How do you express #cos( (5 pi)/6 ) * cos (( 5 pi) /12 ) # without using products of trigonometric functions?

1 Answer
Nghi N.
Mar 18, 2016

#P = - (sqrt3/4)(sqrt(2 - sqrt3))#

Explanation:

#P = cos ((5pi)/6).cos ((5pi)/12)#
Trig table --> #cos ((5pi)/6) = -sqrt3/2#
Find #cos ((5pi)/12)# by the identity: #cos 2a = 2cos^2a - 1#
#cos ((5pi)/6) = -sqrt3/2 = 2cos^2 ((5pi)/12) - 1#
#2cos^2 ((5pi)/12) = 1 - sqrt3/2 = (2 - sqrt3)/2#
#cos^2 ((5pi)/12) = (2 - sqrt3)/4#
#cos ((5pi)/12) = sqrt(2 - sqrt3)/2# -->#cos ((5pi)/12)# is positive.
Finally,
#P = (-sqrt3/2)(sqrt(2 - sqrt3)/2) = - (sqrt3/4)(sqrt(2 - sqrt3))#

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