What is the inverse of f(x)=1+sin^2(x-pi/2) ?

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

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inverse of f(x)=1+sin^2(x-pi/2)

inverse

f (x) = 1 + sin^{2} (x - \frac{π}{2})

arcsin (x - 1) + \frac{π}{2}, arcsin (- x - 1) + \frac{π}{2}

Solution steps

Solve for

sin (x - \frac{π}{2}), y = 1 + sin^{2} (x - \frac{π}{2})

sin (x - \frac{π}{2}) = y - 1, sin (x - \frac{π}{2}) = - y - 1

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

x - \frac{π}{2} = arcsin (y - 1), x - \frac{π}{2} = arcsin (- y - 1)

Solve for

x, x - \frac{π}{2} = arcsin (y - 1)

Solve for

x, x - \frac{π}{2} = arcsin (- y - 1)

arcsin (y - 1) + \frac{π}{2}, arcsin (- y - 1) + \frac{π}{2}

Substitute

y = x

arcsin (x - 1) + \frac{π}{2}, arcsin (- x - 1) + \frac{π}{2}

f (x) = e^{x^{2} - 1}

f (x) = e^{x^{2} - 9}

f (x) = lo g_{3} (x + 1) - 2

f (x) = 4 y = 5 (x + 2)

f (x) = \frac{1}{3} (x - 4)

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