Why is cos(pi) and cos (-pi) both equal to -1?

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Why is cos(pi) and cos (-pi) both equal to -1?

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Answer 1 · 2015-04-22T19:57:35

One of the reasons is because $cos$ $cos$ is an even function.

This means that $cos θ = cos (- θ)$ $costheta=cos(-theta)$

An example is the present case $cos π = cos (- π) = - 1$ $cospi=cos(-pi)=-1$

Answer 2 · 2015-04-22T19:59:33

A full circle is $2 π$ $2pi$

$π$ $pi$ is half way around the circle counter-clockwise.

$- π$ $-pi$ is half way around the circle clockwise.

$cos (π) = cos (- π)$ $cos(pi) = cos(-pi)$ are the both $cos$ $cos$ values for the same place

They are both equal to $(- 1)$ $(-1)$
because
if viewed as a unit circle centered on the Cartesian origin
the $cos$ $cos$ is the $x$ $x$ value
and, at halfway around the unit circle,
$x = - 1$ $x=-1$

Answer 3 · 2017-03-30T00:36:19

This is what a cosine graph looks like. Since $π$ $pi$ in degrees is 180, it is exactly the same distance from 0 degrees.

Explanation:

graph{cos(x) [-4.76, 4.84, -2.64, 2.16]}

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Why is cos(pi) and cos (-pi) both equal to -1?

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