Problem 10 - Precalculus with Calculus Previews

Question

If \(2 \sin t-\cos t=0,\) find all possible values of \(\sin\) \(t\) and \(\cos t\)

Step-by-Step Solution

Answer

The possible values are \(\sin t = \frac{1}{\sqrt{5}}\) and \(\cos t = \frac{2}{\sqrt{5}}\).

Step 1: Isolate one function

To solve the equation \(2 \sin t - \cos t = 0\), we first isolate one trigonometric function. We will isolate \(\cos t\) on one side by adding \(\cos t\) to both sides: \(2 \sin t = \cos t\).

Step 2: Express using trigonometric identity

With \(2 \sin t = \cos t\), divide both sides by \(\cos t\): \(2 \tan t = 1\). This implies that \(\tan t = \frac{1}{2}\).

Step 3: Find general solutions to tangent

From \(\tan t = \frac{1}{2}\), \(t\) could be any angle that satisfies this tangent. The general solution for \(t\) is given by \(t = \tan^{-1}(\frac{1}{2}) + n\pi\) where \(n\) is any integer.

Step 4: Use trigonometric ratios for specific solutions

For \(t = \tan^{-1}(\frac{1}{2})\), compute \(\sin t\) and \(\cos t\) using the identity \(\tan t = \frac{\sin t}{\cos t}\). We have \(\frac{\sin t}{\cos t} = \frac{1}{2}\) which leads to \(\sin t = \frac{1}{\sqrt{5}}\) and \(\cos t = \frac{2}{\sqrt{5}}\), ensuring the equation \((\sin^2 t + \cos^2 t = 1)\) holds true.

Step 5: Verify the range of solutions

Since trigonometric functions can have multiple outputs over different intervals, check that these solutions satisfy all original conditions over the range, confirming the solutions \(\sin t = \frac{1}{\sqrt{5}}\) and \(\cos t = \frac{2}{\sqrt{5}}\) for all \(t = \tan^{-1}(\frac{1}{2}) + n\pi\).

Key Concepts

Solving Trigonometric Equations

Trigonometric equations are mathematical expressions involving trigonometric functions like sine, cosine, and tangent. Solving these equations means finding the values of the variable (often an angle) that make the equation true. For the equation \(2 \sin t - \cos t = 0\), the goal is to find all values of \(t\) that satisfy it.
To start, you might want to rearrange the equation to isolate one trigonometric function. This makes it easier to handle mathematically. In our example, \(\cos t\) was isolated to form \(2 \sin t = \cos t\). Isolating functions is a common first step as it simplifies the problem to a more familiar form, like a single trigonometric ratio which we can solve directly for the angle.

Tangent Function

The tangent of an angle, \(t\), is defined as the ratio of the sine function to the cosine function. Written as \(\tan t = \frac{\sin t}{\cos t}\), it is a crucial part of solving equations where sine and cosine are related.
In our problem, by manipulating the given trigonometric equation, we arrived at \(2 \tan t = 1\), which simplifies to \(\tan t = \frac{1}{2}\). The tangent function is periodic with a period of \(\pi\), meaning it repeats every \(\pi\) units. This property helps in finding the general solutions when solving equations involving tangent, as specific solutions can be generalized to account for all cyclical repeats.

Trigonometric Ratios

Trigonometric ratios are relationships between the sides of a right triangle and its angles. The main trigonometric ratios are sine, cosine, and tangent, defined respectively as \(\sin t = \frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos t = \frac{\text{adjacent}}{\text{hypotenuse}}\), and \(\tan t = \frac{\text{opposite}}{\text{adjacent}}\).
In solving our equation, this allowed us to directly determine the values of \(\sin t\) and \(\cos t\) given the tangent relation: \(\tan t = \frac{1}{2}\). Specifically, \(\sin t\) and \(\cos t\) were computed such that their ratio remains consistent with the tangent value, leading to \(\sin t = \frac{1}{\sqrt{5}}\) and \(\cos t = \frac{2}{\sqrt{5}}\). Together, these values satisfy the identity: \(\sin^2 t + \cos^2 t = 1\), which confirms their correctness in the unit circle framework.

General Solutions of Trigonometric Equations

The general solutions of trigonometric equations provide all possible solutions for the angle \(t\), which includes an infinite set of solutions due to the periodic nature of trigonometric functions.
For a given equation like \(\tan t = \frac{1}{2}\), the initial solution \(t = \tan^{-1}(\frac{1}{2})\) is extended to a general solution using the equation \(t = \tan^{-1}(\frac{1}{2}) + n\pi\), where \(n\) is any integer.
This accounts for all possible values of \(t\) that satisfy the original equation. Periodicity plays a vital role here, as it ensures that adding integer multiples of the period (\(\pi\) for tangent) will continue to satisfy the equation. Understanding this concept helps in covering all scenarios where the equation can hold true, beyond just the primary solution.