/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 134 Write the trigonometric expressi... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 134

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

Short Answer

Expert verified
The algebraic expression of \(\sin (2 \arccos x)\) is \(2x\sqrt{1 - x^2}\).

Step by step solution

01

Identify the formula

The first step it's to recognize the relevant formula, in this case, we have a double angle formula for sine, which states that \(\sin 2胃 = 2\sin胃 \cos胃\). Since we have \(\sin(2胃)\) where \(胃 = \arccos x\), we substitute into the formula.
02

Substitution

Substitute \(胃\) by \(\arccos x\) in the formula \(\sin 2胃 = 2\sin胃 \cos 胃\). Thus, it becomes \(\sin(2 \arccos x) = 2 \sin (\arccos x) \cos (\arccos x)\).
03

Compute \(\sin(\arccos x)\) and \(\cos(\arccos x)\)

We know from Pythagoras' theorem that \(\sin^2(胃) + \cos^2(胃) = 1\). So, we can express \(\sin(\arccos x)\) as \(\sqrt{1 - \cos^2(\arccos x)}\), which simplifies to \(\sqrt{1 - x^2}\). From the definition of arccos, we know that the \(\cos(\arccos x) = x\). Thus, our equation becomes \(2x\sqrt{1 - x^2}\).
04

Finalize the result

Our final algebraic expression for \(\sin(2 \arccos x)\) is \(2x\sqrt{1 - x^2}\). This is the final solution to the problem.

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Most popular questions from this chapter

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$2 \sin ^{2} x-7 \sin x+3=0$$

(a) use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval \([0,2 \pi),\) and (b) solve the trigonometric equation and demonstrate that its solutions are the \(x\) -coordinates of the maximum and minimum points of \(f .\) (Calculus is required to find the trigonometric equation.) Function $$f(x)=\sin x \cos x$$ Trigonometric Equation $$-\sin ^{2} x+\cos ^{2} x=0$$

Use a graphing utility to approximate the solutions (to three decimal places) of the equation in the given interval. $$4 \cos ^{2} x-2 \sin x+1=0, \quad\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

Solve the multiple-angle equation. $$\cos \frac{x}{2}=\frac{\sqrt{2}}{2}$$

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$2 \cos ^{2} x-5 \cos x+2=0$$
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