How do you express #cos( (5 pi)/6 ) * cos (( 7 pi) /12 ) # without using products of trigonometric functions?

2 Answers
Nghi N.
Mar 2, 2016

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

Explanation:

Trig table --> #cos ((5pi)/6) = -sqrt3/2.#
Trig unit circle -->
#cos ((7pi)/12) = cos (pi/12 + pi/2) = - sin (pi/12)#
Product #P = sqrt3/2.sin (pi/12)#.
Find #sin (pi/12)# by the identity: #cos (pi/6) = 1 - 2sin^2 (pi/12)#
#2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2#
#sin^2 (pi/12) = (2 - sqrt3)/4#
Take square root of both sides:
#sin (pi/12) = sqrt(2 - sqrt3)/2# (because #sin (pi/12)# is positive)
Finally,
#P = sqrt3/4(sqrt(2 - sqrt3))#

Shwetank Mauria
Mar 2, 2016

#1/2*cos(pi/4)-1/2*cos(5pi/12)#

Explanation:

As #cos(A+-B)=cosA*cosB-+sinA*sinB#, we can write

#cosA*cosB=1/2[(cos(A+B)+cos(A-B)]#

Hence, #cos(5pi/6)â‹…cos(7pi/12)# =

1/2[(cos(5pi/6+7pi/12)+cos(5pi/6-7pi/12)]#

= #1/2(cos(17pi/12)+cos(3pi/12))#

= #1/2(-cos(5pi/12)+cos(pi/4))#

=#1/2*cos(pi/4)-1/2*cos(5pi/12)#

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