Please help with Using Period to Evaluate Sin and Cosine
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?
Please help me to understand what's going on with this one.
Thanks a lot for any help.
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."
= ((-8 +1)pi)/2
= -8pi/2 +pi/2
= -4pi +pi/2
That is easy.
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)cos(pi/2) -sin(-4pi)sin(pi/2)
= (1)cos(pi/2) -(0)sin(pi/2)
= cos(pi/2) -0
I am sure you know that cos(-4pi) = 1, and sin(-4pi) = 0.