Chapter 7, Problem 10CT

To determine

The solution of ${\cot }^{-1}5$ using the calculator. Express the answer in radian rounded to two decimal places.

Answer to Problem 10CT

Solution:

The solution of ${\cot }^{-1}5$ using the calculator is $\text{0}\text{.20}$.

Explanation of Solution

Given information:

The expression, ${\cot }^{-1}5$

Explanation:

Consider the expression ${\cot }^{-1}5$.

Let $\theta ={\cot }^{-1}5$

$\Rightarrow \cot \theta =5$

Then $\theta$ is the angle $0<\theta <\pi$ such that $\cot \theta =5$.

Since $\cot \theta >0$, it follows that $0<\theta \le \frac{\pi }{2}$.

So $\theta$ lies in quadrant $\text{I}$.

Now find $\cos \theta$ since $y={\cos }^{-1}x$ has the same range as $y={\cot }^{-1}x$, except where undefined.

Since $\cot \theta =\frac{5}{1}=\frac{x}{y}$, let $x=5$ and $y=1$.

point $P=\left(x,y\right)=\left(5,1\right)$ is on a circle of radius $r$ $x^2+y^2=r^2$.

$\Rightarrow 5^2+1^2=r^2$

$\Rightarrow 25+1=r^2$

$\Rightarrow r^2=26$

$\Rightarrow r=\sqrt{26},r>0$

Use the figure to find $\cos \theta$.

Since $x=5$, $y=1$, $r=\sqrt{26}$

$\Rightarrow \cos \theta =\frac{x}{r}=\frac{5}{\sqrt{26}}$, $0<\theta \le \frac{\pi }{2}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{5}{\sqrt{26}}\right)$,

$\Rightarrow {\cot }^{-1}5=\theta ={\cos }^{-1}\left(\frac{5}{\sqrt{26}}\right)$

By using the calculator,

$\Rightarrow {\cos }^{-1}\left(\frac{5}{\sqrt{26}}\right)=0.1973955598$

Rounding to two decimal places,

$\Rightarrow {\cot }^{-1}5=\text{0}\text{.20}$

Therefore, the solution of ${\cot }^{-1}5$ using calculator is $\text{0}\text{.20}$ radian.