Question

How do you evaluate `Sin ((-13pi)/4)`?

Answer 1

`1/sqrt2`.

Explanation:

First, we use `sin(-theta)=-sintheta`, so,

`sin(-13pi/4)=-sin(13pi/4)=-sin(3pi+pi/4)`.

This shows that, `13pi/4=3pi+pi/4` lies in the Third Quadrant in

which, `sin` is `-ve.`

Finally, `sin(3pi+pi/4)=-sin(pi/4)=-1/sqrt2`.

Hence, `sin(-13pi/4)-(-1/sqrt2)=1/sqrt2`.

Answer 2

`sqrt2/2`

Explanation:

Trig table of special arcs, unit circle, and property of supplementary arcs -->
`sin ((-13pi)/4) = sin (-pi/4 - (12pi)/4) = sin (-pi/4 - 3pi) =`
`= sin (-pi/4 - pi) = - sin (pi/4 + pi) = - (- sin pi/4) =`
`= sin (pi/4) = sqrt2/2`