/*! 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 73 Use the trigonometric substituti... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the trigonometric substitution \(u=a \tan \theta,\) where \(-\pi / 2<\theta<\pi / 2\) and \(a>0,\) to simplify the expression \(\sqrt{a^{2}+u^{2}}\).

Short Answer

Expert verified
By using the trigonometric substitution \(u = a \tan\theta\), the expression \(\sqrt{a^{2}+u^{2}}\) simplifies to \(a \sec \theta\).

Step by step solution

01

- Substitute u

Substitute \(u = a\tan \theta\) into the expression \(\sqrt{a^{2}+u^{2}}\), this will give: \(\sqrt{a^2 + a^2\tan^2 \theta}\)
02

- Use Pythagorean identity

Recall the Pythagorean identity \(1 + \tan^{2} \theta = \sec^{2} \theta\). Apply it to the expression from step 1 to get \(\sqrt{a^2(1+\tan^2\theta)} = \sqrt{a^2 \sec^2 \theta}\).
03

- Simplify expression

Note that the square root of a square is the absolute value, and since \(a>0\) and \(-\pi /2 < \theta < \pi /2\) (meaning \(\sec \theta > 0\)), we can simplify \(\sqrt{a^2 \sec^2 \theta}\) as \(a \sec \theta\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pythagorean identity
Understanding the Pythagorean identity is crucial for simplifying expressions involving trigonometric functions. It is derived from the Pythagorean theorem, which relates the lengths of the sides of a right triangle. The basic Pythagorean identity is \(1 + \tan^2 \theta = \sec^2 \theta\). This relates the tangent and secant trigonometric functions back to the radius of the unit circle.

When a trigonometric substitution involving the tangent is made, like \(u = a\tan \theta\), this identity allows us to express the tan term in terms of secant, which often simplifies the algebra involved. By replacing \(\tan^2 \theta\) with \(\sec^2 \theta - 1\), any expression with \(\tan^2 \theta\) can be re-framed in terms of secant, which may simplify the calculations, particularly when dealing with square roots or other complex functions.
Simplifying radical expressions
Radical expressions often appear in mathematical problems, especially within geometry and algebra. A radical term is an expression that includes a root, such as a square root, cube root, etc. Simplifying these expressions is typically done to make the equations easier to understand and work with.

One approach to simplify radicals is to identify and extract powers of the terms under the radical, as we did with the trigonometric substitution \(\sqrt{a^2 + u^2}\). In the exercise mentioned, once the Pythagorean identity was applied, the radical expression \(\sqrt{a^2 \sec^2 \theta}\) could be simplified because \(\sec \theta\) is itself a square when squared. This simplification was possible because we knew that \(a\) and \(\sec \theta\) were both positive, thus removing the need for absolute value signs, and leaving us with the much simpler \(a \sec \theta\).

Understanding how to manipulate and simplify radicals allows us to elegantly solve problems that would otherwise be too cumbersome.
Trigonometric functions
Trigonometric functions are fundamental in mathematics, defining relationships between the angles and sides of triangles, and serving as the basis for describing periodic phenomena. The six primary trigonometric functions are sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot).

Each function has a specific geometric interpretation in the unit circle and can be defined in terms of the position of a point rotating about the origin. The tangent function represents the ratio of the opposite side to the adjacent side of a right-angled triangle, and the secant function is the reciprocal of the cosine function.

In our trigonometric substitution exercise, the tangent function transforms an algebraic expression into a trigonometric one that can be easily simplified using identities. Trigonometric functions are also essential for solving problems involving oscillations, waves, and other cyclical processes in various fields such as physics, engineering, and even finance. Being proficient in utilizing these functions is key to succeeding in many areas of advanced mathematics.

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Most popular questions from this chapter

Solving a Trigonometric Equation In Exercises \(69-74,\) find all solutions of the equation in the interval \([0,2 \pi)\).$$\tan (x+\pi)+2 \sin (x+\pi)=0$$

(a) use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval \(\mathbf{0}, \mathbf{2} \boldsymbol{\pi}\) ), 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.) $$\begin{array}{cc}\text{Function} && \text {Trigonometric Equation} \\\ f(x)=\sin x+\cos x && \cos x-\sin x=0 \end{array}$$

Prove the identity.$$\sin \left(\frac{\pi}{2}+x\right)=\cos x$$

The Mach number \(M\) of a supersonic airplane is the ratio of its speed to the speed of sound. When an airplane travels faster than the speed of sound, the sound waves form a cone behind the airplane. The Mach number is related to the apex angle \(\theta\) of the cone by \(\sin (\theta / 2)=1 / M\). (a) Use a half-angle formula to rewrite the equation in terms of cos \(\theta\). (b) Find the angle \(\theta\) that corresponds to a Mach number of 1. (c) Find the angle \(\theta\) that corresponds to a Mach number of 4.5. (d) The speed of sound is about 760 miles per hour. Determine the speed of an object with the Mach numbers from parts (b) and (c).

Use inverse functions where needed to find all solutions of the equation in the interval \(\mathbf{0}, \mathbf{2} \boldsymbol{\pi}\) ). $$\cot ^{2} x-9=0$$

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