Problem 55 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 53-64: Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(2 \tan ^{2} t+9 \tan t+3=0 ; \quad\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Step-by-Step Solution

Answer

The solution to the equation is approximately \( t = -0.4142 \) within the interval.

Step 1: Rearrange the Quadratic Equation

First, identify the form of the equation, which is a quadratic in terms of \( \tan t \): \( 2(\tan t)^2 + 9(\tan t) + 3 = 0 \). Let's solve this as a standard quadratic equation \( ax^2 + bx + c = 0 \), where \( x = \tan t \), \( a = 2 \), \( b = 9 \), \( c = 3 \).

Step 2: Apply the Quadratic Formula

The solutions for \( \tan t \) can be found using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 2 \), \( b = 9 \), and \( c = 3 \): \[ \tan t = \frac{-9 \pm \sqrt{9^2 - 4(2)(3)}}{4} = \frac{-9 \pm \sqrt{81 - 24}}{4} = \frac{-9 \pm \sqrt{57}}{4} \].

Step 3: Calculate the Tangent Solutions

Calculate the approximate values for \( \tan t \):- With the positive sign: \( \tan t = \frac{-9 + \sqrt{57}}{4} \approx -0.4377 \).- With the negative sign: \( \tan t = \frac{-9 - \sqrt{57}}{4} \approx -3.4377 \).

Step 4: Use Inverse Tangent Function

To find \( t \), use the inverse tangent function, defined as \( t = \arctan(x) \), and ensure the solutions fall within the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). For \( \tan t = -0.4377 \), \( t \approx \arctan(-0.4377) \approx -0.4142 \). The second solution for \( \tan t = -3.4377 \) is not within the interval, because this corresponds to a value that yields \( t \approx -1.2925 \), which is outside the specified interval.

Step 5: Verify and Conclude

Verify the solution \( t \approx -0.4142 \) by checking if substituting \( \tan(-0.4142) \) into the original equation approximately satisfies it. Confirm for each term that the equality holds, ensuring that any computational approximation errors are minor. The suitable solution for \( t \) thus remains \( -0.4142 \) within the given interval.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in mathematics and frequently appear in various fields such as engineering, physics, and economics. A quadratic equation is typically in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In the original exercise, we have a specialized quadratic equation involving tangent, where \(\tan t\) plays the role of \(x\).
Breaking it down:

• \(a = 2\)
• \(b = 9\)
• \(c = 3\)

To solve such equations, we utilize the **quadratic formula**:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For this exercise, applying the quadratic formula to \(2(\tan t)^2 + 9(\tan t) + 3 = 0\) helps find the values of \(\tan t\) that satisfy the equation. These values are critical to finding the angle \(t\) using inverse trigonometric functions.

Tangent Function

The tangent function, often written as \(\tan\), is one of the primary trigonometric functions used to relate angles in right-angled triangles. In the unit circle, it is defined as the ratio of the sine to the cosine of an angle:
\[ \tan t = \frac{\sin t}{\cos t} \]
Tangent is particularly unique because it has an undefined point whenever \(\cos t = 0\), leading to vertical asymptotes in its graph. This periodic behavior repeats every \(\pi\) radians. In problems involving quadratic forms like ours, solving the equation initially gives us potential tangent values. The real challenge is determining the correct angle, \(t\), which results from these tangent values. By using the inverse tangent function, we can extract \(t\) from any known tangent value.

Arctan

The inverse tangent function, also known as \(\arctan\) or \(\tan^{-1}\), is used to determine the angle whose tangent is a given number. It is the reverse process of the tangent function. When you are given a tangent, \(x\), you can find the corresponding angle, \(t\), in radians by calculating \(t = \arctan(x)\).In the solved problem:

• We found \(\tan t = -0.4377\), resulting in \(t \approx \arctan(-0.4377) \approx -0.4142\).

Keep in mind that \(\arctan\) will return values within its principal range, \((-\frac{\pi}{2}, \frac{\pi}{2})\). This range restriction is essential when ensuring that solutions fall within any given specific interval, such as the one stated in the exercise. If a calculated angle is outside the desired interval, as was the case with \(\tan t = -3.4377\), it should be excluded from the set of acceptable solutions.

Intervals in Trigonometry

When working with trigonometric functions, it's often necessary to consider specific intervals. These intervals define the possible values that angles can take and are crucial when using inverse trigonometric functions.

• The primary interval for \(\arctan\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\).
• This is because \(\tan t\) spans from negative to positive infinity within each interval of \(\pi\).

In the original problem, the range \((-\frac{\pi}{2}, \frac{\pi}{2})\) implies that we're looking for angles \(t\) that also reside within this interval. Ensuring solutions fall within this interval is vital to accuracy, as solutions outside it, such as a calculated \(t \approx -1.2925\), do not satisfy the given limits.
By always checking computed values against specified intervals, we maintain consistency and correctness in trigonometric problems.