Chapter 5.4, Problem 52E

To determine

To write: The given trigonometric expression as an algebraic expression.

Answer to Problem 52E

The given trigonometric expression as an algebraic expression as follows:

  $\text{2}x\sqrt{1-x^2}$

Explanation of Solution

Given information:

The following expression:

  $\cos (\arccos x-\arcsin x)$

Formula used:

The following trigonometric relations and identities:

  $\begin{array}{l}\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\\ \sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\\ {\text{Hypotenuse}}^2={\text{Adjacent}}^{\text{2}}{\text{+opposite}}^{\text{2}}\end{array}$

The following difference formula:

  $\cos (u-v)=\cos u\cos v+\sin u\sin v$

Calculation:

This expression fits the formula:

  $\cos (u-v)$

Angles $u=\arccos x$ and $v=\arcsin x$ are shown in the following figures respectively.

  $\begin{array}{l}\text{For figure 1,}\\ u=\arccos x\\ x=\cos u\text{ [Inverse function]}\\ \cos u=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{1}\text{ [Pythagoras relation]}\\ \text{Opposite = }\sqrt{1-x^2}\text{ [Pythagoras identity]}\\ \\ \text{For figure 2,}\\ v=\arcsin x\\ x=\sin v\text{ [Inverse function]}\\ \sin v=\frac{\text{Opposite}}{\text{hypotenuse}}=\frac{x}{1}\text{ [Pythagoras relation]}\\ \text{Adjacent=}\sqrt{1-x^2}\text{ [Pythagoras identity]}\\ \end{array}$

Therefore, the given equation can be written as:

  $\begin{array}{l}\cos (u-v)=\cos (\arccos x)\cos (\arcsin x)+\sin (\arccos x)\sin (\arcsin x)\text{ [Using difference formula]}\\ \text{ =}x\cdot \sqrt{1-x^2}+\sqrt{1-x^2}\cdot x\text{ [Derived from the repective diagrams]}\\ \text{ =2}x\sqrt{1-x^2}\text{ [Simplify]}\end{array}$