Question

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

Answer 1

We should first expand the parentheses. Recall that `cscx = 1/sinx`:

`1/sin(pi/2- theta) = 9`

Expand using the identity `sin(A - B) = sinAcosB - sinBcosA`.

`1/(sin(pi/2)costheta - sinthetacos(pi/2)) = 9`

`1/(1(costheta) - sintheta(0)) = 9`

`1/costheta = 9`

`sectheta = 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 `sectheta = "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 = 4sqrt(5)`

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

`sintheta = "opposite"/"hypotenuse" = (4sqrt(5))/9`

`csctheta = "hypotenuse"/"opposite" = 9/(4sqrt(5)) = (36sqrt(5))/(16(5)) = (9sqrt(5))/20`

`costheta = "adjacent"/"hypotenuse" = 1/9`

`tantheta = "opposite"/"adjacent" = (4sqrt(5))/1 = 4sqrt(5)`

`cottheta = "adjacent"/"opposite" = 1/(4sqrt(5)) = sqrt(5)/20`

Hopefully this helps!