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Chapter 7: Problem 88

\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u .\) Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \text { and } \cos x=2 \cos ^{2} \frac{x}{2}-1$$

Short Answer

Expert verified
Question: Verify the given relations A, B, and C. Relation A: \(dx = \frac{2}{1+u^2}du\) Relation B: \(\sin{x} = \frac{2u}{1+u^2}\) Relation C: \(\cos{x} = \frac{1-u^2}{1+u^2}\) where \(x = 2\tan^{-1}u\). Answer: The given relations A, B, and C are verified using differentiation, right-triangle diagrams, and double-angle formulas.

Step by step solution

01

Verify relation A:

Differentiate \(x=2\tan^{-1}u\) with respect to \(u\): $$\frac{dx}{du} = \frac{d(2\tan^{-1}u)}{du}$$ Now, we know the derivative of \(\tan^{-1}u\) is \(\frac{1}{1+u^2}\). So, $$\frac{dx}{du} = 2\cdot\frac{1}{1+u^2}$$ Finally, multiply both sides by \(du\) to get $$dx = \frac{2}{1+u^2}du$$ Hence, relation A is verified.
02

Verify relations B and C with right-triangle diagrams:

Let's draw a right triangle with one angle as \(\frac{x}{2}\), and let the side opposite to it be called \(b\), the adjacent side be called \(a\), and the hypotenuse be called \(c\). We know \(\tan{\frac{x}{2}} = \frac{b}{a}\), so $$b = a\cdot\tan{\frac{x}{2}}$$ Define, \(u=\tan{\frac{x}{2}}\), substitute \(u\) in the above expression to get $$b = au$$ Now, the Pythagorean theorem relates the sides of the right triangle as $$a^2 + b^2 = c^2$$ Substitute \(b=au\) in the above equation to get, $$a^2 + a^2u^2 = c^2$$ or $$c^2 = a^2(1+u^2)$$ Solving for \(\sin{x}\) and \(\cos{x}\), using the definition of sine and cosine, we get $$\sin{x} = \frac{2ab}{c^2} = \frac{2au}{1+u^2}$$ and $$\cos{x} = \frac{a^2-b^2}{c^2} = \frac{1-u^2}{1+u^2}$$ Hence, relations B and C are verified using right-triangle diagrams.
03

Verify relation B and C with double-angle formulas:

Using the double-angle formulas, we have $$\sin{x} = 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}$$ and $$\cos{x} = 2\cos^2{\frac{x}{2}}-1$$ Since \(u=\tan{\frac{x}{2}}\), we get $$\sin{\frac{x}{2}}=\frac{u}{\sqrt{1+u^2}}$$ and $$\cos{\frac{x}{2}}=\frac{1}{\sqrt{1+u^2}}$$ Substitute these values in the double-angle formulas to get $$\sin{x}=2\cdot\frac{u}{\sqrt{1+u^2}}\cdot\frac{1}{\sqrt{1+u^2}} = \frac{2u}{1+u^2}$$ and $$\cos{x}=2\cdot\left(\frac{1}{\sqrt{1+u^2}}\right)^2 -1= \frac{1-u^2}{1+u^2}$$ Thus, relations B and C are also verified with the double-angle formulas.

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Key Concepts

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

Right Triangle
A right triangle is a triangle where one of the angles measures exactly 90 degrees. This makes it distinct because the other two angles will always sum to 90 degrees as per the angle sum property of triangles. The sides of a right triangle are conventionally referred to as:
  • The adjacent side: next to the angle of interest.
  • The opposite side: opposite to the angle of interest.
  • The hypotenuse: the longest side, opposite the right angle.
In right triangles, trigonometric functions such as sine, cosine, and tangent are commonly used, with their relationships defined as:
  • \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
  • \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
  • \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)
These functions are crucial in solving problems that involve angled measurements and modeling our world in a geometric setting. Right triangles also often employ the Pythagorean theorem, establishing a vital link between geometry and algebra.
Double-Angle Formulas
Double-angle formulas are extremely useful in trigonometry for simplifying expressions involving angles. These formulas relate the trigonometric functions of double angles (like \(2x\)) to functions of single angles (\(x\)).For sine and cosine, the double-angle identities are:
  • \( \sin(2x) = 2 \sin(x) \cos(x) \)
  • \( \cos(2x) = \cos^2(x) - \sin^2(x) \)
Using these formulas allows you to express functions of \(2x\) in terms of \(x\), which can often simplify solving equations or integrating functions.For instance, relation \(B\) from the exercise utilizes the double-angle sine formula to confirm the identity \( \sin(x) = \frac{2u}{1+u^2} \) by substituting \( \sin\left(\frac{x}{2}\right)\) and \( \cos\left(\frac{x}{2}\right)\) in terms of \(u\). These formulas enable deeper understanding and discovery in trigonometric manipulation.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry, named after the ancient Greek mathematician Pythagoras. It applies specifically to right triangles and states:\[ a^2 + b^2 = c^2 \]Here, \(a\) and \(b\) are the lengths of the two shorter sides of the triangle, and \(c\) is the length of the hypotenuse.This theorem is exceptionally important because it provides a way to calculate the length of one side of a triangle when the other two are known. In the original step-by-step solution, this theorem is used to verify relations \(B\) and \(C\) by inserting the expression for \(u = \tan\left(\frac{x}{2}\right)\) and solving expressions relating the sides.The Pythagorean theorem often paves the way for solving real-world problems, proving invaluable in fields like construction, physics, and numerous others that involve space and measurements.

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Most popular questions from this chapter

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}\) a. Find the work required to launch an object in terms of \(m\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18 th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2}\). For Earth to be a black hole, what would its radius need to be?

Consider the general first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=y_{0},\) for \(t \geq 0,\) where \(a, b,\) and \(y_{0}\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to horizontal line segments in the direction field. b. Draw a representative direction field in the case that \(a>0\). Show that if \(y_{0}>-b / a,\) then the solution increases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\). Show that if \(y_{0}>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution increases for \(t \geq 0\).

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\).

Let \(y(t)\) be the population of a species that is being harvested. Consider the harvesting model \(y^{\prime}(t)=0.008 y-h, y(0)=y_{0},\) where \(h>0\) is the annual harvesting rate and \(y_{0}\) is the initial population of the species. a. If \(y_{0}=2000,\) what harvesting rate should be used to maintain a constant population of \(y=2000\) for \(t \geq 0 ?\) b. If the harvesting rate is \(h=200 /\) year, what initial population ensures a constant population for \(t \geq 0 ?\)

The reaction of chemical compounds can often be modeled by differential equations. Let \(y(t)\) be the concentration of a substance in reaction for \(t \geq 0\) (typical units of \(y\) are moles/L). The change in the concentration of the substance, under appropriate conditions, is \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1),\) the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\). c. Graph and compare the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\).
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