Question

Find all 6 trigonometric ratios of the angle θ =7∏/ 6

Answer 1

Check the picture below, those are the x,y pairs or cosine, sine values pair.

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\ cos\left( \frac{7\pi }{6} \right)=\cfrac{\stackrel{adjacent}{-\sqrt{3}}}{\stackrel{hypotenuse}{2}}\qquad \qquad sin\left( \frac{7\pi }{6} \right)=\cfrac{\stackrel{opposite}{1}}{\stackrel{hypotenuse}{2}}$

now, this is from the Unit Circle, and therefore the hypotenuse or radius wil be 1.

$\bf \begin{cases} adjacent=&-\frac{\sqrt{3}}{2}\\ opposite=&\frac{1}{2}\\ hypotenuse=&1 \end{cases}\\\\ -------------------------------\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \quad % tangent tan(\theta)=\cfrac{opposite}{adjacent} \\\\\\ % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \qquad % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \quad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}$

so, just plug in those values... hmmm lemme do the tangent, so you see the division of two fractions.

$\bf tan\left( \frac{7\pi }{6} \right)=\cfrac{sin\left( \frac{7\pi }{6} \right)}{cos\left( \frac{7\pi }{6} \right)}\implies tan\left( \frac{7\pi }{6} \right)=\cfrac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}\implies tan\left( \frac{7\pi }{6} \right)=\cfrac{1}{2}\cdot \cfrac{2}{-\sqrt{3}}$

$\bf tan\left( \frac{7\pi }{6} \right)=\cfrac{1}{-\sqrt{3}}\impliedby \textit{now, let's \underline{rationalize} the denominator} \\\\\\ \cfrac{1}{-\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{-(\sqrt{3})^2}\implies -\cfrac{\sqrt{3}}{3}$