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

##### 1 Answer

We should first expand the parentheses. Recall that

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

Expand using the identity

#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

We now calculate the side opposite

#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!