Question

Evaluate trigonometic function using its period as an aid: sin5(pie)

Answer 1

Question is \[\sin5\pi\]

Answer 2

Do you know that the sine function has a period of 2pi?

Answer 3

No. We use this chart called the Unit Circle. And I go off of that.

Answer 4

So, Sin 5pi is equalivant to sin pi

Answer 5

Since the unit circle is a period of 2pi you can subtract 4pi from 5pi.

Answer 6

Thus, you will be at the same point on the unit circle which is pi.

Answer 7

Then you know that sin 0 is 0, sin pi/2 is 1, sin pi is 0, sin 3pi/2 is -1 and sin 2pi is 0 and we are all the way around the circle and ready to start over. If we go around again, that is 4 pi and we still have pi units to go. So since 5pi = 2pi+2pi+pi the sin of 5pi is the same as the sin of pi which is 0.

Answer 8

Okay. I guess that makes a little more sense.