# How do you use half angle identities and the fact that sin(pi/6)=1/2 to find cos(-pi/12) and sin (-pi/12)?

Aug 3, 2018

#### Explanation:

We have , $\sin \left(\frac{\pi}{6}\right) = \sin {30}^{\circ} = \frac{1}{2}$

Now , $\theta = \frac{\pi}{12} = {15}^{\circ} \to {1}^{s t} Q u a \mathrm{dr} a n t$

$\text{Using"color(blue)" half angle formula :}$

color(blue)(sin^2 (theta/2)=(1-costheta)/2=(1-sqrt(1-sin^2theta))/2

Let , $\theta = {30}^{\circ} \implies \frac{\theta}{2} = {15}^{\circ} \to {1}^{s t} Q u a \mathrm{dr} a n t$

$\therefore {\sin}^{2} {15}^{\circ} = \frac{1 - \sqrt{1 - {\sin}^{2} {30}^{\circ}}}{2}$

$\therefore {\sin}^{2} {15}^{\circ} = \frac{1 - \sqrt{1 - \frac{1}{4}}}{2}$

$\therefore {\sin}^{2} {15}^{\circ} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4} = \frac{4 - 2 \sqrt{3}}{8}$

$\therefore {\sin}^{2} {15}^{\circ} = \frac{3 + 1 - 2 \sqrt{3}}{8}$=$\frac{{\left(\sqrt{3}\right)}^{2} - 2 \left(\sqrt{3}\right) \left(\sqrt{1}\right) + {\left(\sqrt{1}\right)}^{2}}{8}$

$\therefore {\sin}^{2} {15}^{\circ} = {\left(\sqrt{3} - 1\right)}^{2} / \left(4 \times 2\right)$

$\therefore \sin {15}^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \ldots \ldots \to \left[\because {1}^{s t} Q u a \mathrm{dr} a n t\right]$

$\therefore \sin {15}^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$

color(green)(-pi/12=-15^circtoIV^(th)Quadrant =>sin(-pi/12) < 0

:.color(red)(sin(-pi/12)=-sin(pi/12)=-(sqrt6-sqrt2)/4

For $\cos \left(- \frac{\pi}{12}\right)$ ,please see next answer.

Aug 3, 2018

#### Explanation:

We have , $\sin \left(\frac{\pi}{6}\right) = \sin {30}^{\circ} = \frac{1}{2}$

Now , $\theta = \frac{\pi}{12} = {15}^{\circ} \to {1}^{s t} Q u a \mathrm{dr} a n t$

$\text{Using"color(blue)" half angle formula :}$

color(blue)(cos^2 (theta/2)=(1+costheta)/2=(1+sqrt(1-sin^2theta))/2

Let , $\theta = {30}^{\circ} \implies \frac{\theta}{2} = {15}^{\circ} \to {1}^{s t} Q u a \mathrm{dr} a n t$

$\therefore {\cos}^{2} {15}^{\circ} = \frac{1 + \sqrt{1 - {\sin}^{2} {30}^{\circ}}}{2}$

$\therefore {\cos}^{2} {15}^{\circ} = \frac{1 + \sqrt{1 - \frac{1}{4}}}{2}$

$\therefore {\cos}^{2} {15}^{\circ} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4} = \frac{4 + 2 \sqrt{3}}{8}$

$\therefore {\cos}^{2} {15}^{\circ}$=$\frac{3 + 1 + 2 \sqrt{3}}{8}$=$\frac{{\left(\sqrt{3}\right)}^{2} + 2 \left(\sqrt{3}\right) \left(\sqrt{1}\right) + {\left(\sqrt{1}\right)}^{2}}{8}$

$\therefore {\cos}^{2} {15}^{\circ} = {\left(\sqrt{3} + 1\right)}^{2} / \left(4 \times 2\right)$

$\therefore \cos {15}^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \ldots \ldots \to \left[\because {1}^{s t} Q u a \mathrm{dr} a n t\right]$

$\therefore \cos {15}^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$

color(green)(-pi/12=-15^circtoIV^(th)Quadrant =>cos(-pi/12) > 0

:.color(red)(cos(-pi/12)=+cos(pi/12)=(sqrt6+sqrt2)/4

For $\sin \left(- \frac{\pi}{12}\right)$ , please see next answer.