Problem 42 - 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. $$ \sin x+\cos x=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 135° + 180°k\) and \(x = 315° + 180°k\) where \(k\) is an integer.

Step 1: Rewrite the Trigonometric Equation

Start by rewriting the equation, \[ sin x + cos x = 0 \] To solve it, rearrange the terms: \[ sin x = -cos x \] This implies that the sine of the angle is the negative cosine of the angle.

Step 2: Analyze the Sine and Cosine Relationship

By recognizing that \( \sin x = - \cos x \), we can use the sine and cosine identity that tells us these functions are equal with a phase shift. Notice that:\[ \sin x = \cos \left(90° - x\right) \] but we want it to be \(- \cos\), so:\[ \sin x = \cos (180° + x) \] Also, the equation holds true when they are negatives of each other, such as in: \[ \sin x = \cos (90° + x) \] Thus, we need to use the property that if \(\sin x = -\cos x\), then:\[ x = 135° + 180°k \] or \[ x = 315° + 180°k \] where \(k\) is an integer.

Step 3: Derive General Solution

Based on the above analyses, our general solutions in degrees for the equation \( \sin x + \cos x = 0 \) are given by the formulas:\[ x = 135° + 180°k \] and \[ x = 315° + 180°k \] where \(k\) is any integer. These solutions account for the periodic nature of trigonometric functions.

Step 4: Verify the Solutions

It's important to verify our derived expressions actually satisfy the equation for any integer \(k\). Consider:\[ sin(135°) + cos(135°) = \frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) = 0 \] \[ sin(315°) + cos(315°) = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 0 \] Therefore, both solutions \(x = 135° + 180°k\) and \(x = 315° + 180°k\) satisfy the original equation for any integer \(k\).

Key Concepts

Sine and Cosine Relationship

The relationship between the sine and cosine functions is a fundamental aspect of trigonometry. These functions are closely connected as they represent projections of a point on a unit circle onto the y-axis and x-axis, respectively. Mathematically, this connection is described through identities like \(\sin x = \cos (90° - x)\) and \(\cos x = \sin (90° - x)\).
In this particular exercise, we explore a scenario where \(\sin x = -\cos x\). This means that the sine function is equal in magnitude but opposite in sign to the cosine function. Such conditions intimate symmetry around the origin of the unit circle, and inspire us to express this relationship using trigonometric identities that involve phase shifts. Therefore, understanding how these functions transition into one another through a phase shift, particularly at 90 degrees intervals, is key to solving trigonometric equations.

General Solutions

General solutions in trigonometry offer a way to express all potential solutions for an equation. When solving an equation like \(\sin x + \cos x = 0\), finding specific angles \(x\) that satisfy this equation helps to determine how these angles recur over a full cycle of the unit circle.
The solutions derived here, \(x = 135° + 180°k\) and \(x = 315° + 180°k\), are expressed to cover all possibilities by including the entire set of integer values for \(k\).

• \(135°\) and \(315°\) are specific angles where the sine and cosine functions achieve the condition \(\sin x = -\cos x\).
• By adding \(180°k\), where \(k\) is any integer, we account for the periodic repetition of these conditions due to the characteristics of the functions, covering infinitely many solutions.

This process of expressing solutions acknowledges the cyclical nature of trigonometric functions, guaranteeing that no potential solution is overlooked.

Periodic Nature of Trigonometric Functions

Trigonometric functions like sine and cosine are inherently periodic, meaning they repeat their values in regular intervals. This periodicity is a crucial characteristic that underpins the behavior of these functions in many practical applications.
For the functions sine and cosine, this periodic interval is typically \(360°\) or \(2\pi\) radians. However, when dealing with derived relationships, as in this case where \(\sin x = -\cos x\), the recurrence can manifest at half the usual interval. Hence, the pattern repeats every \(180°\), encompassing a half-turn of the unit circle.

• Each solution to a trigonometric equation like \(\sin x + \cos x = 0\) comes in pairs of angles that are \(180°\) apart, hence the solutions \(x = 135° + 180°k\) and \(x = 315° + 180°k\).
• Recognizing this property allows us to generalize solutions efficiently, capturing all instances where the equation holds true.

This periodic nature simplifies what could otherwise be complex computations, thereby aiding in solving numerous problems across mathematics and engineering efficiently.