Question

How do you express `sin(pi/ 6 ) * cos( ( 15 pi) / 8 )` without using products of trigonometric functions?

Answer 1

`-(1/4)sqrt(2 + sqrt2)`

Explanation:

Trig Table of Special arcs --> `sin (pi/6) = 1/2`
`cos ((15pi)/8) = cos (-pi/8 + 2pi) = -cos (pi/8)`
Find (cos (pi/8)) by using trig identity:
`cos (pi/4) = sqrt2/2 = 2cos^2 (pi/8) - 1`
`2cos^2 (pi/8) = 1 + sqrt2/2 = (2 + sqrt2)/2`
`cos^2 (pi/8) = (2 + sqrt2)/4.`
`cos (pi/8) = + sqrt(2 + sqrt2)/2`. (because cos (pi/8) is positive)
Finally,
`sin (pi/6).cos ((15pi)/8) = -1/2cos (pi/8) =`
`= -(1/4)sqrt(2 + sqrt2)`