Question

Find an exact value.

sin(-11/12)

Answer 1

You have to use a sum identity.

Think of your specials:
\[\frac{ \pi }{ 4 } = \frac{ 3\pi }{ 12 }\]\[\frac{ \pi }{ 3 } = \frac{ 4\pi }{ 12 }\]\[\frac{ \pi }{ 6 } = \frac{ 2\pi }{12}\]

11 isn't going to result from adding just two of these together. However, doesn't 8 + 3 = 11? and 8 = 2*4. Therefore, you can use \[\frac{ 8\pi }{ 12 } + \frac{ 3\pi }{ 12 }\] \[= \frac{ 2\pi }{ 3 } + \frac{ \pi }{ 4 }\]

So, recall that \[\sin(a-b) = \sin(a)*\cos(b) - \cos(a)*\sin(b)\]
So, you would set this up like \[\sin(-\frac{ 2\pi }{ 3 }-\frac{ \pi }{ 4 }) = \sin(-\frac{ 2\pi }{ 3 })*\cos(\frac{ \pi }{ 4 }) - \cos(-\frac{ 2\pi }{ 3 })*\sin(\frac{ \pi }{ 4 })\] Can you go from there?