How do you find the exact value of all 6 trigonometric functions for the angle pi/6?

1 Answer
George C.
Sep 26, 2015

Construct a right angled triangle with angles #pi/6#, #pi/3#, #pi/2# and look at the ratios of the lengths of the sides.

Explanation:

Consider an equilateral triangle with sides of length #2#. It will have angles #pi/3#, #pi/3#, #pi/3#.

Bisect it to make two right angled triangles, with sides of length #1#, #sqrt(3)# and #2#, since #1^2 + sqrt(3)^2 = 1 + 3 = 4 = 2^2#.

The smallest angle of one of these right angled triangles will be #pi/6# (the others being #pi/3# and #pi/2#).

Hence:

#sin (pi/6) = 1/2# (from: sin = opposite/hypotenuse)

#cos (pi/6) = sqrt(3)/2# (from: cos = adjacent/hypotenuse)

#tan (pi/6) = 1/sqrt(3)# (from: tan = opposite/adjacent)

Then the reciprocal trig functions:

#csc (pi/6) = 1/sin (pi/6) = 2/1 = 2#

#sec (pi/6) = 1/cos (pi/6) = 2/sqrt(3)#

#cot (pi/6) = 1/tan (pi/6) = sqrt(3)/1 = sqrt(3)#

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