How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle #67^circ 30'#?

1 Answer
Mar 10, 2017

#sin t = sqrt(2 + sqrt2)/2#
#cos t = sqrt(2 - sqrt2)/2#
#tan t = sqrt(2 + sqrt2)/(sqrt(2 - sqrt2))#

Explanation:

Call #67^@30'= t# --> #2t = 135^@#
Use trig table, and trig identity:
#2sin^2 t = 1 - cos 2t#, and #2cos^2 t = 1 + cos 2t#
#2sin^2 t = 1 + sqrt2/2 = (2 + sqrt2)/2#
#sin^2 t = (2 + sqrt2)/4#
#sin t = sin (67.50) = +- sqrt(2 +- sqrt2)/2#
sin (67.50) is positive. Take the positive answer
#sin (67.50) = sqrt(2 + sqrt2)/2#
#2cos^2 t = 1 + cos 2t = 1 - sqrt2/2 = (2 - sqrt2)/2#
#cos t = cos (67.50) = sqrt(2 - sqrt2)/2#
#tan t = sin/(cos) = sqrt(2 + sqrt2)/(sqrt(2 - sqrt2))#