Find the Inverse of a Function - P...
Find the Inverse of a Function - Part I

# Evaluate Inverse cos Function

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 ${-\infty}< x < {\infty}$ Range $-1\leq y \leq 1$ $f(x)=cos^{-1}x$ $-1\leq x \leq 1$ $0 \leq y \leq {\prod}$

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 = $82.7^{0}$

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

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

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

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

∴ y = $cos^{-1}(\frac{\sqrt{2}}{2})$

∴ y = $45^{0}$

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

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

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

⇒ y = $180^{0}$

precalculus

Home

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.