Question

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

Answer 1

`=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)]`

Answer 2

`- (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)`