Question

How do you evaluate `sin ((-5pi)/6)`?

Answer 1

`sin(-(5pi)/6)=-1/2`

Explanation:

As `sin(-theta)=-sintheta` and `sin(pi-theta)=sintheta`

`sin(-(5pi)/6)=-sin((5pi)/6)`

= `-sin((6pi-pi)/6)=-sin(pi-pi/6)`

= `-sin(pi/6)`

= `-1/2`