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While converting angles in trigonometric function I was taught (or I understood) the following

We take coordinate axes and a ray which has it’s pivot at the origin. Like so

enter image description here

So considering an angle, we rotate the ray in the anti clockwise direction. And we see if the particular trigonometric function is positive or negative in the quadrant that we get after rotation. This decides the sign of the trigonometric function that we are converting it into.

My question is, if we don’t know if $x$ is an acute angle or not, we shouldn’t be able to say where the rotating ray will end up in. It could be in the fourth quadrant for an angle of $-30°$, or in the third quadrant for an angle of $-120°$. $\cos$ is positive in 4th quadrant and negative in the third. So it should be $\cos 30°$ in the first case and $-\cos(120°)$ in the second. But still $\cos(-x)=\cos x$ seems to be universal since $\cos$ is an even function.

Why is it so?

Zoltan Zimboras
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Natru
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    Because no matter what, $x$ and $-x$ will have "traveled" the same way but not in the same direction and will be aligned on the $y-$axis, so they have the same cosinus. – Atmos Sep 12 '21 at 10:41
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    Related – Arctic Char Sep 12 '21 at 10:43
  • Make a circle and draw an angle "x" then return to the definition of the function "cos" – Z. Alfata Sep 12 '21 at 10:49
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    Take any angle $\theta$ and its opposite $-\theta$ (which is the same angle but measured clockwise). You'll easily see that they share the same $x$ coordinate. It doesn't matter if the angle is acute or not. – David Sep 12 '21 at 10:51
  • The other comments have pretty much covered it. The words that I would use is that in your diagram, the graph of $\cos(-x)$ is nothing more than a reflection around the $x$-axis of the graph of $\cos(x)$. Therefore, for any given value of $x$, the positions on the real number line (AKA the $x$-axis) of $\cos(x)$ and $\cos(-x)$ must always be identical. – user2661923 Sep 12 '21 at 11:05
  • If you'd like to work with power series, there's a proof for this identity – SPARSE Sep 12 '21 at 11:09
  • If $x$ is in I then $-x$ is in IV and vice versa, if $x$ is in II then $-x$ is in III and vice versa. – kingW3 Sep 12 '21 at 11:14

3 Answers3

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You can also write:

$$\begin{align}\cos (-x)&=\cos (0-x)\\ &=\cos 0\cos x+\sin 0\sin x\\ &=\cos x.\end{align}$$

lone student
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If we take the unit circle definition of $\cos(−x)$,

$$\cos(−x)=\cos(2π−x) $$

Assume that $x ∈[0,π/2]$.

Thus, $2π−x$ lies in the fourth quadrant.

The cosine of an angle in the fourth quadrant is always positive.

Thus, $$\cos(2π−x)=\cos(x)$$

Hence,

$$\cos(−x)=\cos(x)$$

Note:

This derivation exists for only those angles $θ ∈[0,π/2]$.

Sebastiano
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SSS
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If $\DeclareMathOperator{\deg}{°}-180\deg\le x\le 180\deg$, then just by considering the unit circle you can see that $\cos(x)=-\cos(x)$. Since $\cos$ is a periodic function (i.e. it repeats itself every $360\deg)$, it follows that $\cos(x)=-\cos(x)$ for all values of $x$. Try sketching the graph of $\cos$ to see this.

Joe
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