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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