Question

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

Answer 1

`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))`