How do you express sin(pi/ 6 ) * cos( ( 5 pi) / 12 ) sin(π6)cos(5π12) without using products of trigonometric functions?

1 Answer
Jan 25, 2016

(1/4)(sqrt(2 - sqrt3))(14)(23)

Explanation:

a. Find sin (pi/6).
Trig Table of special arcs --> sin pi/6 = 1/2sinπ6=12
b. Find cos (5pi)/2
cos (10pi)/12 = cos (5pi/6) = -sqrt3/2cos(10π)12=cos(5π6)=32
Use trig identity: cos 2x = 2cos^2 x - 1cos2x=2cos2x1
cos (5pi/6) = -sqrt3/2 = 2cos^2 (5pi/12) - 1cos(5π6)=32=2cos2(5π12)1
2cos^2 (5pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/22cos2(5π12)=132=232
cos (5pi/12) = sqrt(2 - sqrt3)/2cos(5π12)=232
Finally,
sin pi/6.cos ((5pi)/12) = (1/4)(sqrt(2 - sqrt3)) sinπ6.cos(5π12)=(14)(23)