In this section of ask-math, we will discuss how to evaluate inverse cosine function.
The inverse cosine function is represented as $f(x)=cos^{-1}x$. Here the -1 is not an exponent, it represents the inverse.The another name for $cos^{-1}x$ is arccosx.
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 cosine inverse function is not one-to-one function so it is restricted sine function.
The another name for inverse sine function is :
$cos^{-1}(x)$ = arccos (x)
To evaluate inverse cosine functions remember that the following statement is equivalent.
$\Theta = cos^{-1}(x) \Leftrightarrow x = cos(\Theta)$
In other words, when we evaluate cosine inverse function we are asking what angle,θ, did we plug into the cosine function (regular, not inverse!) to get x.
f(x)= cos(x)
Domain
$\frac{-\prod }{2}\leq x \leq \frac{\prod}{2}$
Range
$-1\leq y \leq 1$
$f(x)=cos^{-1}x$
$-1\leq x \leq 1$
$\frac{-\prod }{2}\leq y \leq \frac{\prod}{2}$
In cosine function, we know that $cos \Theta =\frac{adjacent side }{hypotenuse}$
The inverse sine function $cos{-1}$ takes the ratio adjacent/hypotenuse and gives the angle θ.
∴ θ = $cos^{-1}(\frac{adjacent side }{hypotenuse})$
For example, $cos 60^{0} =\frac{1}{2}$
∴ θ = $cos^{-1}(\frac{1}{2})$
⇒ θ = $60^{0}$
Evaluate inverse Cosine function
In the unit circle given above, the first coordinate gives you cosine value and second coordinate gives you sine value.
1) Evaluate f(x) = $cos^{-1}(0)$
Solution : f(x) = $cos^{-1}(0)$
y = $cos^{-1}(0)$
⇒ y = $90^{0}$ ( For the degree you can use a unit circle or calculator)
2) Evaluate f(x) = $cos^{-1}(\frac{\sqrt{3}}{2})$
Solution : f(x) = $cos^{-1}(\frac{\sqrt{3}}{2})$
y = $cos^{-1}(\frac{\sqrt{3}}{2})$
⇒ y = $30^{0}$
3) Find y, when cos(y) = 0.1276. Solution : cos(y) = 0.1276.
∴ y = $cos^{-1}(0.1276) $
⇒ y = $82.6690^{0}$
⇒ y = $82.7^{0}$
