How do you express cos(pi/ 3 ) * sin( ( 9 pi) / 8 ) cos(π3)sin(9π8) without using products of trigonometric functions?

1 Answer
Jun 24, 2016

P = - (1/2)sin (pi/8)P=(12)sin(π8)

Explanation:

Product P = cos (pi/3).sin ((9pi)/8)P=cos(π3).sin(9π8)
Trig table --> cos (pi/3) = 1/2cos(π3)=12
sin ((9pi)/8) = sin (pi/8 + pi) = - sin (pi/8)sin(9π8)=sin(π8+π)=sin(π8)
P can be expressed as:
P = - (1/2)sin (pi/8).P=(12)sin(π8).
Note. We can evaluate sin (pi/8)sin(π8) by using the trig identity:
cos 2a = 1 - 2sin^2 a.cos2a=12sin2a.
cos ((2pi)/8) = cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)cos(2π8)=cos(π4)=22=12sin2(π8)
sin^2 (pi/8) = 1 - sqrt2 = (2 - sqrt2)/4sin2(π8)=12=224
sin (pi/8) = sqrt(2 - sqrt2)/2sin(π8)=222.
Take the positive value since sin (pi/8) is positive.
P = - (1/4)sqrt(2 - sqrt2)P=(14)22