Think about the graph of y = cos(x) over the interval [0, 2pi].
You have the following values:
y(0) = 1
y(pi/2) = 0
y(pi) = -1
y(3pi/2) = 0
y(2pi) = 1
As you can see, cosine has a period of 2pi, and the argument (the value inside cosine) doesn't need to have a denominator, as 0, pi, 2pi, 3pi, etc... clearly equal either 1 or -1.
Okay, basically, cosine has a period of 2pi, so you can add/subtract any interval of 2pi from a large argument (like -33pi) to find an easy to work with reference number. You can do this because angles larger than 360 degrees = 2pi simply revolve around the unit circle more than once.
So add 17(2pi) = 34pi to to -33pi and you get just pi. Remember what I said earlier? Any angle larger than 2pi simply generations more revolutions.
So:
cos(-33pi + 34pi) = -1
cos(pi) = -1
TRUE
Hope I helped! If you need additional explanation just say so.
Comments
-33pi = -32pi - pi
You can ignore multiples of 2pi like -32pi, since they don't change the value.
cos(-33pi) = cos(-pi)
NOW do you know why it's -1? -pi is -180 degrees.
Think about the graph of y = cos(x) over the interval [0, 2pi].
You have the following values:
y(0) = 1
y(pi/2) = 0
y(pi) = -1
y(3pi/2) = 0
y(2pi) = 1
As you can see, cosine has a period of 2pi, and the argument (the value inside cosine) doesn't need to have a denominator, as 0, pi, 2pi, 3pi, etc... clearly equal either 1 or -1.
Okay, basically, cosine has a period of 2pi, so you can add/subtract any interval of 2pi from a large argument (like -33pi) to find an easy to work with reference number. You can do this because angles larger than 360 degrees = 2pi simply revolve around the unit circle more than once.
So add 17(2pi) = 34pi to to -33pi and you get just pi. Remember what I said earlier? Any angle larger than 2pi simply generations more revolutions.
So:
cos(-33pi + 34pi) = -1
cos(pi) = -1
TRUE
Hope I helped! If you need additional explanation just say so.
Start by taking arccos(-1).