Question

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

Answer 1

`(1/4)(sqrt(2 - sqrt3))`

Explanation:

a. Find sin (pi/6).
Trig Table of special arcs --> `sin pi/6 = 1/2`
b. Find cos (5pi)/2
`cos (10pi)/12 = cos (5pi/6) = -sqrt3/2`
Use trig identity: `cos 2x = 2cos^2 x - 1`
`cos (5pi/6) = -sqrt3/2 = 2cos^2 (5pi/12) - 1`
`2cos^2 (5pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2`
`cos (5pi/12) = sqrt(2 - sqrt3)/2`
Finally,
`sin pi/6.cos ((5pi)/12) = (1/4)(sqrt(2 - sqrt3))`