Question

How do you express `sin(pi/ 8 ) * cos(( 7 pi)/6 )` without using products of trigonometric functions?

Answer 1

`-(sqrt3)(sqrt(2 - sqrt2)/4)`

Explanation:

First, find `sin (pi/8)` and `cos ((7pi)/6)` separately.
Call `sin (pi/8) = sin t`
Use the trig identity: `cos (2t) = 1 - 2sin^2 t`
`cos (pi/4) = sqrt2/2 = 1 - sin^2 t`
`2sin^2 t = 1 - sqrt2 = (2 - sqrt2)/2`
`sin^2 t = (2 - sqrt2)/4`
`sin (pi/8) = sin t = (sqrt(2 - sqrt2))/2`

`cos ((7pi)/6) = cos (pi/6 + pi) = - cos (pi/6) = - sqrt3/2`

Finally, `sin (pi/8).cos ((7pi)/6) = - (sqrt3)(sqrt(2 - sqrt2)/4)`