How do you find the other trigonometric functions given csc(pi/2-theta)=9, ?

Feb 28, 2017

We should first expand the parentheses. Recall that $\csc x = \frac{1}{\sin} x$:

$\frac{1}{\sin} \left(\frac{\pi}{2} - \theta\right) = 9$

Expand using the identity $\sin \left(A - B\right) = \sin A \cos B - \sin B \cos A$.

$\frac{1}{\sin \left(\frac{\pi}{2}\right) \cos \theta - \sin \theta \cos \left(\frac{\pi}{2}\right)} = 9$

$\frac{1}{1 \left(\cos \theta\right) - \sin \theta \left(0\right)} = 9$

$\frac{1}{\cos} \theta = 9$

$\sec \theta = 9$

This would signify that if we drew a triangle, the hypotenuse would measure $9$ units and the side adjacent $\theta$ would measure $1$, because $\sec \theta = \text{hypotenuse"/"adjacent}$.

We now calculate the side opposite $\theta$ (call it $b$):

${b}^{2} = {9}^{2} - {1}^{2}$

${b}^{2} = 80$

$b = \sqrt{80}$

$b = 4 \sqrt{5}$

We now apply the definitions for the five other trig functions.

$\sin \theta = \text{opposite"/"hypotenuse} = \frac{4 \sqrt{5}}{9}$

$\csc \theta = \text{hypotenuse"/"opposite} = \frac{9}{4 \sqrt{5}} = \frac{36 \sqrt{5}}{16 \left(5\right)} = \frac{9 \sqrt{5}}{20}$

$\cos \theta = \text{adjacent"/"hypotenuse} = \frac{1}{9}$

$\tan \theta = \text{opposite"/"adjacent} = \frac{4 \sqrt{5}}{1} = 4 \sqrt{5}$

$\cot \theta = \text{adjacent"/"opposite} = \frac{1}{4 \sqrt{5}} = \frac{\sqrt{5}}{20}$

Hopefully this helps!