Question

How do you express `sin(pi/ 6 ) * cos( ( 5 pi) / 4 )` without using products of trigonometric functions?

Answer 1

`- sqrt2/4`

Explanation:

`cos ((5pi)/4) = cos (pi/4 + pi) = - cos (pi/4)`
Trig table of special arcs -->
`sin (pi/6) = 1/2`, and `cos (pi/4) = sqrt2/2`
There for,
`sin (pi/6)cos ((5pi)/4) = (1/2)(- sqrt2/2) = - sqrt2/4`