How do you express sin(pi/ 8 ) * cos(( ( 5 pi) / 12 ) sin(π8)cos((5π12) without using products of trigonometric functions?

1 Answer
Feb 9, 2016

((sqrt(2 - sqrt2)).(sqrt(2 + sqrt3))]/4(22).(2+3)4

Explanation:

Product P = sin (pi/8).cos ((5pi)/12)P=sin(π8).cos(5π12).
a. Find sin (pi/8)sin(π8) by trig identity: cos 2x = 1 - 2sin^2 xcos2x=12sin2x
cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)cos(π4)=22=12sin2(π8)
sin^2 (pi/8) = (2 - sqrt2)/4sin2(π8)=224
sin (pi/8) = sqrt(2 - sqrt2)/2sin(π8)=222 --> (sin pi/8π8 is positive)
b. Find cos ((5pi)/12) = cos tcos(5π12)=cost by identity: cos 2t = 2cos^2 t - 1cos2t=2cos2t1
cos 2t = cos ((10pi)/12) = cos ((5pi)/6) = -sqrt3/2cos2t=cos(10π12)=cos(5π6)=32
sqrt3/2 = 2cos^2 t - 132=2cos2t1
2cos^2 t = 1 + sqrt3/2 = (2 + sqrt3)/22cos2t=1+32=2+32
cos^2 t = (2 + sqrt3)/4cos2t=2+34
cos t = cos ((5pi)/12) = sqrt(2 + sqrt3)/2cost=cos(5π12)=2+32--> (cos t is positive)

Finally: P = [(sqrt(2 - sqrt2}).(sqrt(2 + sqrt3))]/4P=(22).(2+3)4