Circular Functions

[PopcornTime]

What is the easiest method to work out questions like:

Evaluate:

sin(2pi/3), sin(7pi/6), cos(7pi/4), cos(-21pi/3)

I'm a little confused about how to located these on the unit circle but understand how the table of exact values works.

[S200]

What is the easiest method to work out questions like:

Evaluate:

sin(2pi/3), sin(7pi/6), cos(7pi/4), cos(-21pi/3)

I'm a little confused about how to located these on the unit circle but understand how the table of exact values works.

Ok. Major edit here, 'cause a kind user pointed out that my working didn't make sense (Btw; It was hopelessly wrong...)

Updated working

So, first find the point \(\frac {2\pi} {3}\). A little knowledge of the unit circle tells us that angles in the second quadrant are found by \(\pi - \theta\). Using this, we find that \(\frac {2\pi} {3}\) is actually \(\pi - \frac {\pi} {3}\). We know that \(\sin\) is positive in Q1 and Q2, so \(\sin {\frac {2\pi} {3}} = \sin {\frac {\pi} {3}}\).

For the others, it's a similar story.
\(\frac {7\pi}{6} = \pi + \frac {\pi}{6} \quad \therefore \) Third quarter.
\(\frac {7\pi}{4} = 2\pi - \frac {\pi} {4} \quad \therefore\) Fourth Quarter

The fourth one has been answered elsewhere, so I'll just link to that...

Here is OG's answer to that question.