Question

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

Answer 1

`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))`

Answer 2

`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)⋅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)`