Question

How do you express `sin(pi/ 6 ) * cos( ( 13 pi) / 8 )` without using products of trigonometric functions?

Answer 1

`=>1/2*sqrt(1/2(1-1/sqrt2)`

Explanation:

`sin(pi/6)*cos(3pi/8)`
`=>sin(pi/6)*sqrt(1/2(1+cos(cancel2*3pi/cancel8^4))`
since `costheta =sqrt(1/2(1+cos2theta))`
`=>1/2*sqrt(1/2(1+cos(3pi/4))`
`=>1/2*sqrt(1/2(1+cos(pi-pi/4))`
`=>1/2*sqrt(1/2(1-cos(pi/4))`
`=>1/2*sqrt(1/2(1-1/sqrt2)`

Answer 2

`sin (pi/8)/2`

Explanation:

P = sin (pi/6).cos ((13pi)/8)
sin (pi/6) = 1/2
On the trig unit circle,
`cos ((13pi)/8) = cos (-(3pi)/8 + 2pi) = cos ((-3pi)/8) =`
`= cos ((3pi)/8) = cos (-pi/8 + pi/2) = sin (pi/8).`
Therefor, the product can be expressed as:
`P = sin (pi/8)/2`
Next, we can find the product P by evaluating `sin (pi/8)`