Question

How do you express `sin(pi/ 6 ) * cos( ( 9 pi) / 8 )` without using products of trigonometric functions?

Answer 1

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

Explanation:

`sin(pi/6)*cos(pi+pi/8)= -1/2*cos(pi/8)`
`= -1/2sqrt(1/2*(1+cos(pi/4))) =-1/2sqrt(1/2*(1+1/sqrt2))`

It may also be expressed as sum of two trigonometric functions
using formula
`sinAcosB=1/2*(sin(A+B)+sin(A-B))`

`sin(pi/6)*cos(pi+pi/8)= -sin(pi/6)*cos(pi/8)`

`=-1/2*(sin(pi/6+pi/8)+sin(pi/6-pi/8))`

`=-1/2*(sin(7pi/24)+sin(pi/24)`

Answer 2

`- (1/2)(cos pi/8)`

Explanation:

`P = sin (pi/6).cos ((9pi)/8)`
Trig table --> `sin (pi/6) = 1/2`
`cos ((9pi)/8) = cos (pi/8 + pi) = - cos (pi/8)`
The product P can be expressed as:
`P = - (1/2) (cos pi/8)`

If being asked, we can find P by evaluating `cos (pi/8)`, using the trig identity: `cos 2a =2cos^2a - 1.`