# How do you find the exact values of sin^3(pi/6) using the half angle formula?

Aug 4, 2015

${\sin}^{3} \left(\frac{\pi}{6}\right) = \frac{1}{8}$

#### Explanation:

Using the half-angle formula (as requested)
$\textcolor{w h i t e}{\text{XXXX}}$sin(pi/6)= sqrt((1-cos(pi/3))/2

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \sqrt{\frac{1 - \frac{1}{2}}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

So ${\sin}^{3} \left(\frac{\pi}{6}\right) = {\left(\frac{1}{2}\right)}^{3} = \frac{1}{8}$

...although I don't understand why you would know $\cos \left(\frac{\pi}{3}\right)$ and not $\sin \left(\frac{\pi}{6}\right)$; $\frac{\pi}{3}$ and $\frac{\pi}{6}$ are both standard angles

Aug 4, 2015

Find ${\sin}^{3} \left(\frac{\pi}{6}\right)$

Ans: 1/8

#### Explanation:

Use trig identity:${\sin}^{2} x = \frac{1 - \cos 2 x}{2}$

Trig table --> cos (2pi/6) = cos (pi/3) = 1/2

${\sin}^{2} \left(\frac{\pi}{6}\right) = \frac{1 - \frac{1}{2}}{2} = \frac{2 - 1}{4} = \frac{1}{4}$
$\sin \left(\frac{\pi}{6}\right) = \sqrt{\frac{1}{4}} = \frac{1}{2}$

Finally: $\sin \left(\frac{\pi}{6}\right) . {\sin}^{2} \left(\frac{\pi}{6}\right) = {\sin}^{3} \left(\frac{\pi}{6}\right) = \frac{1}{2} \left(\frac{1}{4}\right) = \frac{1}{8}$