Inverse trig functions?

Question

If I have f(x) = sin1/2x + a
How do I work out it's inverse function?

Answer 1

f(x) = sin½x + a
So x = sin(½f^(-1)( x)) + a
x - a = sin(½f^(-1)( x))
so ½f^(-1)( x) = arcsin(x - a)
So f^(-1)( x) = 2 arcsin(x - a)
Domain: a -1 ≤ x ≤ a + 1
Range: - π ≤ f^(-1)( x) ≤ π

Answer 2

if f(x)=y=sin1/2x + a

then simply swap x and y and solve for y.Thus

x=sin(1/2y) + a

Solving for y=0.5/sin^-1(x-a) or you may use arcsin for sin^-1

Note I have taken your sin 1/2x to be equivalent to sin(0.5/x). If I have this wron use the above technique with the correct function and you will gwt the correct answer.

Answer 3

f(x)=sin1/2x+a
Let y=f(x) and rearrange to make x the subject
y=sin1/2x+a
sin1/2x=y-a
1/2x=sin^-1(y-a)
x=2sin^-1(y-a)
Hence, f(x)^-1=2sin^-1(x-a)

Hope that makes sense.
I find it quite difficult to use the keyboard rather than write it on a piece of paper!

Answer 4

y = sin1/2x + a
y - a = sin1/2x
1/2x = sin-1(y-a)
x = 2sin-1(y-a)

f-1(x) = 2sin-1(x-a)

Answer 5

Not to be redundant, I'll introduce a check on the answer.
If f^(-1)(x)=2sin^(-1)(x-a) then we must be able to show
that this leads to f(f^(-1)(x))=x.

f(2sin^(-1)(x-a))=

sin((1/2)2sin^(-1)(x-a))+a=

sin(sin^(-1)(x-a))+a =

x-a+a=x OK

Answer 6

let y=f(x)

sin(1/2x)=y-a

1/2x =arcsin(y-a)
x =2*arcsin(y-a)
>>>>x =2*arcsin(sin(1/2x))

therefore,

f^(-1)(x) =2*arcsin(sin(1/2x))

i hope that this helps