Problem 53 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ \cos \alpha+\sin \alpha=1 $$

Step-by-Step Solution

Answer

The solutions are \(\alpha = 0\) and \(\alpha = \frac{\pi}{2}\).

Step 1: Rewrite the Equation

Start with the given equation: \(\cos \alpha + \sin \alpha = 1\). Try expressing one trigonometric function in terms of the other. Subtract \(\cos \alpha\) from both sides: \(\sin \alpha = 1 - \cos \alpha\).

Step 2: Use Trigonometric Identities

Square both sides of the equation: \((\sin \alpha)^2 = (1 - \cos \alpha)^2\). Using the Pythagorean identity, \(\sin^2 \alpha = 1 - \cos^2 \alpha\), and the binomial expansion, the equation becomes \(1 - \cos^2 \alpha = 1 - 2\cos \alpha + \cos^2 \alpha\).

Step 3: Simplify the Equation

Rearrange the equation from Step 2 to isolate terms. Combine like terms to get \(2\cos^2 \alpha - 2\cos \alpha = 0\).

Step 4: Factor the Quadratic Equation

Factor out the common term, resulting in the equation \(2\cos \alpha(\cos \alpha - 1) = 0\). This gives two potential solutions: \(2\cos \alpha = 0\) or \(\cos \alpha - 1 = 0\).

Step 5: Solve Each Factor

Solve each equation separately. For \(2\cos \alpha = 0\), we have \(\cos \alpha = 0\). The solutions in the interval \([0, 2\pi)\) are \(\alpha = \frac{\pi}{2}, \frac{3\pi}{2}\). For \(\cos \alpha - 1 = 0\), we have \(\cos \alpha = 1\). The solution is \(\alpha = 0\).

Step 6: Verify Solutions

Substitute the solutions back into the original equation to verify. For \(\alpha = 0\), \(\cos(0) + \sin(0) = 1 + 0 = 1\), which matches. For \(\alpha = \frac{\pi}{2}\), \(\cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = 0 + 1 = 1\). For \(\alpha = \frac{3\pi}{2}\), \(\cos\left(\frac{3\pi}{2}\right) + \sin\left(\frac{3\pi}{2}\right) = 0 - 1 = -1\), which does not match. Only \(\alpha = 0\) and \(\alpha = \frac{\pi}{2}\) are valid solutions.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental in solving equations involving angles and sides of triangles. They serve as tools to simplify complex trigonometric expressions.
One of the key identities used in this problem is the Pythagorean identity, which states:

• \( an^2 \alpha + 1 = \sec^2 \alpha\)
• \(1 + \cot^2 \alpha = \csc^2 \alpha\)
• Most importantly for us, \(\sin^2 \alpha + \cos^2 \alpha = 1\)

Using these identities, we can rewrite one trigonometric function in terms of another. This simplifies the process of solving the equation by expressing all terms in a common trigonometric function.
In our example, knowing that \(\sin^2 \alpha + \cos^2 \alpha = 1\) helped transform the equation into a solvable form.

Quadratic Trigonometric Equations

Quadratic trigonometric equations, like any quadratic equation, can often be factored or solved by substitution. In our case, squaring both sides of the original equation transformed it into a quadratic one. Here's the process we followed:

• The equation was initially assumed as \((\sin \alpha)^2 = (1 - \cos \alpha)^2\).
• By using the identity \(\sin^2 \alpha = 1 - \cos^2 \alpha\), we replaced and expanded the square \((1 - \cos \alpha)^2\).
• After expansion, it resulted in a quadratic form \(2\cos^2 \alpha - 2\cos \alpha = 0\).

This approach helps resolve equations where trigonometric functions are squared. Factoring becomes the next step, which reveals potential solutions for \(\alpha\).

Interval Solutions

Interval solutions focus on finding specific angles within a given range \([0, 2\pi)\). This requires evaluating potential solutions of the equation, then narrowing them down to those in the specified domain of angles.

• In our solution, we derived two conditions: \(2\cos \alpha = 0\) and \(\cos \alpha - 1 = 0\).
• Solving the first equation, \(\cos \alpha = 0\), gives us angles \(\alpha = \frac{\pi}{2}\) and \(\alpha = \frac{3\pi}{2}\).
• For \(\cos \alpha = 1\), we find \(\alpha = 0\).
• By checking these solutions against the original equation, only \(\alpha = 0\) and \(\alpha = \frac{\pi}{2}\) prove to be correct.

Careful checking ensures that solutions fit both the mathematical conditions and the problem's domain.

Pythagorean Identity

The Pythagorean identity \(\sin^2 \alpha + \cos^2 \alpha = 1\) is a cornerstone in trigonometry. This relationship is derived from the Pythagorean theorem applied to the unit circle, where any point is given by coordinates \((\cos \alpha, \sin \alpha)\) on a circle with a radius of 1.
In our context, it allowed simplification by converting one squared function into another form, which then led to the quadratic equation. Employing this identity simplifies the equation-solving process considerably. It helps in reducing the degree of terms and connects various trigonometric expressions directly. This identity is frequently utilized in problems involving trigonometric equations, making it an essential concept to master.