Problem 58 - 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 \([0, \pi)\) $$\cos 5 t \cos 3 t=\frac{1}{2}+\sin (-5 t) \sin 3 t$$

Step-by-Step Solution

Answer

The solutions are \(t = \frac{\pi}{24}\) and \(t = \frac{5\pi}{24}\).

Step 1: Use Trigonometric Identity

Start by using the identity for the cosine of a sum: \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]Given the equation\[ \cos 5t \cos 3t = \frac{1}{2} + \sin(-5t) \sin 3t \]use the identity by rewriting \[ \cos(5t - 3t) = \cos 5t \cos 3t + \sin 5t \sin 3t \] or equivalently for \[ \cos 5t \cos 3t + \sin(-5t) \sin 3t = \cos(5t - (-3t)) = \cos(8t). \]

Step 2: Simplify the Equation

Substituting the identity into the equation we get:\[ \cos (8t) = \frac{1}{2} \]

Step 3: Solve for t

We now solve for \(t\) by finding the angles where \(\cos \theta = \frac{1}{2}\):- The general solutions for \(\cos \theta = \frac{1}{2}\) are \(\theta = \frac{\pi}{3} + 2k\pi\) and \(\theta = \frac{5\pi}{3} + 2k\pi\) where \(k\) is an integer.For our equation \(8t = \theta\), therefore:- \(8t = \frac{\pi}{3} + 2k\pi\)- \(8t = \frac{5\pi}{3} + 2k\pi\)

Step 4: Find Solutions within Interval

We need only solutions for \(t\) in the interval \([0, \pi)\).- For \(8t = \frac{\pi}{3}\): \[ t = \frac{\pi}{24} \]- For \(8t = \frac{5\pi}{3}\): \[ t = \frac{5\pi}{24} \] Test increments \(k\) to ensure all solutions are in interval \([0, \pi)\). The increments for \(k\) do not provide new solutions within this offered interval for \(t\).

Step 5: Verifying Solutions

Verify whether the obtained values satisfy the original equation in the necessary interval.Checking for \( t = \frac{\pi}{24} \) and \( t = \frac{5\pi}{24} \), since they satisfy \(\cos (8t) = \frac{1}{2}\), both are valid.

Key Concepts

Cosine of a Sum Formula

The cosine of a sum formula is a fundamental concept in trigonometry. It is beautifully expressed through the identity: \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] This formula allows us to express the cosine of the sum of two angles, \(A\) and \(B\), in terms of the individual cosines and sines of those angles. It's particularly helpful in solving equations like the one in the exercise above, where you need to rewrite complex trigonometric expressions in simpler terms. By rearranging and substituting back, you can solve for unknowns or equate parts of an equation to find solutions. Let's look at an example:

• If \( \cos 5t \cos 3t + \sin(-5t) \sin 3t \) appears, you can rewrite it using \( \cos(5t + 3t) \), making the problem easier to navigate.

This transformation makes it possible to simplify and solve complex trigonometric problems effectively.

General Solution of Trigonometric Equations

The goal when solving trigonometric equations is often finding all possible angles that satisfy the equation. These are what we refer to as the general solutions. In trigonometry, the periodic nature of sine and cosine functions leads to multiple solutions for any given equation.For instance, if you have an equation like \( \cos \theta = \frac{1}{2} \), the solutions are:

• \( \theta = \frac{\pi}{3} + 2k\pi \)
• \( \theta = \frac{5\pi}{3} + 2k\pi \)

where \(k\) is any integer. This represents an infinite set of solutions, distributed periodically over angles differing by full circle rotations.Knowing these patterns enables you to understand and calculate all possible solutions of a trigonometric equation rather than just a single one. For applications or when specified, you may be tasked with narrowing these down to relevant ones, such as when dealing with specified intervals like \([0, \pi)\).

Interval Notation Solutions

Solving trigonometric equations often requires providing solutions in a particular range or interval. In this exercise, the interval of interest is \([0, \pi)\).Interval notation helps succinctly express a range of numbers. Here:

• \([0, \pi)\) means all numbers from 0 to \(\pi\), including 0 but excluding \(\pi\).

This focused interval requires you to examine solutions to ensure they fit within the boundaries. Begin by finding candidates for solutions through the general approach, then test them against the interval.Returning to our exercise: post simplification, solutions like \( t = \frac{\pi}{24} \) and \( t = \frac{5\pi}{24} \) are checked to see if they reside within \([0, \pi)\). These values fit, thus validating them as appropriate solutions to share when considering the specified interval. This efficient organization of solutions ensures clarity and preciseness in communicating results.