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 :
To evaluate inverse sine functions remember that the following statement is equivalent.
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 = $90^{0}$
precalculus
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