reference angle to find the exact value of the expression

Question

Use reference angle to find the exact value of the expression. Show your work.

1. sin (-300°)
2. cos 1200°

Answer 1

sin-300 = sin(360+300) = sin60 = (sqr3)/2

cos1200 = cos(1200-3(360)) = cos(1200-1080) = cos120 = -1/2

Answer 2

The reference angle is the positive acute angle the terminal side makes with the x-axis. In the case of -300° here is a picture showing where the reference angle is:

So the reference angle is 60°

sin(-300°) = sin(reference angle) adjusted for the quadrant = sin(60°) = √3/2

In the case of 1200°, we can subtract 360° three times and still be in the same location so, 1200° - 360° - 360° - 360° = 120°

Then, the reference angle for 120° is shown below:

So the reference angle is again 60°.

cos(1200°) = cos(reference angle) adjusted for the quadrant = cos(60°) but since we are in quadrant II, and the cosine is negative in quadrant II, then the value is -1/2

Answer 3

I am not sure what "reference angle" means, but

the trig functions of -300° are the same as those for 60° (add 360°) and

cos 1200° is the same cos 220° (1200-3*360) which is that same as -cos 40°.

The functions of 60° are known since in a 30-60-90 right triangle the sides are in the ratio 2:1:√3.

The functions of 40° can only be calculated by using the Taylor series for 40° (or some other angle with an angle known from other considerations, e.g. 30°+10°)