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Chapter 5: Problem 54

Write the trigonometric expression as an algebraic expression. $$\sin (\arctan 2 x-\arccos x)$$

Short Answer

Expert verified
The trigonometric expression \( \sin (\arctan 2 x-\arccos x) \) simplifies to an algebraic expression \( \frac{2x^2-\sqrt{1-x^2}}{\sqrt{1+4x^2}} \)

Step by step solution

01

Understanding Identities

Understand identities for inverse trigonometric functions, particularly for arctan and arccos. The identities we need here are: 1) \( \sin(\arctan x) = \frac{x}{\sqrt{1+x^2}} \), 2) \( \cos(\arccos x) = x \).
02

Sin Difference Identity

Recall the sin difference identity: \( \sin (a - b) = \sin a \cos b - \cos a \sin b \). We'll apply this identity to the given expression \( \sin (\arctan 2x-\arccos x) \). After applying, the expression becomes \( \sin(\arctan 2x)\cos(\arccos x) -\cos(\arctan 2x)\sin(\arccos x) \).
03

Substituting Identities

Now, substitute the identities from Step 1 into the expression. The expression becomes \( \frac{2x}{\sqrt{1+(2x)^2}} * x - \frac{1}{sqrt{1+(2x)^2}}*\sqrt{1-x^2} \).
04

Simplifying the expression

The expression simplifies to \( \frac{2x^2}{\sqrt{1+4x^2}} -\frac{\sqrt{1-x^2}}{\sqrt{1+4x^2}} \). We can rewrite both the terms over the same square root denominator and combine them.
05

Final simplification

After simplifying and combining, we get the final expression as \(F(x) = \frac{2x^2 -\sqrt{1-x^2}}{\sqrt{1+4x^2}}\)

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Most popular questions from this chapter

Let \(x=\pi / 3\) in the identity in Example 8 and define the functions \(f\) and \(g\) as follows. \begin{array}{l}f(h)=\frac{\sin [(\pi / 3)+h]-\sin (\pi / 3)}{h} \\\g(h)=\cos \frac{\pi}{3}\left(\frac{\sin h}{h}\right)-\sin \frac{\pi}{3}\left(\frac{1-\cos h}{h}\right)\end{array} (a) What are the domains of the functions \(f\) and \(g ?\) (b) Use a graphing utility to complete the table.$$\begin{array}{|l|l|l|l|l|l|l|}\hline h & 0.5 & 0.2 & 0.1 & 0.05 & 0.02 & 0.01 \\\\\hline f(h) & & & & & & \\\\\hline g(h) & & & & & & \\\\\hline\end{array}$$. (c) Use the graphing utility to graph the functions \(f\) and \(g\). (d) Use the table and the graphs to make a conjecture about the values of the functions \(f\) and \(g\) as \(h \rightarrow 0^{+}\).

Determine whether the statement is true or false. Justify your answer. Because the sine function is an odd function, for a negative number \(u, \sin 2 u=-2 \sin u \cos u\).

Find the exact value of the expression. $$\sin \frac{\pi}{12} \cos \frac{\pi}{4}+\cos \frac{\pi}{12} \sin \frac{\pi}{4}$$

Use the product-to-sum formulas to rewrite the product as a sum or difference. $$\sin 5 \theta \sin 3 \theta$$

Verify the identity. $$a \sin B \theta+b \cos B \theta=\sqrt{a^{2}+b^{2}} \sin (B \theta+C)\( where \)C=\arctan (b / a)\( and \)a>0$$
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