Sin[ArcCos(-5/13)]

Question

Use Principal Values to evaluate. Show all work.

How do I make the graph that goes along with this?

Answer 1

NOTE that the exact values of x, y, and the hypotenuse are not needed. Only the ratios are.

So you can use x = - 5. You are dealing with right triangles, so if the hypotenuse is 13, the y value must be ± 12.

(The 13 can be used for the denominator of your ratio.)

Cosine = x / hypotenuse

AND cosine inverse is defined only in quadrants 1 and 2. Because -5<0, this means we are in quadrant 2. y>0 in quadrant 2, so the answer is 12/13.

Answer 2

sin[Arcos(-5/13)] = √1 - (-5/13)² = 12/13