 - \cos 18 * \sin 15 sin 3 = sin ( 18 − 15 ) = sin 18 ∗ cos 15 − cos 18 ∗ sin 15 sin 3 = ( 5 − 1 4 ) ( 6 + 2 4 ) − ( 10 + 2 5 4 ) ( 6 − 2 4 ) \sin 3 = \left(\frac{\sqrt{5} - 1}{4}\right) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) - \left(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\right) \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) sin 3 = ( 4 5 − 1 ) ( 4 6 + 2 ) − ( 4 10 + 2 5 ) ( 4 6 − 2 )
**This is exact value of sin 3 \sin 3 sin 3
If we need find sin 6 \sin 6 sin 6
sin 6 = 2 sin 3 ∗ cos 3 \sin 6 = 2 \sin 3 * \cos 3 sin 6 = 2 sin 3 ∗ cos 3
So we need find value of cos 3 \cos 3 cos 3
cos 3 = cos ( 18 − 15 ) = cos 18 ∗ cos 15 + sin 18 ∗ sin 15 \cos 3 = \cos(18 - 15) = \cos 18 * \cos 15 + \sin 18 * \sin 15 cos 3 = cos ( 18 − 15 ) = cos 18 ∗ cos 15 + sin 18 ∗ sin 15 cos 3 = ( 10 + 2 5 4 ) ( 6 + 2 4 ) + ( 5 − 1 4 ) ( 6 − 2 4 ) \cos 3 = \left(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\right) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) + \left(\frac{\sqrt{5} - 1}{4}\right) \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) cos 3 = ( 4 10 + 2 5 ) ( 4 6 + 2 ) + ( 4 5 − 1 ) ( 4 6 − 2 ) sin 6 = 2 sin 3 ∗ cos 3 = 2 [ ( 5 − 1 4 ) ( 6 + 2 4 ) − ( 10 + 2 5 4 ) ( 6 − 2 4 ) ] ∗ [ ( 10 + 2 5 4 ) ( 6 + 2 4 ) + ( 5 − 1 4 ) ( 6 − 2 4 ) ] \begin{aligned}
\sin 6 &= 2 \sin 3 * \cos 3 \\
&= 2 \left[ \left(\frac{\sqrt{5} - 1}{4}\right) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) - \left(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\right) \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) \right] \\
&* \left[ \left(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\right) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) + \left(\frac{\sqrt{5} - 1}{4}\right) \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) \right]
\end{aligned} sin 6 = 2 sin 3 ∗ cos 3 = 2 [ ( 4 5 − 1 ) ( 4 6 + 2 ) − ( 4 10 + 2 5 ) ( 4 6 − 2 ) ] ∗ [ ( 4 10 + 2 5 ) ( 4 6 + 2 ) + ( 4 5 − 1 ) ( 4 6 − 2 ) ]
**This is exact value of sin 6 \sin 6 sin 6
If we need find sin 9 \sin 9 sin 9 sin 3 α = 3 sin α − 4 sin 3 α \sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha sin 3 α = 3 sin α − 4 sin 3 α
sin 9 = 3 sin 3 − 4 sin 3 3 \sin 9 = 3 \sin 3 - 4 \sin^3 3 sin 9 = 3 sin 3 − 4 sin 3 3 sin 9 = 3 [ ( 5 − 1 4 ) ( 6 + 2 4 ) − ( 10 + 2 5 4 ) ( 6 − 2 4 ) ] − 4 [ ( 5 − 1 4 ) ( 6 + 2 4 ) − ( 10 + 2 5 4 ) ( 6 − 2 4 ) ] 3 \sin 9 = 3 \left[ \left(\frac {\sqrt {5} - 1}{4}\right) \left(\frac {\sqrt {6} + \sqrt {2}}{4}\right) - \left(\frac {\sqrt {10 + 2\sqrt {5}}}{4}\right) \left(\frac {\sqrt {6} - \sqrt {2}}{4}\right) \right] - 4 \left[ \left(\frac {\sqrt {5} - 1}{4}\right) \left(\frac {\sqrt {6} + \sqrt {2}}{4}\right) - \left(\frac {\sqrt {10 + 2\sqrt {5}}}{4}\right) \left(\frac {\sqrt {6} - \sqrt {2}}{4}\right) \right] ^ {3} sin 9 = 3 [ ( 4 5 − 1 ) ( 4 6 + 2 ) − ( 4 10 + 2 5 ) ( 4 6 − 2 ) ] − 4 [ ( 4 5 − 1 ) ( 4 6 + 2 ) − ( 4 10 + 2 5 ) ( 4 6 − 2 ) ] 3
This is exact value of sin 9 \sin 9 sin 9
NOTE:
If you don't know how to find sin 15 \sin 15 sin 15 cos 15 \cos 15 cos 15
sin 15 = sin ( 45 − 30 ) = sin 45 ∗ cos 30 − cos 45 ∗ sin 30 \sin 15 = \sin (45 - 30) = \sin 45 * \cos 30 - \cos 45 * \sin 30 sin 15 = sin ( 45 − 30 ) = sin 45 ∗ cos 30 − cos 45 ∗ sin 30 cos 15 = cos ( 45 − 30 ) = cos 45 ∗ cos 30 + sin 45 ∗ sin 30 \cos 15 = \cos (45 - 30) = \cos 45 * \cos 30 + \sin 45 * \sin 30 cos 15 = cos ( 45 − 30 ) = cos 45 ∗ cos 30 + sin 45 ∗ sin 30 
#340153 on Dec 2023