What is the inverse of f(x)=sin(sqrt(16-x^2)) ?

The inverse of f(x)=sin(sqrt(16-x^2)) is sqrt(-arcsin^2(x)+16),-sqrt(-arcsin^2(x)+16)

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inverse of f(x)=sin(sqrt(16-x^2))

inverse

f (x) = sin (16 - x^{2})

- arcsin^{2} (x) + 16, - - arcsin^{2} (x) + 16

Solution steps

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

16 - x^{2} = arcsin (y)

Solve for

x, 16 - x^{2} = arcsin (y)

- arcsin^{2} (y) + 16, - - arcsin^{2} (y) + 16

Substitute

y = x

- arcsin^{2} (x) + 16, - - arcsin^{2} (x) + 16

sin (\frac{2}{6})

f (x) = x^{2} - 4 x - 16

lo g_{6} (4 x + 4)

f (x) = \frac{- 4 x + 1}{9 + 5 x}

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

What is the inverse of f(x)=sin(sqrt(16-x^2)) ?
The inverse of f(x)=sin(sqrt(16-x^2)) is sqrt(-arcsin^2(x)+16),-sqrt(-arcsin^2(x)+16)