How do you express #cos(pi/ 3 ) * sin( ( 11 pi) / 8 ) # without using products of trigonometric functions?

2 Answers
Apr 15, 2016

#=1/2 [sin ((41pi)/24) + sin ((25pi)/24)]#

Explanation:

Use formula

# sin (A+B) - sin (A-B) = 2 cos A sin B#

#cosAsinB=1/2 [sin (A+B) - sin (A-B)]#

#A=pi/3 and B=(11pi)/8#

#cos(pi/3)sin((11pi)/8)=1/2 [sin (pi/3+(11pi)/8) - sin (pi/3-(11pi)/8)]#

#=1/2 [sin ((41pi)/24) - sin ((-25pi)/24)]#

#=1/2 [sin ((41pi)/24) + sin ((25pi)/24)]#

Apr 16, 2016

#- (1/2)sin ((3pi)/8)#

Explanation:

#P = cos (pi/3).sin ((11pi)/8) #
Trig table --># cos (pi/3) = 1/2#.
#sin ((11pi)/8) = sin ((3pi)/8 + pi) = - sin ((3pi)/8)#
The product can be expressed as
#P = -(1/2)sin((3pi)/8)#

We can evaluate P by applying the identity: #cos 2a = 1 - 2sin^2 a#
#cos ((6pi)/8) = cos ((3pi)/4) = - sqrt2/2 = 1 - 2sin^2 ((3pi)/8)#
#2sin^2 ((3pi)/8) = 1 + sqrt2/2 = (2 + sqrt2)/2#
#sin^2 ((3pi)/4) = (2 + sqrt2)/4#
#sin ((3pi)/8) = sqrt(2 + sqrt2)/2# --> #sin ((3pi)/8)# is positive.
Finally,
#P = - (1/2).sin ((3pi)/8) = -(1/4)sqrt(2 + sqrt2)#