How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle #(7pi)/12#?

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Answer 1 · 2017-05-01T05:30:36

#sin ((7 pi)/12) = sqrt(2+sqrt(3))/2#
#cos ((7 pi)/12) = -sqrt(2-sqrt(3))/2#
#tan ((7 pi)/12) = -sqrt( (2+sqrt(3))/(2-sqrt(3)))#

Given: Use the half angle formulas and angle #(7 pi)/12 = 105^@#

This angle is found in the second quadrant where sine is positive, cosine is negative and tangent is negative.

Let #theta = (7 pi)/6 " since "((7 pi)/6)/2 = ((7 pi)/6)/(2/1) = (7 pi)/6 * 1/2 = (7 pi)/12#

#sin (theta/2) = +-sqrt((1-cos theta)/2) " sign based on quadrant"#

#cos (theta/2) = +-sqrt((1+cos theta)/2) " sign based on quadrant"#

#tan (theta/2) = +- sqrt((1-cos theta)/(1+ cos theta))" sign based on quadrant"#

From a trigonometric circle or a #30-60-90^@# in the 3rd quadrant where #(7 pi)/6 = 210^@#:

#cos ((7 pi)/6) = (-sqrt(3))/2#

#sin ((7 pi)/12) = sqrt((1- (-sqrt(3))/2)/2) = sqrt((2+sqrt(3))/2 * 1/2) #

#= sqrt((2+sqrt(3))/4) = sqrt(2+sqrt(3))/2#

#cos ((7 pi)/12) = -sqrt((1+ (-sqrt(3))/2)/2) = -sqrt((2-sqrt(3))/2 * 1/2) #

#= -sqrt((2-sqrt(3))/4) = -sqrt(2-sqrt(3))/2#

#tan ((7 pi)/12) = - sqrt((1-(-sqrt(3))/2)/(1-sqrt(3)/2)) = - sqrt(((2+sqrt(3))/2)/((2-sqrt(3))/2)) #

#= - sqrt((2+sqrt(3))/cancel(2) * cancel(2)/(2-sqrt(3))) = -sqrt( (2+sqrt(3))/(2-sqrt(3)))#