Parker Pitts

2022-09-06

Use the fact that the terminal point associated with moving a distance $t=\pi /6$units anti-clockwise around the unit circle, ${x}^{2}+{y}^{2}=1$, has coordinates $\left(x,y\right)=\left(\sqrt{3}/2,1/2\right)$,and the method from Example 63 in the lectures to determine the exact coordinates of theterminal point associated with moving a distance $t=pi/12$ units ant-clockwise around theunit circle. Hence determine the exact values of

### Answer & Explanation

cegukwt

Step 1

$\mathrm{sin}\left(\pi /12\right)=\sqrt{\frac{1-\mathrm{cos}\left(\pi /6\right)}{2}}$
$⇒\mathrm{sin}\left(\pi /12\right)=\sqrt{\frac{1-\left(\sqrt{3}/2\right)}{2}}$
$⇒\mathrm{sin}\left(\pi /12\right)=\sqrt{\frac{\left(2-\sqrt{3}\right)/2}{2}}$
$⇒\mathrm{sin}\left(\pi /12\right)=\sqrt{\frac{2-\sqrt{3}}{4}}$
$⇒\mathrm{sin}\left(\pi /12\right)=\sqrt{\frac{2-\sqrt{3}}{2}}$
$\mathrm{cos}\left(\pi /12\right)=\sqrt{\frac{1+\mathrm{cos}\left(\pi /6\right)}{2}}$
$⇒\mathrm{cos}\left(\pi /12\right)=\sqrt{\frac{1+\left(\sqrt{3}/2\right)}{2}}$
$⇒\mathrm{cos}\left(\pi /12\right)=\sqrt{\frac{\left(2+\sqrt{3}\right)/2}{2}}$
$⇒\mathrm{cos}\left(\pi /12\right)=\sqrt{\frac{2+\sqrt{3}}{4}}$
$⇒\mathrm{cos}\left(\pi /12\right)=\sqrt{\frac{2+\sqrt{3}}{2}}$
Step 2
$\mathrm{tan}\left(\pi /12\right)=\frac{\mathrm{sin}\left(\pi /12\right)}{\mathrm{cos}\left(\pi /12\right)}$
$⇒\mathrm{tan}\left(\pi /12\right)=\frac{\frac{\sqrt{2-\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}}$
$⇒\mathrm{tan}\left(\pi /12\right)=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$
$⇒\mathrm{tan}\left(\pi /12\right)=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}$
$⇒\mathrm{tan}\left(\pi /12\right)=\sqrt{\frac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}$
$⇒\mathrm{tan}\left(\pi /12\right)=\sqrt{\frac{\left(2-\sqrt{3}{\right)}^{2}}{\left({2}^{2}-{\sqrt{3}}^{2}\right)}}$
$⇒\mathrm{tan}\left(\pi /12\right)=\sqrt{\frac{\left(2-\sqrt{3}{\right)}^{2}}{4-3}}$
$⇒\mathrm{tan}\left(\pi /12\right)=\sqrt{\frac{\left(2-\sqrt{3}{\right)}^{2}}{1}}$
$⇒\mathrm{tan}\left(\pi /12\right)=2-\sqrt{3}$
coordinates of the terminal point associated with $t=\pi /12$ are $\left(\frac{\sqrt{2+\sqrt{3}}}{2},\frac{\sqrt{2-\sqrt{3}}}{2}\right)$

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