Question

(-2/3, sqrt5/3) is a point on a unit circle. Find the cosine, cosecant, and sine of the angle.

Answer 1

Look at the picture.

$\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{\frac{y}{r}}=\dfrac{r}{y}$

We have the right triangle x, y and r. From the Pythagorean theorem we have:

$r^2=x^2+y^2\to r=\sqrt{x^2+y^2}$

We have the point

$\left(-\dfrac{2}{3};\ \dfrac{\sqrt5}{3}\right)$

Substitute:

$r=\sqrt{\left(-\dfrac{2}{3}\right)^2+\left(\dfrac{\sqrt5}{3}\right)^2}\\\\r=\sqrt{\dfrac{4}{9}+\dfrac{5}{9}}\\\\r=\sqrt{\dfrac{9}{9}}\\\\r=1$

$\csc\theta=\dfrac{1}{\frac{\sqrt5}{3}}=\dfrac{3}{\sqrt5}=\dfrac{3\cdot\sqrt5}{\sqrt5\cdot\sqrt5}=\boxed{\dfrac{3\sqrt5}{5}}$