What is the inverse of sin^2(2α) ?

The inverse of sin^2(2α) is (arcsin(sqrt(α)))/2 ,(arcsin(-sqrt(α)))/2

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inverse of sin^2(2α)

inverse

sin^{2} (2 α)

\frac{arcsin ( α )}{2}, \frac{arcsin ( - α )}{2}

Solution steps

Solve for

sin (2 α), y = sin^{2} (2 α)

sin (2 α) = y, sin (2 α) = - y

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

2 α = arcsin (y), 2 α = arcsin (- y)

Solve for

α, 2 α = arcsin (y)

Solve for

α, 2 α = arcsin (- y)

\frac{arcsin ( y )}{2}, \frac{arcsin ( - y )}{2}

Substitute

y = α

\frac{arcsin ( α )}{2}, \frac{arcsin ( - α )}{2}

f (x) = \frac{( e ^{x} - 8 )}{2 - e ^{x}}

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

\frac{4 x - 3}{2 - x}

f (x) = 3 x + 4 y = 7

f (x) = y - 2

What is the inverse of sin^2(2α) ?
The inverse of sin^2(2α) is (arcsin(sqrt(α)))/2 ,(arcsin(-sqrt(α)))/2