Problem 29 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Solve the equation for \(0 \leq x<2 \pi\). \(\cos \left(x+\frac{\pi}{6}\right)-\cos \left(x-\frac{\pi}{6}\right)=1\)

Step-by-Step Solution

Answer

The solution to the equation \(\cos \left(x+\frac{\pi}{6}\right)-\cos \left(x-\frac{\pi}{6}\right)=1\) within the interval \(0 \leq x < 2 \pi\) is \(x = \frac{3\pi}{2}\).

Step 1: Apply Trigonometric Identity

We use the trigonometric identity: \(\cos(A) - \cos(B) = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)\). After applying the identity, the equation can be rewritten as: \(-2 \sin \left(x\right) \sin \left(\frac{\pi}{6}\right) = 1\).

Step 2: Simplify the Equation

The value of \(\sin(\frac{\pi}{6})\) is known to be \(\frac{1}{2}\). By substituting this value into the equation, we get \( -2 \sin \left(x\right) * \frac{1}{2} = 1\). This simplifies to: \(- \sin(x)=1\).

Step 3: Solve for x

To find 'x', we take the arcsine (inverse sine) on both sides of the equation. This yields: \(x = \sin^{-1}(-1)\). The arcsine function tells us the angle whose sine is -1 is -\(\frac{\pi}{2}\). However, this value is not within the specified interval. So, we add \(2\pi\) (a full circle) to -\(\frac{\pi}{2}\), to bring the angle to the first equivalent positive angle within our stated interval. The solution therefore is \(x = \frac{3\pi}{2}\).

Step 4: Confirm the Solution lies within the Interval

Finally, confirm that the solution \(x = \frac{3\pi}{2}\) lies within the interval \(0 \leq x < 2 \pi\), which it does. This is therefore the valid solution to the equation within the given range.

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is key to solving many trigonometric equations. Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They help simplify complex trigonometric expressions into more manageable forms.

For example, in the given problem, the identity used is:

• \( \cos(A) - \cos(B) = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right) \).

This identity allows us to transform the original expression \( \cos \left(x+\frac{\pi}{6}\right) - \cos \left(x-\frac{\pi}{6}\right) \) into a product of sines, helping us isolate and solve for \( x \).

Mastering these identities is essential for solving equations efficiently and accurately, as they simplify the process significantly.

Arcsine Function

The arcsine function, often denoted as \( \sin^{-1} \), is the inverse of the sine function. It answers the question: what angle has a given sine value? However, arcsine only returns values within its principal range, which is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

In our problem, after simplifying the equation to \( -\sin(x) = 1 \), we need to find the angle \( x \) where the sine is \(-1\). The arcsine of \(-1\) gives us \(-\frac{\pi}{2}\), but this isn't in the given interval \([0, 2\pi)\).

• To adjust, we add \(2\pi\), converting it to the equivalent positive angle \( \frac{3\pi}{2} \).

This usage of the arcsine function and adjusting for the interval ensures we solve the equation correctly within the given range.

Interval Notation

Interval notation is a way of describing a set of numbers along an interval on a line. It's crucial for ensuring that our solutions are within the provided boundaries. In this exercise, our interval is \(0 \leq x < 2\pi\).

• The square bracket "[" means that the endpoint is included, while ")" means it's not.
• So \([0, 2\pi)\) includes 0 and all numbers up to, but not including, \(2\pi\).

When solving trigonometric equations, the interval helps us determine the correct angle solutions. In the exercise, we confirmed that \( x = \frac{3\pi}{2} \) correctly lies within this interval.

Knowing how to use interval notation ensures that your solutions make sense in the context of the problem.