# How to find exact value of sin(pi/24)?

Nov 1, 2017

sin(pi/24)=1/2sqrt(2-sqrt(2+sqrt3)

#### Explanation:

As $\frac{\pi}{24} = {180}^{\circ} / 24 = {\left(7 \frac{1}{2}\right)}^{\circ}$, let us first work out $\cos \left(\frac{\pi}{12}\right)$ or $\cos {15}^{\circ}$

$\cos {15}^{\circ} = \cos \left({45}^{\circ} - {30}^{\circ}\right)$

= $\cos {45}^{\circ} \cos {30}^{\circ} + \sin {45}^{\circ} \sin {30}^{\circ}$

= $\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}$

= $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Now as $\sin \theta = \sqrt{\frac{1 - \cos 2 \theta}{2}}$

$\sin \left(\frac{\pi}{24}\right) = \sin {\left(7 \frac{1}{2}\right)}^{\circ}$

= $\sqrt{\frac{1 - \cos {15}^{\circ}}{2}}$

= $\sqrt{\frac{1 - \frac{\sqrt{3} + 1}{2 \sqrt{2}}}{2}}$

= $\sqrt{\frac{2 - \frac{\sqrt{3} + 1}{\sqrt{2}}}{4}}$

= $\frac{1}{2} \sqrt{2 - \frac{\sqrt{3} + 1}{\sqrt{2}}}$

= $\frac{1}{2} \sqrt{2 - \frac{\sqrt{6} + \sqrt{2}}{2}}$

= 1/2sqrt(2-sqrt((sqrt6+sqrt2)^2/4)

= 1/2sqrt(2-sqrt((8+2sqrt12)/4)

= 1/2sqrt(2-sqrt((8+4sqrt3)/4)

= 1/2sqrt(2-sqrt(2+sqrt3)