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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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    cos(x) is even function – 2017-01-29
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    If your $f(x)=\cos(x-90)$, then $f(-x)$ would be $\cos(-x-90)$, not $\cos(-(x-90))$. I think this is the source of confusion. – 2017-01-29
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    It's true that $y = f(-x)$ is a reflection of $y = f(x)$ across the $y$-axis. But it's a bit more subtle here: $y = f(-(x - 90))$ is a reflection of $y = f(x - 90)$ across the line $x = [\text{where ever } x - 90 = 0]$. Cosine has a lot of symmetry... – 2017-01-29

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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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    I like this answer very much! Thank you! Wait, so in the last line, you are saying that cos(x-90) is not the same as cos(-x-90)? – 2017-01-29
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    @BobSmith I'm glad you found it helpful ^^ – 2017-01-29
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    I updated the comment! Apologies. – 2017-01-29
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    @BobSmith I updated the answer! – 2017-01-29
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    OH I GOT IT! Using the even function, if my x is x-90, then cos(x-90) = cos(-x+90) don't worry now it makes full sense! :) – 2017-01-29
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    Don't worry now your post becomes sense! Thanks so so much! Gladly appreciate it! Man I love StackExchange. – 2017-01-29
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    @BobSmith I know how you feel :D MathsSE is lovely. – 2017-01-29
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The reason is the cosine function is even: $\cos(-\alpha)=\cos\alpha$.