Question

Need help with homework

Answer 1

For this case we have the following angle:

`\theta=-(23\pi)/(6)=-690º`

Part a:

Quadrant I

Part b:

We can do this 2*360 -690 = 30º

`(\pi)/(6)=30º`

Part c

for this case we can do this:

`x=1\cdot\cos 30=\frac{\sqrt[]{3}}{2},y=1\cdot\sin 30=(1)/(2)`

Then the point where theta intersect the unit circle is:

`(\frac{\sqrt[]{3}}{2},(1)/(2))`