sin^-1(sin(200))

[vital dragon]

what exactly ?

[low roost]

.solved

use that sin¯¹(sin(x)) = x only if -π/2 ≤ x ≤ π/2
so sin¯¹(sin(1)) = 1 but sin¯¹(sin(2)) isn't 2

however you know that you can subtract multiples of 2π and still get the same sin
or that sin(200) = sin(200 - 2π) = sin(200 - 4π) = etc.

so subtract enough 2πs to get 200 to be inside [-π/2, π/2]
if you need to subtract n many 2πs,
200 - 2πn ≤ π/2
200 - π/2 ≤ 2πn
(200 - π/2) / 2π ≤ n
31.6 ≤ n
youre not subtracting a fraction of a 2π here, so that means at least 32 many 2πs
so sin(200) = sin(200 - 32 * 2π) = sin(-1.0619298297...)

therefore,
sin^-1(sin(200)) = sin^-1(sin(200 - 64π)) = 200 - 64π