$\cos(x-90)$ and $\cos(-(x-90))$ how can they be the same?

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I plotted two lines in Desmos Calculator.

$\cos(x-90)$ which looks exactly like a $\sin x$ graph.

$\cos (-(x-90))$ which looks exactly the same.

However, I thought that $f(-x)$ reflects the entire line in the $y$-axis.

So why do the two lines above look the same, based on graph transformations?

I might be stupid, but just a question.

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2 Answers

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Don't make confusion with the parity of a function. The parity transformation only acts on the variable.

$$F(x) ~~~ \text{becomes} ~~~ F(-x)$$

And it is said to be even when $$F(-x) = F(x)$$

But again: the parity does act only on the variable.

For example: take $F(x) = 3x^2 - 4$

Then

$$ F(-x) = 3(-x)^2 - 4 = 3x^2 - 4 $$

The function is then identical to the original one hence it's even .

Notice that we do not change the sign of the constant $4$.

Cosine is an even function :

$$\cos(x) = \cos(-x)$$

But in your case the parity does not attack $90$. Under the parity your function becomes:

$$\boxed{\cos(x - 90) \to \cos(-x - 90)}$$

And they are identical.

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The reason is the cosine function is even: $\cos(-\alpha)=\cos\alpha$.

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