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

2 Answers
Bdub
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)]

Nghi N.
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)

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