Question

How do you evaluate `sin((19pi)/12)cos(-pi/6) - cos((19pi)/12)sin(-pi/6)`?

Answer 1

I found: `-sqrt(2)/2`

Explanation:

I think we can use the identity:
`sin(x-y)=sin(x)cos(y)-sin(y)cos(x)`
where in our case:
`x=19/12pi`
`y=-pi/6`
so we get:
`sin(19/12pi)cos(-pi/6)-sin(-pi/6)cos(19/12pi)=sin(19/12pi+pi/6)=`
`=sin((19+2)/12pi)=sin(21/12pi)=sin(7/4pi)=-sqrt(2)/2`