Chapter 5.4, Problem 56E

To determine

To find: The algebraic form of the given trigonometric expression.

Answer to Problem 56E

The algebraic form of the trigonometric expression $\cos \left(\arccos x-\arctan x\right)$ is $\frac{x\left(1-\sqrt{1-x^2}\right)}{\sqrt{1+x^2}}$ .

Explanation of Solution

Given information:

The trigonometric expression $\cos \left(\arccos x-\arctan x\right)$ .

Formula used:

Difference rule for cosine function is $\cos \left(x-y\right)=\cos x\cos y+\sin x\sin y$ .

Calculation:

Consider trigonometric expression $\cos \left(\arccos x-\arctan x\right)$ .

The above expression is of the form $\cos \left(v-u\right)$ , where $u=\arctan x$ and $v=\arccos x$ .

Recall difference rule for cosine function is $\cos \left(x-y\right)=\cos x\cos y+\sin x\sin y$ .

Trigonometric and its inverse trigonometric function cancel each other.

Apply it,

  $\begin{array}{c}\cos \left(\arccos x-\arctan x\right)=cos\left(\arccos x\right)\cos \left(\arctan x\right)-\sin \left(\arccos x\right)\sin \left(\arctan x\right)\\ =x\cdot \cos \left(\arctan x\right)-\sin \left(\arccos x\right)\sin \left(\arctan x\right)\end{array}$

For $u=\arctan x$ .

Recall that $\tan x=\frac{\text{perpendicular}}{\text{base}}$

Construct a right angle triangle with perpendicular p as $x$ and base b as 1.

Since, $h^2=p^2+b^2$ , where h is hypotenuse of triangle.

So, hypotenuse h of triangle is $\sqrt{1+x^2}$ .

  

For $v=\arccos x$ .

Recall that $\cos x=\frac{\text{base}}{\text{hypotenuse}}$

Construct a right angle triangle with base b as x and hypotenuse h as 1.

Since, $h^2=p^2+b^2$ , where p is perpendicular of triangle.

So, perpendicular p of triangle is $\sqrt{1-x^2}$ .

  

The expression is reduced to,

  $\begin{array}{c}\cos \left(\arccos x-\arctan x\right)=x\cdot \cos \left(\arccos \frac{1}{\sqrt{1+x^2}}\right)-\sin \left(\arcsin \frac{\sqrt{1-x^2}}{1}\right)\sin \left(\arcsin \frac{x}{\sqrt{1+x^2}}\right)\\ =\frac{x}{\sqrt{1+x^2}}-\sqrt{1-x^2}\frac{x}{\sqrt{1+x^2}}\\ =\frac{x-x\sqrt{1-x^2}}{\sqrt{1+x^2}}\\ =\frac{x\left(1-\sqrt{1-x^2}\right)}{\sqrt{1+x^2}}\end{array}$

Thus, the algebraic form of the trigonometric expression $\cos \left(\arccos x-\arctan x\right)$ is $\frac{x\left(1-\sqrt{1-x^2}\right)}{\sqrt{1+x^2}}$ .