# Thread: why is cos(t) -1 as opposed to 1 when P = 3pi

1. ## why is cos(t) -1 as opposed to 1 when P = 3pi

find the coordinates of P and the exact values of the trigonomic functions at t whenever possible.

given

My answers
Sin(t) = 0 csc(t)= 1/0 or Undefined
cos(t) = 1 sec(t) = 1/1 or 1
tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined

The book

Sin(t) = 0 csc(t)= 1/0 or Undefined
cos(t) =- 1 sec(t) = - 1/1 or - 1
tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined

Given that the the point is why is cos(t) -1 as opposed to 1?

2. Originally Posted by rasczak
Given that the the point is why is cos(t) -1 as opposed to 1?
Because $-3\pi$ is equivalent to $\pi$
Two numbers, $t~\&~s$, are equivalent if $t=s+2n\pi$ for some integer $n$.

3. Originally Posted by rasczak
find the coordinates of P and the exact values of the trigonomic functions at t whenever possible.

given

My answers
Sin(t) = 0 csc(t)= 1/0 or Undefined
cos(t) = 1 sec(t) = 1/1 or 1
tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined

The book

Sin(t) = 0 csc(t)= 1/0 or Undefined
cos(t) =- 1 sec(t) = - 1/1 or - 1
tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined

Given that the the point is why is cos(t) -1 as opposed to 1?
Hi rasczak,

A rotation of $\pi$ radians is an anti-clockwise rotation of $180^o$ or a half circle.

A rotation of $2{\pi}$ is a full circle, back to the starting point.

Therefore, a rotation of $3{\pi}$ brings you to the same position that a rotation of $\pi$ does.

This position is the extreme left of the unit circle.
The point here is (-1,0).

The horizontal co-ordinate is $Cos(angle)$ and the vertical co-ord is $Sin(angle)$ for any point on the circumference of the unit circle centred at (0,0).

If the rotation is done in a clockwise direction, we refer to this using negative angles.

Hence, a rotation of $3{\pi}$ ends up at the same position that a rotation of $-3{\pi}$ ends up in.

This corresponds to an angle $\pi$