If #sin theta = 1/4#, theta in quadrant 2, how do you find the exact value of #sin(theta - pi/3)#?

1 Answer
Nghi N.
Jun 18, 2018

#sin (t - pi/3) = (1 + 3sqrt5)/8#

Explanation:

#sin t = 1/4#. Find cos t
#cos^2 t = 1 - sin^2 t = 1 - 1/16 = 15/16#
#cos t = - sqrt15/4# (because t lies in Quadrant 2)
From trig identity, we get:
#sin (t - pi/3) = sin t.cos (pi/3) - sin (pi/3).cos t# (1)
Note.
#sin t = 1/4# --> #cos t = - sqrt15/4# -->
#cos (pi/3) = 1/2# --> #sin (pi/3) = sqrt3/2#
Equation (1) becomes:
#sin (t - pi/3) = 1/8 + (sqrt3.sqrt15)/8 = (1 + 3sqrt5)/8#
Check by calculator.
#sin t = 0.25# --> #t = 162^@52# --> #(t - 60) = 105^@22#
#sin (t - 60) = 0.96#
#(1 + 3sqrt5)/8 = 0.96#. Proved.

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