What is the f(θ)=arcsin((sqrt(2))/2),-pi/2 <θ< pi/2 ?

The f(θ)=arcsin((sqrt(2))/2),-pi/2 <θ< pi/2 is

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f(θ)=arcsin((sqrt(2))/2),-pi/2 <θ< pi/2

f (θ) = arcsin (\frac{2}{2}), - \frac{π}{2} < θ < \frac{π}{2}

Y Intercepts : (0, \frac{π}{4})

Slope : m = 0

Inverse : None

Domain : - \frac{π}{2} < θ < \frac{π}{2}

Range : f (θ) = arcsin (\frac{2}{2})

Interval Notation

Domain : (- \frac{π}{2}, \frac{π}{2})

Decimal

Range : f (θ) = 0.78539 \dots

Solution steps

Axis interception points of

arcsin (\frac{2}{2}), - \frac{π}{2} < θ < \frac{π}{2} :

Y Intercepts

: (0, \frac{π}{4})

Slope of

arcsin (\frac{2}{2}) : m = 0

Inverse of

arcsin (\frac{2}{2}), - \frac{π}{2} < θ < \frac{π}{2} :

None

Domain of

arcsin (\frac{2}{2}), - \frac{π}{2} < θ < \frac{π}{2} : - \frac{π}{2} < θ < \frac{π}{2}

Range of

arcsin (\frac{2}{2}), - \frac{π}{2} < θ < \frac{π}{2} : f (θ) = arcsin (\frac{2}{2})

f (x) = ∣ 4 x + 2 ∣

f (x) = \frac{( x + 2 )}{x ^{2}}

h (x) = - lo g_{10} (3 x - 2) + 4

f (x) = 3 x + 1 - 2

f (x) = \frac{4}{2 + 8 e ^{- 2 x}}

What is the f(θ)=arcsin((sqrt(2))/2),-pi/2 <θ< pi/2 ?
The f(θ)=arcsin((sqrt(2))/2),-pi/2 <θ< pi/2 is