Chapter 5.4, Problem 54E

To determine

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

Answer to Problem 54E

The algebraic form of the trigonometric expression $\sin \left(\arctan 2x-\arccos x\right)$ is $\frac{2x^2-\sqrt{1-x^2}}{\sqrt{1+4x^2}}$ .

Explanation of Solution

Given information:

The trigonometric expression $\sin \left(\arctan 2x-\arccos x\right)$ .

Formula used:

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

Calculation:

Consider trigonometric expression $\sin \left(\arctan 2x-\arccos x\right)$ .

The above expression is of the form $\sin \left(u-v\right)$ , where $u=\arctan 2x$ and $v=\arccos x$ .

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

Trigonometric and its inverse trigonometric function cancel each other.

Apply it,

  $\begin{array}{c}\sin \left(\arctan 2x-\arccos x\right)=\sin \left(\arctan 2x\right)cos\left(\arccos x\right)-cos\left(\arctan 2x\right)\sin \left(\arccos x\right)\\ =\sin \left(\arctan 2x\right)\cdot x-cos\left(\arctan 2x\right)\sin \left(\arccos x\right)\end{array}$

For $u=\arctan 2x$ .

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

Construct a right angle triangle with perpendicular p as $2x$ and base b as 1.

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

So, hypotenuse h of triangle is $\sqrt{1+4x^2}$ .

  

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,

  $\begin{array}{c}\sin \left(\arctan 2x-\arccos x\right)=\sin \left(\arcsin \frac{2x}{\sqrt{1+4x^2}}\right)\cdot x-cos\left(\arccos \frac{1}{\sqrt{1+4x^2}}\right)\sin \left(\arcsin \frac{\sqrt{1-x^2}}{1}\right)\\ =\frac{2x^2}{\sqrt{1+4x^2}}-\frac{\sqrt{1-x^2}}{\sqrt{1+4x^2}}\\ =\frac{2x^2-\sqrt{1-x^2}}{\sqrt{1+4x^2}}\end{array}$

Thus, the algebraic form of the trigonometric expression $\sin \left(\arctan 2x-\arccos x\right)$ is $\frac{2x^2-\sqrt{1-x^2}}{\sqrt{1+4x^2}}$ .