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Chapter 6: Problem 111

Use a right triangle to write \(\sin \left(2 \sin ^{-1} x\right)\) as an algebraic expression. Assume that \(x\) is positive and in the domain of the given inverse trigonometric function.

Short Answer

Expert verified
The algebraic expression for \(\sin \left(2 \sin ^{-1} x\right)\) is \(2x\sqrt{1 - x^2}\).

Step by step solution

01

Interpret the Inverse Sine Function

The function \(\sin^{-1} x\) gives us the angle whose sine is \(x\). Let's call this angle \(A\). That is, \(A = \sin^{-1} x\). Therefore, the function \(\sin \left(2 \sin ^{-1} x\right)\) becomes \(\sin(2A)\).
02

Draw the Right Triangle and Calculate Cos(A)

Let's draw a right triangle with angle \(A\), where \(A = \sin^{-1} x\). From the definition of sine function, we can infer that the opposite side is \(x\) and the hypotenuse is \(1\). Now, using the Pythagorean theorem, we can calculate the length of the adjacent side (which is \(\sqrt{1 - x^2}\)), and consequently, \(\cos A = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}\).
03

Apply the Double Angle Identity for Sine

To rewrite \(\sin(2A)\) in algebraic terms, we use the double-angle identity for sine, that is, \(\sin(2A) = 2\sin(A)\cos(A)\). But we already know that \(\sin(A) = x\) and \(\cos(A) = \sqrt{1 - x^2}\). Replacing these values in, we obtain: \(\sin(2A) = 2x\sqrt{1 - x^2}\).

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Most popular questions from this chapter

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$ 5 \sin x=2 \cos ^{2} x-4 $$

Graph each side of the equation in the same viewing rectangle. If the graphs appear to coincide, verify that the equation is an identity. If the graphs do not appear to coincide, this indicates that the equation is not an identity. In these exercises, find a value of \(x\) for which both sides are defined but not equal. $$ \sin 1.2 x \cos 0.8 x+\cos 1.2 x \sin 0.8 x=\sin 2 x $$

Will help you prepare for the material covered in the next section. In each exercise, use exact values of trigonometric functions to show that the statement is true. Notice that each statement expresses the product of sines and/or cosines as a sum or a difference. $$ \sin \pi \cos \frac{\pi}{2}=\frac{1}{2}\left[\sin \left(\pi+\frac{\pi}{2}\right)+\sin \left(\pi-\frac{\pi}{2}\right)\right] $$

Suppose you are solving equations in the interval \([0,2 \pi)\) Without actually solving equations, what is the difference between the number of solutions of \(\sin x=\frac{1}{2}\) and \(\sin 2 x=\frac{1}{2} ?\) How do you account for this difference?

Exercises \(110-112\) will help you prepare for the material covered in the next section. Use the appropriate values from Exercise 110 to answer each of the following. a. Is \(\sin \left(2 \cdot 30^{\circ}\right),\) or \(\sin 60^{\circ},\) equal to \(2 \sin 30^{\circ} ?\) b. Is \(\sin \left(2 \cdot 30^{\circ}\right),\) or \(\sin 60^{\circ},\) equal to \(2 \sin 30^{\circ} \cos 30^{\circ} ?\)
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