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

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

inverse

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}

y = \frac{2 ^{x} - 1}{2 ^{x} + 1}

f (x) = 34 e^{- 0.00625 x}

f (x) = 0.03 x + 0.63

(4 x - 1)^{2}

g (x) = 10 x - 4

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