Problem 61 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 57-62: Use an addition or subtraction formula to find the solutions of the equation that are in the interval \([0, \pi\) ). $$ \tan 2 t+\tan t=1-\tan 2 t \tan t $$

Step-by-Step Solution

Answer

The solutions are \( t = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4} \) within \([0, \pi)\).

Step 1: Recognize and Apply Trigonometric Identities

The equation given is \( \tan 2t + \tan t = 1 - \tan 2t \tan t \). Recognize that this can be simplified using the tangent addition formula: \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \). Here, we realize that the left side resembles \( \tan(2t + t) \). Apply the identity:\[\tan(3t) = \tan(2t + t) = \frac{\tan 2t + \tan t}{1 - \tan 2t \tan t}\] This simplifies our expression to \( \tan(3t) = 1 \).

Step 2: Solve for the angles where \( \tan(3t) = 1 \)

Since \( \tan(3t) = 1 \), we need to find values of \( 3t \) such that this is true. The general solution for \( \tan \theta = 1 \) is \( \theta = \frac{\pi}{4} + n\pi \), where \( n \) is an integer. Thus, for \( 3t \), we have:\[3t = \frac{\pi}{4} + n\pi\]Where \( n \) is an integer.

Step 3: Determine \( t \) Values in the Given Interval

We find the values of \( t \) by solving \( t = \frac{\pi}{12} + \frac{n\pi}{3} \). We need these values within the interval \([0, \pi)\). Start by setting \( n = 0 \):\[ t = \frac{\pi}{12} \].Next, set \( n = 1 \):\[ t = \frac{\pi}{12} + \frac{\pi}{3} = \frac{5\pi}{12} \].With \( n = 2 \):\[ t = \frac{\pi}{12} + \frac{2\pi}{3} = \frac{9\pi}{12} = \frac{3\pi}{4} \].Lastly, with \( n = 3 \):\[ t = \frac{\pi}{12} + \pi = \frac{13\pi}{12} \] (which is not in \([0, \pi)\)).So the suitable values of \( t \) are \( \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4} \).

Key Concepts

Using the Tangent Addition Formula

The tangent addition formula is a fundamental trigonometric identity used to combine the tangents of two angles. It's expressed as: \[\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \] In our exercise, this formula helps us simplify the equation \( \tan 2t + \tan t = 1 - \tan 2t \tan t \). Recognizing that the left side of the equation looks like \( \tan(2t + t) \), we can use the formula to express this as \( \tan(3t) \). This simplification step is crucial. It transforms a complex equation into a more manageable form, allowing us to apply known solutions for tangent equations, ultimately reducing our problem-solving effort.

Solving Trigonometric Equations

To solve trigonometric equations, especially ones involving tangent, it's important to first simplify the equation as much as possible. Once we used the tangent addition formula, our equation became: \\( \tan(3t) = 1 \). The next step is to identify when the tangent of an angle equals 1. This occurs at specific values of \( \theta \), such as:

• \( \theta = \frac{\pi}{4} \)
• Or more generally, \( \theta = \frac{\pi}{4} + n\pi \), where \( n \) is an integer

By allowing for multiple solutions through integer \( n \), we can fully explore all scenarios that satisfy the condition. Hence, solving \( 3t = \frac{\pi}{4} + n\pi \) helps us find all possible solutions for \( 3t \). Each value of \( n \) provides a potential solution for \( t \), which we can later check against the given interval.

Finding Interval Solutions

When solving trigonometric equations where the solutions must fit within a specific interval, it's essential to ensure all potential solutions are checked against the interval's limits. For this exercise, we're interested in values for \( t \) that lie within \([0, \pi)\). Once we've expressed \( 3t\) as \( \frac{\pi}{4} + n\pi \), solving for \( t \) gives us \( t = \frac{\pi}{12} + \frac{n\pi}{3} \). We then substitute integer values for \( n \), checking each result:

• For \( n = 0 \), \( t = \frac{\pi}{12} \)
• For \( n = 1 \), \( t = \frac{5\pi}{12} \)
• For \( n = 2 \), \( t = \frac{3\pi}{4} \)

These values all fit the interval \([0, \pi)\). When \( n = 3 \), \( t = \frac{13\pi}{12} \), which exceeds \( \pi \) and is therefore outside our interval of interest. Only the valid interval solutions are considered as the final answers, providing the correct set of possible values for \( t \).