Problem 60 - Precalculus: Functions and Graphs

Question

Exer. \(57-62:\) Use an addition or subtraction formula to find the solutions of the equation that are in the Interval \([0, \pi)\) $$\sin 3 t \cos t+\cos 3 t \sin t=-\frac{1}{2}$$

Step-by-Step Solution

Answer

The solutions for \( t \) are \( t = \frac{7\pi}{24} \) and \( t = \frac{11\pi}{24} \).

Step 1: Recognize the Formula

The expression given, \( \sin 3t \cos t + \cos 3t \sin t \), matches the formula for the sine of a sum: \( \sin(a + b) = \sin a \cos b + \cos a \sin b \). Here, we identify \( a = 3t \) and \( b = t \). Thus, \( \sin(3t + t) = \sin 4t \).

Step 2: Simplify the Equation

Using the identity from Step 1, we can simplify the original equation to \( \sin 4t = -\frac{1}{2} \).

Step 3: Find General Solutions

We know that \( \sin x = -\frac{1}{2} \) at specific angles. \( \sin x = -\frac{1}{2} \) when \( x = \frac{7\pi}{6} + 2k\pi \) or \( x = \frac{11\pi}{6} + 2k\pi \) for any integer \( k \). Let \( x = 4t \), then those are our general solutions for \( 4t \).

Step 4: Solve for \( t \) in the Interval

Solve \( 4t = \frac{7\pi}{6} + 2k\pi \) and \( 4t = \frac{11\pi}{6} + 2k\pi \). This gives \( t = \frac{7\pi}{24} + \frac{k\pi}{2} \) and \( t = \frac{11\pi}{24} + \frac{k\pi}{2} \). We need to find solutions where \( t \) is within the interval \([0, \pi)\).

Step 5: Find Valid Solutions

Substitute different integer values for \( k \) until \( t \) falls within the interval \([0, \pi)\). For \( t = \frac{7\pi}{24} \), set \( k = 0 \), giving \( t = \frac{7\pi}{24} \). Next, for \( t = \frac{11\pi}{24} \), set \( k = 0 \), giving \( t = \frac{11\pi}{24} \). Since \( k = 1 \) produces solutions greater than \( \pi \), these are the only solutions within the desired interval.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental mathematical equations involving trigonometric functions. These equations are always true for all values of the included variables. They help simplify and manipulate expressions in trigonometry.
For instance, one crucial identity used here is the sine addition formula, expressed as:

• \(\sin(a + b) = \sin a \cos b + \cos a \sin b\)

This identity allows us to combine products of sine and cosine terms into a single sine function. It simplifies complex trigonometric equations by reducing their terms, thus making it easier to solve them.
An understanding of trigonometric identities is essential when solving equations where expressions need to be rewritten or rearranged using these identities. They are the backbone for solving many trigonometric equations efficiently.

Sine Function

The sine function is one of the primary trigonometric functions and is often notated as \(\sin\). It represents the y-coordinate of a point on the unit circle where an angle \(\theta\) meets the circle, describing a wave-like pattern.
The sine function is periodic, with a period of \(2\pi\). It oscillates between -1 and 1.
Key properties of the sine function include:

• It is an odd function: \(\sin(-x) = -\sin(x)\).
• It has symmetry about the origin.
• The value \(\sin x = -\frac{1}{2}\) can happen at specific angles such as \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\) in one full cycle.

Understanding these properties helps in solving trigonometric equations when given specific sine values. Recognizing these angles quickly allows us to identify solutions in various intervals or transform expressions using known identities.

Interval Notation

Interval notation is a shorthand method for writing subsets of the real number line and is particularly useful in calculus and algebra.
It uses parentheses \((...)\) and brackets \([...])\) to describe sets of numbers that represent parts of the real line.
Brackets \([...])\) indicate that the endpoint is included \("closed" interval\), while parentheses \((...)\) indicate it is not \("open" interval\).
For example:

• \([0, \pi)\) includes 0 but does not include \(\pi\).

In the context of solving trigonometric equations, defining a specific interval like \([0, \pi)\) helps limit the solutions to only those within the desired range. This is particularly helpful when dealing with periodic functions such as sine and cosine, which repeat over intervals.

Equation Solving

Equation solving involves finding values that satisfy the given mathematical equation. In trigonometric equation solving, the challenge often lies in handling periodic functions and identities to find all possible solutions.
For example, when solving an equation like \(\sin 4t = -\frac{1}{2}\), first identify intervals and angles where the equation holds true by using known angles from the unit circle, such as \(\frac{7\pi}{6} + 2k\pi\) and \(\frac{11\pi}{6} + 2k\pi\).
With these general solutions, we move forward to solve for the variable \(t\) in the desired interval \([0, \pi)\). Substitute integers for \(k\) to find particular solutions within this interval. This methodical approach ensures all possible solutions are considered and correctly identified within the predetermined bounds.