Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

We need the identities $$ \begin{array}{l} \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \\ \cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \end{array} $$ in the next two exercises and in the next section. It is unlikely that you remember them from a trigonometry class. We hope you do remember, however, the double angle formulas 7.6 , $$ \sin (A+B)=\sin A \cos B+\cos A \sin B \quad \cos (A+B)=\cos A \cos B-\sin A \sin B $$ a. Use \(\sin (A+B)=\sin A \cos B+\cos A \sin B\) and the identities, \(\sin (-A)=-\sin A\) and \(\cos (-A)=\cos A,\) to show that $$ \sin (A-B)=\sin A \cos B-\cos A \sin B $$ b. Use the equations $$ \begin{array}{l} \sin (A+B)=\sin A \cos B+\cos A \sin B \\ \sin (A-B)=\sin A \cos B-\cos A \sin B \end{array} $$ to show that $$ \sin (A+B)-\sin (A-B)=2 \cos A \sin B $$ c. Solve for \(A\) and \(B\) in $$ \begin{array}{l} A+B=x \\ A-B=y \end{array} $$ d. Substitute the values for \(A+B, A-B, A,\) and \(B\) into Equation 7.7 to obtain $$ \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} $$ e. Use \(\cos (A+B)=\cos A \cos B-\sin A \sin B\) to show that $$ \cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) $$ The argument will be similar to the previous steps.

Step-by-Step Solution

Answer

The given identities are proven using trigonometric formulas and substitutions.

Step 1: Prove Identity for \( \sin(A-B) \)

Recall the identity \( \sin(-A) = -\sin A \) and \( \cos(-A) = \cos A \). Thus, \( \sin(A-B) = \sin(A+(-B)) \). Use the formula for \( \sin(A+B) \): \( \sin(A-B) = \sin A \cos(-B) + \cos A \sin(-B) \). Simplifying using the identities gives \( \sin(A-B) = \sin A \cos B - \cos A \sin B \).

Step 2: Derive \( \sin(A+B) - \sin(A-B) \) Identity

Using the two identities: \( \sin(A+B) = \sin A \cos B + \cos A \sin B \) and \( \sin(A-B) = \sin A \cos B - \cos A \sin B \), subtract them: \( \sin(A+B) - \sin(A-B) = (\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B) \). This simplifies to \( 2 \cos A \sin B \).

Step 3: Solve for \( A \) and \( B \) given \( A+B=x \) and \( A-B=y \)

From the equations \( A+B = x \) and \( A-B = y \), solve for \( A \) and \( B \): \( A = \frac{x+y}{2} \) and \( B = \frac{x-y}{2} \).

Step 4: Substitute Values for \( A \), \( B \)

Substitute \( A = \frac{x+y}{2} \) and \( B = \frac{x-y}{2} \) into the derived formula \( \sin(A+B) - \sin(A-B) = 2 \cos A \sin B \). It becomes \( \sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \), which is the required identity.

Step 5: Prove Identity for \( \cos x - \cos y \)

Use the formula \( \cos(A+B) = \cos A \cos B - \sin A \sin B \). From the expressions \( A = \frac{x+y}{2} \) and \( B = \frac{x-y}{2} \), substitute into \( \cos x - \cos y = -2 \sin A \sin B \). This simplifies to \( -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \), proving the identity \( \cos x - \cos y = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \).

Key Concepts

Double Angle Formulas

The double angle formulas are an important tool in trigonometry that allow you to express trigonometric functions of sums as products of functions of single angles. Specifically, for sine and cosine, the double angle formulas are given by:

• \( \sin(2A) = 2\sin(A)\cos(A) \)
• \( \cos(2A) = \cos^2(A) - \sin^2(A) = 2\cos^2(A) - 1 = 1 - 2\sin^2(A) \)

These formulas are useful because they can simplify expressions and solve equations involving trigonometric functions.
To understand these formulas, recall that they derive from the sum of angles identities: \( \sin(A+B) = \sin A \cos B + \cos A \sin B \) and \( \cos(A+B) = \cos A \cos B - \sin A \sin B \). By replacing \( B \) with \( A \) in these identities, we arrive at the double angle identities.
In situations where calculating the sine or cosine of a double angle is necessary, these identities offer a simplification, which is especially handy in proving other trigonometric identities or solving equations.

Sine and Cosine Difference

The difference identities for sine and cosine are derived from the angle addition identities. They help to find the sine or cosine of the difference between two angles. The relevant formulas are:

• \( \sin(A-B) = \sin A \cos B - \cos A \sin B \)
• \( \cos(A-B) = \cos A \cos B + \sin A \sin B \)

These identities are similar to angle sum identities but adjusted for differences.
The understanding of these identities is crucial for many calculus and trigonometry problems. They allow one to express more complex trigonometric expressions in simpler forms and are pivotal in proving further identities.
In practice, the formulas help to break down the sine or cosine of a compound angle into more manageable parts. This is quite applicable in solving problems where direct computation isn't straightforward. When working with trigonometric identities, recognizing these formulas can aid in simplifying and reworking expressions.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals using trigonometric identities. It involves replacing variables with trigonometric functions to take advantage of the Pythagorean identity. This method is particularly useful in solving integrals involving square roots and can often turn a difficult integral into a more solvable form.

• For an integral involving \( \sqrt{a^2 - x^2} \), use \( x = a\sin(\theta) \).
• For \( \sqrt{a^2 + x^2} \), use \( x = a\tan(\theta) \).
• For \( \sqrt{x^2 - a^2} \), use \( x = a\sec(\theta) \).

The substitution transforms the variable into a trigonometric function, which can drastically simplify the integration process.
Trigonometric substitution takes advantage of the relationships between angles and lengths in right triangles, representing algebraic expressions in more manageable forms. Once the integral is simplified using a trigonometric identity, the solution can be found using standard integration techniques. After integration, it's necessary to convert back to the original variable using inverse trigonometric functions. This method is powerful, especially in calculus, for tackling otherwise complex integrals.