How do you express #cos( (4 pi)/3 ) * cos (( pi) /6 ) # without using products of trigonometric functions?

2 Answers
Nghi N.
Apr 1, 2016

#- sqrt3/2#

Explanation:

#P = cos ((4pi)/3).cos (pi/6)#
Trig table --> #cos (pi/6) = sqrt3/2#
#cos ((4pi)/3) = cos (pi/3 + (3pi)/3) = cos (pi/3 + pi) = - cos (pi/3) = -1/2#
#P = (-1/2)(sqrt3/2) = - sqrt3/4#

Bdub
Apr 4, 2016

#cos ((4pi)/3) cos (pi/6)=1/2 cos ((9pi)/6) +1/2 cos ((7pi)/6)=-sqrt3/4#

Explanation:

#2 cos A cos B=cos (A+B) + cos (A-B) #
# cos A cos B=1/2 (cos (A+B) + cos (A-B)) #
#A=(4pi)/3, B = pi/6#
# cos ((4pi)/3) cos (pi/6)=1/2 (cos ((4pi)/3+pi/6) + cos ((4pi)/3-pi/6))#
#=1/2 (cos ((9pi)/6) + cos ((7pi)/6))#
#=1/2 cos ((9pi)/6) +1/2 cos ((7pi)/6)#
#=1/2 (0) + 1/2 (-sqrt3/2)#
#=-sqrt3/4#
#cos ((4pi)/3) cos (pi/6)=1/2 cos ((9pi)/6) +1/2 cos ((7pi)/6)=-sqrt3/4#

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