Problem 27 - Algebra and Trigonometry with Analytic Geometry

Question

Use sum-to-product formulas to find the solutions of the equation. $$ \sin 5 t+\sin 3 t=0 $$

Step-by-Step Solution

Answer

The solutions are \( t = \frac{n\pi}{4} \) and \( t = \frac{(2m+1)\pi}{2} \) for integers \( n \) and \( m \).

Step 1: Identify the Problem

The given equation is \( \sin 5t + \sin 3t = 0 \). We need to solve this equation using the sum-to-product identities.

Step 2: Apply the Sum-to-Product Identity

Recall the sum-to-product identity for sine: \( \sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \). Use this identity with \( A = 5t \) and \( B = 3t \).

Step 3: Substitute into Identity

Substitute \( A = 5t \) and \( B = 3t \) into the identity \( \sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \), resulting in: \[ \sin 5t + \sin 3t = 2 \sin \left(\frac{5t + 3t}{2}\right) \cos \left(\frac{5t - 3t}{2}\right) \].

Step 4: Simplify Terms

Simplify to obtain: \( 2 \sin(4t) \cos(t) = 0 \). This product is zero when either \( \sin(4t) = 0 \) or \( \cos(t) = 0 \).

Step 5: Solve \(\sin(4t) = 0\)

The equation \( \sin(4t) = 0 \) implies \( 4t = n\pi \), where \( n \) is an integer. Thus, \( t = \frac{n\pi}{4} \).

Step 6: Solve \(\cos(t) = 0\)

The equation \( \cos(t) = 0 \) implies \( t = \frac{(2m+1)\pi}{2} \), where \( m \) is an integer.

Step 7: General Solution

The general solution combines the two separate solutions: \( t = \frac{n\pi}{4} \) and \( t = \frac{(2m+1)\pi}{2} \), where \( n \) and \( m \) are integers.

Key Concepts

Trigonometric Equations

Trigonometric equations are mathematical statements that involve trigonometric functions like sine, cosine, and tangent. These equations are essential in various fields, such as physics, engineering, and mathematics. Solving trigonometric equations often requires manipulating trigonometric identities to simplify and solve for the variable.

• In our example, the equation \( \sin 5t + \sin 3t = 0 \) involves the sine function of two different angles.
• The goal is to find all possible values of the variable \( t \) that satisfy the equation.

To solve equations like this, you often need to use special identities, like the sum-to-product formulas, to combine or separate trigonometric terms effectively.Using these formulas helps in transforming sums or differences of sines and cosines into products, making these equations easier to solve.

Sine Function

The sine function is one of the primary trigonometric functions, symbolized by \( \sin \). This periodic function is fundamental in the study of waves, oscillations, and circles.

• It is defined as the ratio of the opposite side to the hypotenuse in a right triangle.
• The function cycles every \( 2\pi \) radians, meaning every \( 360^\circ \).

In the given problem, both terms \( \sin 5t \) and \( \sin 3t \) are parts of the same type of function. To solve such an equation, the sum-to-product identity can be applied:\[\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)\]This helps transform the equation into a more solvable product form.

Cosine Function

The cosine function, denoted as \( \cos \), is another fundamental trigonometric function, and it is used alongside the sine function in many equations.

• Cosine measures the ratio of the adjacent side to the hypotenuse in a right triangle.
• Like sine, it also completes a cycle every \( 2\pi \) radians.

In our transformation to solve the equation \( \sin 5t + \sin 3t = 0 \), the cosine function appears because the sum-to-product identity introduces it:\[2 \sin(4t) \cos(t) = 0\]This equation is satisfied when either the sine part or the cosine part of the product equals zero. Therefore, we need to check the solutions for both \( \sin(4t) = 0 \) and \( \cos(t) = 0 \).

General Solution

The goal of solving a trigonometric equation is often to find the general solution, which includes all possible solutions that satisfy the given equation. Trigonometric equations can have infinite solutions due to the periodic nature of sine and cosine functions.

• For \( \sin(4t) = 0 \), the solutions occur at intervals of \( \pi \, \text{(i.e., multiples of 180°)} \). Therefore, we express \( t \) as \( \frac{n\pi}{4} \), where \( n \) is an integer.
• For \( \cos(t) = 0 \), solutions occur at odd multiples of \( \frac{\pi}{2} \). This leads to the expression \( t = \frac{(2m+1)\pi}{2} \), with \( m \) as an integer.

Thus, the comprehensive general solution combines both these sets of solutions to ensure every potential value of \( t \) is covered. By acknowledging that \( n \) and \( m \) are integers, we grasp the full range of periodic solutions that these trigonometric relationships produce.