Hi, I am trying to follow the logic of tis problem and the solution.

cos (- 7pi/2) = cos(-4pi + pi/2) = cos pi/2 = 0.

Why do we use -4pi + pi/2 to get pi/2? How do we know to add -4pi + pi/2?

Thanks a lot for any help.

2. Period of sine or cosine has nothing to do with this.

This is just to explain the posted
"cos (- 7pi/2) = cos(-4pi + pi/2) = cos pi/2 = 0."

-7pi/2
= ((-8 +1)pi)/2
= -8pi/2 +pi/2
= -4pi +pi/2
That is easy.

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The cos(-4pi +pi/2) = cos(pi/2) is not that easy.

1) The author of the answer could have meant that by going all the way to -4pi then going back pi/2, the resulting angle is the same as pi/2.
(In measuring the -4pi, we go two revolutions clockwise. Then, pi/2 is pi/2 by going the normal way, counterclockwise.)

2)Another way is by using the trig identity cos(A+B) = cosAcosB -sinAsinB. (I assume that the author did not mean to use this, though.)
cos(-4pi +pi/2)
= cos(-4pi)cos(pi/2) -sin(-4pi)sin(pi/2)
= (1)cos(pi/2) -(0)sin(pi/2)
= cos(pi/2) -0
= cos(pi/2)

I am sure you know that cos(-4pi) = 1, and sin(-4pi) = 0.

3. Tanks for your help ticbol. I couldn't have solved it without your way of expleaning it.

,

using period to evaluate sine and cosine 3

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