Find the exact value of sin(pi/12) , cos (11pi/12) , and tan (7pi/12)?

1 Answer
Sep 25, 2015

Find sin (pi/12); cos (pi/12) and tan ((7pi)/12)

Explanation:

Call #sin pi/12 = sin t.# Use trig identity: #cos 2a = 1 - sin^2 a#
#cos 2pi/12 = cos pi/6 = sqrt3/2 = 1 - sin^2 t.#
#sin^2 t = 1 - sqrt3/2 = (2 - sqrt3)/4#
#sin t = +- sqrt(2 - sqrt3)/2#. Since #pi/12# has its sin positive, then
#sin (pi/12) = sin t = sqrt(2 - sqrt3)/2#

Call #cos (pi/2) = cos t#
#cos ((2pi)/12) = sqrt3/2 = 2cos^2 t - 1#
#cos^2 t = (2 + sqrt3)/4#
#cos t = +- sqrt(2 + sqrt3)/2#. Since cos #(pi/12)# is positive, then
#cos (pi/12) = cos t = sqrt(2 + sqrt3)/2#

#tan ((7pi)/12) = tan (pi/12 + pi) = tan (pi/12) = sin/(cos) = #
#= (sqrt(2 - sqrt3))/(sqrt(2 + sqrt3))#