Is it possible to find value of $~\cos 180^{\circ}~$ and $~\cos 360^{\circ}~$ with the reference angle method?

Question

I'm confused as to which quadrant $~+180^{\circ}~$ and $~+360^{\circ}~$ lie in.

If I'm able to determine this and their reference angles then I will be able to assign a positive or negative sign to the values of $~\cos 180^{\circ}~$ and $~\cos 360^{\circ}~$ .

Or is there another approach to this?

Answer 1

I would make an image of the function $$y=\cos(x)$$ with the period $2\pi$ then we get $$\cos(180^{\circ})=-1$$ and $$\cos(360^{\circ})=1$$

Answer 2

$~+180^{\circ}~$ lies in the negative $~X~$ axis and $~+ 360^{\circ}~$ lies on the positive $~X~$ axis.

See the picture

In this connection I would like to mention that we take angle $~\theta~$ in anti-clock wise direction.

So we can write

$1.\quad~\theta=+0^{\circ}~$ along positive $~X~$ axis (i.e., along $~OX~$)

$2.\quad~\theta=+90^{\circ}~$ along positive $~Y~$ axis (i.e., along $~OY~$)

$3.\quad~\theta=+180^{\circ}~$ along negative $~X~$ axis (i.e., along $~-OX~$)

$4.\quad~\theta=+270^{\circ}~$ along negative $~Y~$ axis (i.e., along $~-OY~$)

$5.\quad~\theta=+360^{\circ}~$ along positive $~X~$ axis (i.e., along $~OX~$)

and so on.

Similarly,

$1.\quad~\theta=-0^{\circ}~$ along positive $~X~$ axis (i.e., along $~OX~$)

$2.\quad~\theta=-90^{\circ}~$ along negative $~Y~$ axis (i.e., along $~-OY~$)

$3.\quad~\theta=-180^{\circ}~$ along negative $~X~$ axis (i.e., along $~-OX~$)

$4.\quad~\theta=-270^{\circ}~$ along positive $~Y~$ axis (i.e., along $~OY~$)

$5.\quad~\theta=-360^{\circ}~$ along positive $~X~$ axis (i.e., along $~OX~$)

and so on.

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From the picture you can conclude that $~\cos 180^{\circ}~=~-1~$ and $~\cos 360^{\circ}~=~1~$

Answer 3

Why not use the unit circle approach?

Since $180^{\circ}$ terminates in standard position at the point $(-1,0)$, $~\cos 180^{\circ}~ = -1$. Similarly, $~\cos 360^{\circ}~ = 1$ as $360^{\circ}$ terminates at $(1,0)$ in standard position.

Answer 4

For $180^{\circ}$ as well as $360^{\circ}$ in order to evaluate $ \cos \theta $ consider absolute value ratio of x-axis projection and unit radius vector $1$ when it rotates..without regard to sign at first.

Next,

there is no ambiguity in settling the sign of either case. Because

for $180^{\circ}$ cosine is negative in neighboring second and third quadrants anyhow.

And,

for $360^{\circ}$ cosine is positive in neighboring first and fourth quadrants anyhow.

So $ \cos 180^{\circ} =-1 $ and $ \cos 360^{\circ} =+1. $