How do you evaluate  arc cos (1)?

Sep 3, 2016

arcos(1) = 0

Explanation:

$\textcolor{b l u e}{\text{Important introduction}}$

If you take the cosine of an angle you have a numeric value. This number represents the ratio of length of particular sides of a right triangle.

When you evaluate "arccos" you are converting that ratio back to the original angle concerned. By the way, another way of writing "arcos" is ${\cos}^{- 1}$. So arcos(1) could also be written as ${\cos}^{- 1} \left(1\right)$

cosine is the ratio $\left(\text{adjascent")/("hypotenuse}\right)$ where the length of the hypotenuse has been 'converted' to 1

Suppose you had $\left(\text{adjascent")/("hypotenuse}\right) \to \frac{3}{5}$

This is then 'converted' so that $\frac{3}{5} \equiv \frac{3 \div 5}{5 \div 5} = \frac{0.6}{1}$

However, people do not write the 1 so instead of $\frac{0.6}{1}$ we have $0.6$. So the cosine of the angle would be $0.6$

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$\textcolor{b l u e}{\text{Answering the question}}$

$\textcolor{b r o w n}{\text{Consider Figure 1}}$

$\cos \left(\theta\right) = \left(\text{adjascent")/("hypotenuse}\right) = \frac{A C}{A B} < 1$

For this right triangle to exist the length of AB must be more than the length AC.

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$\textcolor{b r o w n}{\text{Consider Figure 2}}$

By writing arcos(1) in the question they are declaring that the length of DE is the same as the length of DF. For a right triangle this can only happen if $\theta$ is 0.

So arcos(1) = 0

Sep 4, 2016

$\arccos \left(1\right) = 0$ (in radians and in degrees - the same answer)

Explanation:

The best way to define trigonometric functions and their inverse is by using the unit circle. Using this approach, we can define not only angles of right triangle, that are only measured in the interval $\left(0 , {90}^{o}\right)$, by any angles, including negative.
For detail description of this we can refer to UNIZOR.COM Trigonometry chapter of the course.

In short, the definitions are as follows.

Imagine a circle of a radius $1$ with a center at the origin of Cartesian coordinates on the XY-plane.
An angle is formed by rotating a radius with an endpoint at $\left(1 , 0\right)$ (the one directed along the X-axis towards its positive end) counterclockwise. Thus, when the radius is directed vertically up (along the Y-axis towars its positive end), it's an angle of ${90}^{0}$ (or $\frac{\pi}{2}$ in radians).
When the radius is directed horizontally to the left towards negative X-coordinates, it's an angle of ${180}^{o}$ ($\pi$ in radians) etc.

For angles measured in negative numbers the rotation of the radius is clockwise.

By definition, for any angle $\phi$, defined by this rotation of a radius,
$\sin \phi$ is the Y-coordinate of the endpoint of the radius that forms this angle
$\cos \phi$ is the X-coordinate
$\tan \phi$ is $\sin \frac{\phi}{\cos} \phi$
$\cot \phi$ is $\cos \frac{\phi}{\sin} \phi$
$\sec \phi$ is $\frac{1}{\cos} \phi$
$\csc \phi$ is $\frac{1}{\sin} \phi$.

Inverse trigonometric function $\arccos$ is define on interval $\left[- 1 , 1\right]$, ranging from $0$ radians ($= {0}^{o}$) to $\pi$ radians ($= {180}^{o}$), which corresponds to monotonic behavior of $\cos$ on the interval $\left[0 , \pi\right]$, taking values from $1$ to $- 1$.

Since $\arccos \left(1\right)$ is an angle $\phi$ such that $\cos \left(\phi\right) = 1$, this angle corresponds to a position of a radius of a unit circle with X-coordinate $1$ (recall, $\cos$ is an X-coordinate). On a unit circle it is the point $\left(1 , 0\right)$ - the original position of a radius, from which we measure the angle of rotation. This original position measures the angle of ${0}^{o} = 0 r a d$,