Problem 38 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Explain how to use the figure to solve the equation \(\sin \left(x+\frac{\pi}{4}\right)-\sin \left(\frac{\pi}{4}-x\right)=0\) for \(0 \leq x<2 \pi\).

Step-by-Step Solution

Answer

The solution to the equation \( \sin \left(x+\frac{\pi}{4}\right)-\sin \left(\frac{\pi}{4}-x\right)=0\) for the interval \(0 \leq x<2\pi\) is \( x= \frac {3\pi} {2}\).

Step 1: Apply trigonometric identity

Start by using the identity \( \sin(a) = \sin(\pi - a)\). This will simplify the equation, yielding \( \sin \left(x+\frac {\pi} {4}\right)= \sin \left(\pi - \left( \frac {\pi} {4}-x \right) \right) = \sin \left(\frac {3\pi} {4} + x\right)\).

Step 2: Adjust the interval

Make use of the property of the sine function being periodic with period \(2\pi\). That is, \( \sin(x+\pi) = \sin(x)\). This gives: \( \sin \left(x+\frac {\pi} {4}\right) = \sin \left(x+\frac {3\pi} {4}\right)\).

Step 3: Set up the equation

As sine is a one-to-one function on the interval \(-\pi/2 \leq x \leq \pi/2\), the argument of the sine functions must be equal in the interval. Let's set up the equation: \( x+\frac {\pi} {4}=x+\frac {3\pi} {4}\).

Step 4: Solve for x

Solving for x gives \( x= -\frac {\pi} {2}\). However, this answer is not in the given interval \(0 \leq x<2 \pi\).

Step 5: Adjust for periodicity of sine function

Repeat the values of \( x\) in the interval \(0 \leq x<2\pi\) using that \(\sin (x+2\pi) = \sin(x)\), resulting in \( x= -\frac {\pi} {2} + 2\pi= \frac {3\pi} {2}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in solving trigonometric equations. These are equations containing trigonometric functions like sine, cosine, and tangent. Identities can simplify complex equations by transforming them into a more manageable form. For example, one of the most commonly used identities is the Pythagorean identity, which relates the sine and cosine functions; \( \text{sin}^2(x) + \text{cos}^2(x) = 1 \).

Another crucial identity used to solve the provided exercise is \( \text{sin}(a) = \text{sin}(\pi - a) \). This identity is based on the sine function's symmetry with respect to the y-axis on the unit circle. By applying this identity, a complex-looking equation can be simplified significantly, allowing us to compare the arguments of sine functions directly. It’s important for students to familiarize themselves with these identities, as they not only aid in solving equations but also deepen the understanding of the trigonometric functions’ properties.

Sine Function Periodicity

The sine function is known for its periodic behavior, meaning it repeats its values at regular intervals. Specifically, the sine function has a period of \(2\pi\), which indicates that \(\text{sin}(x) = \text{sin}(x + n \cdot 2\pi)\), where \(n\) is an integer. This periodicity is pivotal when solving trigonometric equations over specific intervals, as it allows us to find all possible solutions within a given range.

In the context of the exercise, this periodic behavior helped to adjust for solutions that initially fell outside the interval \(0 \leq x < 2\pi\). By adding \(2\pi\) to the solution that was not within this interval, we were able to ‘shift’ it into the desired range. Recognizing and utilizing the periodic nature of the sine function is fundamental for solving trigonometric equations effectively.

Trigonometric Equation Interval Solutions

When dealing with trigonometric equations, it's essential to consider the interval in which we seek solutions. For the sine function, any interval of length \(2\pi\) contains all possible values the function can take. In our exercise, we are interested in solutions for \(x\) within the interval \(0 \leq x < 2\pi\).

We can use a one-to-one relationship between the sine values and their angles in the interval \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \) to determine that for \(\text{sin}(\alpha) = \text{sin}(\beta)\), \(\alpha\) and \(\beta\) must be identical if both are within this specific range. However, because the sine function is periodic, we often find multiple intervals where \(\text{sin}(\alpha)\) can equal \(\text{sin}(\beta)\), and we must find the \(\alpha\) that lies in our given interval, just like we did in Step 5 of the solution. Understanding how to handle the interval solutions makes the process of solving trigonometric equations more systematic and efficient.