Problem 41 - Precalculus with Calculus Previews

Question

Find all solutions of the given trigonometric equation if \(x\) is a real number and \(\theta\) is an angle measured in degrees. $$ \sin \theta+\cos \theta=1 $$

Step-by-Step Solution

Answer

Solutions are when \(\theta = 90^\circ + k \times 360^\circ\), where \(k\) is an integer.

Step 1: Square Both Sides

Start by squaring both sides of the equation \(\sin \theta + \cos \theta = 1\). This gives us \((\sin \theta + \cos \theta)^2 = 1^2\). Expanding the left side, we get \(\sin^2 \theta + 2\sin \theta\cos \theta + \cos^2 \theta = 1\).

Step 2: Use Pythagorean Identity

Apply the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). Substitute this identity into the equation from Step 1: \(1 + 2\sin \theta\cos \theta = 1\). Simplifying, we have \(2\sin \theta\cos \theta = 0\).

Step 3: Solve for Trigonometric Functions

The equation \(2\sin \theta\cos \theta = 0\) implies \(\sin \theta\cos \theta = 0\). Therefore, either \(\sin \theta = 0\) or \(\cos \theta = 0\).

Step 4: Find Solutions for \(\sin \theta = 0\)

Solving \(\sin \theta = 0\), we get \(\theta = n \times 180^\circ\), where \(n\) is any integer.

Step 5: Find Solutions for \(\cos \theta = 0\)

Solving \(\cos \theta = 0\), we get \(\theta = 90^\circ + m \times 180^\circ\), where \(m\) is any integer.

Step 6: Verify Original Equation Solutions

Substitute each of the angle values (\(\theta = 0^\circ, 180^\circ, 270^\circ, 360^\circ, \ldots \)) back into the original equation. Check both results for each value: only \(\theta = 90^\circ + k \times 360^\circ\) satisfies \(\sin 90^\circ + \cos 90^\circ = 1\) and so are the final answers.

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a fundamental concept in trigonometry. It expresses the intrinsic relationship between the sine and cosine functions of an angle. The identity is given by: \[ \sin^2 \theta + \cos^2 \theta = 1 \]This equality holds true for any angle \(\theta\).
To understand this identity, visualize a right triangle. Here, the sine of an angle corresponds to the ratio of the length of the opposite side to the hypotenuse, while the cosine represents the adjacent side over the hypotenuse. Squaring both these ratios and adding them together, you obtain the Pythagorean Identity.
In solving equations like \(\sin \theta + \cos \theta = 1\), the Pythagorean Identity is a handy tool. During step 2 of the exercise solution, it was applied to simplify the equation by substituting \(\sin^2 \theta + \cos^2 \theta\) with 1. This step is crucial for breaking down complex trigonometric equations more simply and uncovering hidden solutions.

Sine and Cosine Functions

Sine and cosine functions are foundational to trigonometry. They describe the primary periodic oscillations for angles in a unit circle.
- **Sine Function (\(\sin \theta\)):** - When \(\theta\) is an angle in a right triangle, sine is the ratio of the opposite side to the hypotenuse. - The sine function oscillates between -1 and 1 as the angle varies from 0° to 360°.- **Cosine Function (\(\cos \theta\)):** - This function gives the ratio of the adjacent side to the hypotenuse. - Like sine, cosine varies between -1 and 1 over a complete 0° to 360° cycle.
These functions can often complement each other to find solutions in complex equations. In our exercise, combining sine and cosine via the equation \(\sin \theta + \cos \theta = 1\) prompts the exploration of possible angle solutions by setting their sum to one. This involves periodic findings such as when one value dominates or balances the other, leading to particular angle values.

Angle Solutions

Finding angle solutions is a key task in solving trigonometric equations. In this exercise, the objective is to solve \(\sin \theta + \cos \theta = 1\) for real number angles. Understanding when sine or cosine reaches specific values allows us to determine possible angle solutions.
During computation:- Solving \(\sin \theta = 0\) gives angles such as 0°, 180°, where the sine function vanishes.- \(\cos \theta = 0\) results in angles like 90°, 270°, as cosine becomes zero at these points.
Specifically, only angles \(\theta = 90° + k \times 360°\), where \(k\) is an integer, satisfy the original equation, because here, sine equals 1 and cosine equals zero, aligning with \(\sin 90° + \cos 90° = 1\). These angle solutions show that, despite the possible intervals, only specific orientations actually solve the equation.