What is the inverse of f(x)=sin(x)+1,0<x<= pi/2 ?

The inverse of f(x)=sin(x)+1,0<x<= pi/2 is arcsin(x-1)

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inverse of f(x)=sin(x)+1,0<x<= pi/2

inverse

f (x) = sin (x) + 1, 0 < x \leq \frac{π}{2}

arcsin (x - 1)

Solution steps

Solve for

sin (x), y = sin (x) + 1

sin (x) = y - 1

sin (a) = b \Rightarrow a = arcsin (b)

arcsin (y - 1)

Substitute

y = x

arcsin (x - 1)

e^{e^{x}} - 2

y = \frac{1}{x - 10}

y = \frac{9 ( 4 x - 9 )}{4} + 17

y = - 8 - 5 x

f (x) = \frac{( 2 x - 4 )}{3}

What is the inverse of f(x)=sin(x)+1,0<x<= pi/2 ?
The inverse of f(x)=sin(x)+1,0<x<= pi/2 is arcsin(x-1)