Evaluate Inverse Sine function

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}

4) Evaluate : f(x) = $sin^{-1}(sin(\frac{\prod }{4}))$

Solution : f(x) = $sin^{-1}(sin(\frac{\prod }{4}))$

y = $sin^{-1}(sin(\frac{\prod }{4}))$

The value of $sin(\frac{\prod }{4}) = (\frac{\sqrt{2}}{2})$

∴ y = $sin^{-1}(\frac{\sqrt{2}}{2})$

∴ y = $45^{0}$

5) Evaluate f(x) = $sin^{-1}(1)$

Solution : f(x) = $sin^{-1}(1)$

y = $sin^{-1}(1)$

⇒ y = $90^{0}$


11th grade math

precalculus

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