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I understand that:

when x = 3pi/2, arcsin(sin x) = -1/2 * pi

When x = pi/2, arcsin (sin x) = 1/2 * pi

When x = 0,pi or 2Pi, arcsin (sin x) = 0

I'm still really confused by why things are straight lines. Literally no other trigonometric function has been a straight line it's all been curves but boom for some reason they decide to switch it up now?! Help please
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I can't tell if you're trolling, but sin and arcsin are inverse functions, at least for a certain domain. Therefore `sin(arcsin(x)) = arcsin(sin(x)) = x`
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arcsin(cos(x)) is basically the same, just moved a little bit. The point of arcsin is to reverse the application of sin (at least near 0) so arcsin(sin(x)) is equal to x for [-π/2,π/2] by definition, the rest happenes by the symmetry and period of sin.
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Arcsine, when used as a function, can only realistically give one possible output despite the fact that there are infinitely many possible arcsines for any given value.

If you want to see all possible outputs you'd be better off plotting implicitly with something like sin(y) = sin(x) rather than y = arcsin(x). Note that these are equivalent.

When you do this, you'll see what looks like a chain-link fence(!) because it's (infinitely) many "single value arcsines" stacked one above the other.

As a comparison, you could ask why square root is only half a parabola (oriented to the right) when x^2 is a full parabola (oriented upwards). This is because square root, when treated as a function, can only give back one of the possible values.

Switching to implicit y^2 = x from y = √x reveals that full parabola.

In both cases, we choose the version of the inverse function that is most useful.

Arccosine is an interesting case, on the other hand, because there are two "single value" versions in common use in calculators, calculating software and textbooks.

One version is discontinuous at 0 to stay in the range -pi/2 ≤ y ≤ pi/2 and the other is continuous because it uses the range 0 ≤ y ≤ pi instead.
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A function is a **mapping**, you plug in something and (usually) get something else in return

       sin : (-∞, +∞) → [-1, +1]

       sin(1) = 0.84
       sin(19.85) = 0.84

    arcsin : [-1, +1] → [-π/2, +π/2]

    arcsin(0.84) = 1

If `x∊[-π/2, +π/2]`, then you can apply the two functions consecutively and return to what you started with

               sin: [-π/2,+π/2]→[-1,+1]
            arcsin:              [-1,+1]→[-π/2,+π/2]
    (arcsin o sin): [-π/2,+π/2]     →    [-π/2,+π/2]

    arcsin(sin(1)) = arcsin(0.84) = 1

            arcsin: [-1,+1]→[-π/2,+π/2]
               sin:          [-π/2,+π/2]→[-1,+1]
    (sin o arcsin): [-1,+1]       →      [-1,+1]

    sin(arcsin(0.84)) = sin(1) = 0.84

However, if x does not lie in the range of arcsin, then you won't get the original value

               sin: [11π/2,13π/2]→[-1,+1]
            arcsin:                [-1,+1]→[-π/2,+π/2]
    (arcsin o sin): [11π/2,13π/2]    →     [-π/2,+π/2]

    arcsin(sin(19.85)) = arcsin(0.84) = 1
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