In this section of ask-math, we will discuss how to evaluate inverse sine function.
The inverse sine function is represented as $f(x)=sin^{-1}x$. Here the -1 is not an exponent, it represents the inverse.The another name for $sin^{-1}x$ is arcsinx.
ask-math has already explains you about the sine graph. The domain for sine is $\frac{-\prod }{2}\leq x \leq \frac{\prod}{2}$. The sine inverse function is not one-to-one function so it is restricted sine function.
The another name for inverse sine function is :
$sin^{-1}(x)$ = arcsin (x)
To evaluate inverse sine functions remember that the following statement is equivalent.
$\Theta = sin^{-1}(x) \Leftrightarrow x = sin(\Theta)$
In other words, when we evaluate sine inverse function we are asking what angle,θ, did we plug into the sine function (regular, not inverse!) to get x.
f(x)= sin(x)
Domain
$\frac{-\prod }{2}\leq x \leq \frac{\prod}{2}$
Range
$-1\leq y \leq 1$
$f(x)=sin^{-1}x$
$-1\leq x \leq 1$
$\frac{-\prod }{2}\leq y \leq \frac{\prod}{2}$
In sine function, we know that $sin \Theta =\frac{opposite side }{hypotenuse}$
The inverse sine function $sin{-1}$ takes the ratio opposite/hypotenuse and gives the angle θ.
∴ θ = $sin^{-1}(\frac{opposite side }{hypotenuse})$
For example, $sin 30^{0} =\frac{1}{2}$
∴ θ = $sin^{-1}(\frac{1}{2})$
⇒ θ = $30^{0}$
Evaluate inverse sine function
1) Evaluate f(x) = $sin^{-1}(\frac{\sqrt{2}}{2})$
Solution : f(x) = $sin^{-1}(\frac{\sqrt{2}}{2})$
y = $sin^{-1}(\frac{\sqrt{2}}{2})$
⇒ y = $45^{0}$ ( For the degree you can use a unit circle or calculator)
2) Evaluate f(x) = $sin^{-1}(\frac{\sqrt{3}}{2})$
Solution : f(x) = $sin^{-1}(\frac{\sqrt{3}}{2})$
y = $sin^{-1}(\frac{\sqrt{3}}{2})$
⇒ y = $60^{0}$
3) Find y, when sin(y) = 0.2384. Solution : sin(y) = 0.2384.
∴ y = $sin^{-1}(0.2384) $
⇒ y = $13.798^{0}
⇒ y = $13.8^{0}
