Problem 80 - Algebra and Trigonometry

Question

Use an identity to solve each equation on the interval \([0,2 \pi)\) $$ \sin \left(x+\frac{\pi}{3}\right)+\sin \left(x-\frac{\pi}{3}\right)=1 $$

Step-by-Step Solution

Answer

Hence, the solutions for \(x\) within the interval \([0,2 \pi)\) for the given equation are \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\), \(\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\).

Step 1: Use Trigonometric Identity

Employing the addition and subtraction identities for sine, \(sin(A + B) = sinAcosB + cosAsinB\) and \(sin(A - B) = sinAcosB - cosAsinB\), the equation can be rewritten: \[\begin{align*}sin(x)cos\left(\frac{\pi}{3}\right) + cos(x)sin\left(\frac{\pi}{3}\right) + sin(x)cos\left(\frac{\pi}{3}\right) - cos(x)sin\left(\frac{\pi}{3}\right) = 1\end{align*}\] This simplifies to \[\begin{align*}2sin(x)cos\left(\frac{\pi}{3}\right) = 1\end{align*}\]

Step 2: Employ Double Angle Identity and Solve

Use the double-angle identity \(2sin(x)cos(B) = sin(2x)\), thus solving for \(x\). The equation now reads:\[\begin{align*}sin(2x) = 1\end{align*}\] This equation means that \(2x = \frac{\pi}{2} + n\pi\) where \(n\) is an integer because \(sin(x)\) equals 1 at \(x = \frac{\pi}{2}\) and the sine value repeats after an interval of \(\pi\). Thus, the general solution becomes \(x = \frac{\pi}{4} + \frac{n\pi}{2}\).

Step 3: Find Values Within Given Interval

Given that \(x\) must be within the interval \([0,2 \pi)\), find the specific solutions by substituting integer values of \(n\) which give values of \(x\) in the required interval. Depending on the values of \(n\), \(x\) becomes \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\), \(\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\).

Key Concepts

Addition and Subtraction Identities

Understanding trigonometric identities is essential when solving equations involving sine and cosine. The addition and subtraction identities allow us to expand expressions such as \( \sin(A + B) \) and \( \sin(A - B) \). These identities are:

• \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
• \( \sin(A - B) = \sin A \cos B - \cos A \sin B \)

For instance, in the given exercise, we applied these identities to rewrite the equation involving \( \sin(x + \frac{\pi}{3}) + \sin(x - \frac{\pi}{3}) \). By using these identities, complex expressions are simplified into more manageable forms, often revealing underlying patterns that assist in solving the equation. This step is crucial as it transforms the original problem into one that can be approached with standard algebraic techniques, enabling us to move forward to find a solution.

Double Angle Identity

The double angle identity is another powerful tool in trigonometry. It relates functions of twice an angle to functions of a single angle. One of the common double angle identities is \( \sin(2x) = 2 \sin(x) \cos(x) \). This identity helps simplify equations by reducing the number of variables and turning them into solvable forms. For example, in our problem, after applying the addition and subtraction identities, we arrived at \( 2 \sin(x) \cos(\frac{\pi}{3}) = 1 \).

Using the Identity

By recognizing \( 2 \sin(x) \cos(\frac{\pi}{3}) \) as a format of \( \sin(2x) \), we can see that our equation becomes \( \sin(2x) = 1 \). This significantly simplifies our work, reducing the complexity of finding solutions. Often, employing the double angle identity can reveal solutions that are otherwise obscured in their original forms.

Sine Function

Sine function plays a major role in trigonometric equations, and understanding its properties is crucial. The sine function, \( \sin(x) \), measures the y-coordinate of a point on the unit circle as it travels from the origin to a point \( (\cos(x), \sin(x)) \).

Characteristics of Sine Function

Some critical features of sine:

• It is periodic with a period of \( 2\pi \), meaning it repeats every \( 2\pi \) radians.
• Its range is from -1 to 1.
• It reaches its maximum value of 1 at \( \frac{\pi}{2} + n\pi \), where \( n \) is an integer.

In the problem, we're asked to solve for \( \sin(2x) = 1 \). Knowing that sine reaches 1 at specific intervals allowed us to set \( 2x = \frac{\pi}{2} + n\pi \) as a premise to find suitable \( x \) values within the defined interval.

Interval Solutions

When solving trigonometric equations, considering the interval is important for finding relevant solutions. In this exercise, we are tasked with finding solutions within \([0, 2\pi)\). Rewriting our equation with the interval in mind, we determine \( x = \frac{\pi}{4} + \frac{n\pi}{2} \), where different values of \( n \) will yield solutions within our defined domain.

Finding Solutions within the Interval

• Set \( n = 0 \), yielding \( x = \frac{\pi}{4} \).
• Set \( n = 1 \), leading to \( x = \frac{3\pi}{4} \).
• Set \( n = 2 \), resulting in \( x = \frac{5\pi}{4} \).
• Set \( n = 3 \), giving \( x = \frac{7\pi}{4} \).

By evaluating these potential solutions, we ensure they fit within the constraints of the interval. It's a vital step, guaranteeing that we don't overlook any viable solutions or include those that fall outside the necessary range.