Question

if y=tan(x+pi),find x as a function of y.

Answer 1

y=tan(x+pi),
y belongs to real.
tan(x+pi)=tanx.
thus, tanx=y
\[x=\tan^{-1} y\]

Answer 2

you can not actually do inverse on both sides unless you are given the domain of the function

Answer 3

well, usually, if nothing is given we take the standard doman and the range.
and y takes all the values, then there is nothing wrong in taking the inverse, since it is a bijective one in its domain.

Answer 4

how about
tan^-1 (y) = x+pi ?

Answer 5

since, tan^-1 (y) is restricted to open -pi/2 to open pi/2, x= tan^-1(y)-pi, doesn't fit the domain of x. we assume x to be in (-pi/2,pi/2) right?
hence, we need to adjust the equation in order to satisfy the characteristics of the function(especially the inverse one).
it happens only when we neglect the -pi from tan^-1(y). to satisfy the equation.
hence it is only tan^(-1)=x.

Answer 6

how can you assume x to be in (-pi/2 , pi/2) ?

Answer 7

we cannot, but since the asker has not mention anything about x, we can assume it to be within -pi/2 to pi/2. the main aim is to know what a inverse is, and it is done.

Answer 8

i'd disagree here, we can not assume it to be (-pi/2 , pi/2)
why not assume it to be (-3pi/2 , -pi/2) so that now we can directly take inverse

Answer 9

there's no harm in it. you can do it.
but then you are going out of the standard domain. tanx is also bijective in (-3pi/2, -pi/2). you are allowed to do that.
but it would be more preferable if it is done in its standard domain. that's why i assumed x to be that way

Answer 10

hmm

Answer 11

@oldrin.bataku in here!

Answer 12

we need to know restrictions on \(x\) as the inverse is only a function after a specific branch cut

Answer 13

the asker has not mentioned any.
tan(x) is always invertible in any
\[(\frac{ (2n-1) \pi }{ 2 }, \frac{ (2n+1) \pi }{ 2 })\]
where n belongs to an integer.
thus, it's inverse exists in any of that period.