Question

How do you express `sin(pi/ 6 ) * cos( (3 pi) / 8 )` without using products of trigonometric functions?

Answer 1

`sqrt(2 - sqrt2)/4`

Explanation:

`P = sin (pi/6).cos ((3pi)/8).`
Trig table --> `sin pi/6 = 1/2`
Find `cos ((3pi)/8)` by the identity: `cos 2a = 2cos^2 a - 1`
`cos ((6pi)/8) = 2cos^2 ((3pi)/8) - 1.`
`cos ((3pi)/4) = -sqrt2/2 = 2cos^2 ((3pi)/8) - 1`
`2cos^2 ((3pi)/8) = 1 - sqrt2/2 = (2 - sqrt2)/2`
`cos^2 ((3pi)/8) = (2 - sqrt2)/4`
`cos ((3pi)/8) = sqrt(2 - sqrt2)/2` (since `cos ((3pi)/8)` is positive.
Finally,

`P = (1/2)(sqrt(2 - sqrt2)/2) = (sqrt(2 - sqrt2)/4)`