Problem 88 - Calculus Early Transcendentals

Question

\(A n\) 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 $$\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \text { and } \cos x=2 \cos ^{2} \frac{x}{2}-1$$

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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.

Step 3: 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.

Key Concepts

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.