How do you find the remaining trigonometric ratios if #tan(alpha) = 9# and #0 < alpha < pi/2#?

2 Answers
Jul 20, 2015

Answer:

Use SOH-CAH-TOA mnemonic and Pythagoras to find:

#sin(alpha) = 9/sqrt(82)# and #cos(alpha) = 1/sqrt(82)#

#csc(alpha) = sqrt(82)/9#, #sec(alpha) = sqrt(82)# and #cot(alpha) = 1/9#

Explanation:

#tan(alpha) = 9# is the #"opposite"/"adjacent"# ratio of a right angled triangle with angle #alpha#.

For our purposes, it does not matter what size the triangle is - just its proportions. So let the length of the opposite side be #9# and the length of the adjacent side be #1#. Then the length of the hypotenuse is #sqrt(9^2+1^2) = sqrt(82)#.

Hence #sin(alpha) = "opposite"/"hypotenuse" = 9/sqrt(82)#

and #cos(alpha) = "adjacent"/"hypotenuse" = 1/sqrt(82)#

Jul 20, 2015

Answer:

Find other trig functions knowing tan x = 9

Explanation:

Use a calculator.
tan x = 9 --> x = 83.66 deg
sin 83.66 = 0.99
cos 83.66 = 0.11
#cot x = 1/9#
#sec x = 1/0.11 = 9.09#
#csc x = 1/0.99 = 1.01#