By definition, `arcsin x = y` if `sin y = x` . Replacing x by sin y, yields:
`arcsin x = arcsin (sin y) = y`
Hence, replacing y by `3pi` , yields:
`arcsin(sin 3pi) = 3pi`
You need to evaluate `arcsin(sin 3pi)` , hence, you need to verify if sin `3pi ` falls in interval [-1,1].
Hence, first you need to evaluate sin 3pi, such that:
`sin 3pi = sin(2pi+pi) = sin 2pi*cos pi + sin pi*cos 2pi = 0*(-1) + 0*1 = 0` Notice that `0 in [-1,1].`
Replacing `sin 3pi` by 0, yields:
`arcsin(sin 3pi) = arcsin 0 = 0, pi, 2pi, 3pi,...,n*pi,..`
Hence, `arcsin 0 = k*pi,` where `k in Z` , so `arcsin (sin 3pi) = 3pi` .
Use the identity
General solutions for sin(x)=0 are,
solutions for the range -`pi/2` `<=x<=pi/2` ` `
Be aware that `3pi` is not in the range of the arcsine function.
The range of the arcsine function is `-pi/2<=y<=pi/2.`
By the properties of inverse functions, a function and its inverse function will 'cancel' each other out and result in simply the input of the function. In this problem, we know that `arcsin` and `sin` are inverse functions. Therefore:``
`arcsin(sin(3pi)) = 3pi`
However, if the value is computed in a calculator as written, the result will be zero:
`sin(3pi) = 0 -> arcsin(sin(3pi)) = arcsin(0) = 0`
This is a unique case because of the domain of arcsin(x).
First, arcsine the inverse sign function. So `arcsin(sin(3pi))` is the same question as `sin^(-1)(sin(3pi))`
On one hand, functions and inverse functions undo each other. So it seems that `sin^(-1)(sin(3pi))` is `3pi` . However, this answer fails to take into account the order in which functions are evaluated i.e. inside to out.
First, evaluate `sin(3pi)` to 0 .
Second, evaluate `sin^(-1)(0)` . This is asking the question, what angle (in radians) when inserted into the sine function will give a result of 0. Since, `sin(0)=0` then `sin^(-1)(0) = 0` .
Putting all of this together, `sin^(-1)(sin(3pi))=0`
sin and arcsin are inverse functions. Whenever you apply a function to it's inverse they cancel out. So in this case, arcsin(sin(3`pi`)) is equal to 3`pi.`