Problem 29 - Algebra 2 and Trigonometry

Question

In \(26-33 :\) a. Rewrite each function value in terms of its cofunction. b. Find the exact value of the function value found in a. $$ \sec \frac{2 \pi}{3} $$

Step-by-Step Solution

Answer

The exact value of \( \sec \frac{2\pi}{3} \) rewritten as its cofunction is \(-2\).

Step 1: Identify the Cofunction Identity

The given function is a secant function. The cofunction identity for secant is the cosecant (csc) identity, which states: \[ \sec(\theta) = \csc\left(\frac{\pi}{2} - \theta\right) \]. We will use this identity to rewrite \( \sec \frac{2\pi}{3} \) in terms of its cofunction.

Step 2: Apply the Cofunction Identity

Substitute the given angle into the cofunction identity: \[ \sec \left(\frac{2\pi}{3}\right) = \csc \left(\frac{\pi}{2} - \frac{2\pi}{3}\right) \]. Simplify the angle inside the cosecant function: \[ \frac{\pi}{2} - \frac{2\pi}{3} = \frac{3\pi}{6} - \frac{4\pi}{6} = -\frac{\pi}{6} \]. Therefore, \[ \sec \left(\frac{2\pi}{3}\right) = \csc\left(-\frac{\pi}{6}\right) \].

Step 3: Simplify the Cofunction Using Even-Odd Properties

The cosecant function is an odd function, meaning \( \csc(-x) = -\csc(x) \). Thus, \[ \csc\left(-\frac{\pi}{6}\right) = -\csc\left(\frac{\pi}{6}\right) \].

Step 4: Find the Exact Value

Using trigonometric tables or known values, \( \csc\left(\frac{\pi}{6}\right) = 2 \). Therefore, \[ -\csc\left(\frac{\pi}{6}\right) = -2 \]. Hence, the exact value of \( \sec\left(\frac{2\pi}{3}\right) \) is \(-2\).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental to understanding angles and their relationships. They relate the angles of a triangle to the lengths of its sides. Here's a quick look at some important trigonometric functions:

• Sine (sin): Opposite side over Hypotenuse
• Cosine (cos): Adjacent side over Hypotenuse
• Tangent (tan): Opposite side over Adjacent side
• Secant (sec): Reciprocal of Cosine
• Cosecant (csc): Reciprocal of Sine
• Cotangent (cot): Reciprocal of Tangent

Secant is particularly interesting because it is defined as the reciprocal of the cosine function:\[\sec(\theta) = \frac{1}{\cos(\theta)}\]It's essential to understand these relationships, as they are the building blocks upon which more advanced concepts, such as identities and equations, are formed. Learning to rewrite a function in terms of another, like secant into cosecant, involves understanding these fundamental properties.

Even-Odd Properties

The even and odd properties of trigonometric functions help simplify expressions and solve equations. Here's what you need to know:

• Even functions: Symmetrical about the y-axis. Example: Cosine, where \(\cos(-x) = \cos(x)\)
• Odd functions: Symmetrical about the origin. Example: Sine and Cosecant, where \(\sin(-x) = -\sin(x)\) and \(\csc(-x) = -\csc(x)\)

For this exercise, knowing that the cosecant function is an odd function is crucial. It tells us that when you take the cosecant of a negative angle, it results in the negative of the cosecant of the positive angle. This property is utilized when simplifying expressions such as \(\csc(-\frac{\pi}{6})\), translating it into \(-\csc(\frac{\pi}{6})\). Understanding these properties helps in solving trigonometric equations and altering function expressions efficiently.

Exact Values

Exact values in trigonometry refer to the precise values of trigonometric functions for specific angles without any approximation. It's often beneficial to memorize or reference these when dealing with common angles, such as \(0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3},\) and \(\frac{\pi}{2}\):

• \(\csc(\frac{\pi}{6}) = 2\)
• \(\csc(\frac{\pi}{4}) = \sqrt{2}\)
• \(\csc(\frac{\pi}{3}) = \frac{2}{\sqrt{3}}\)

Remember, these exact values help provide an accurate solution, unlike decimal approximations, which might not perfectly reflect the theoretical outcomes. In the exercise, using the exact value \(\csc(\frac{\pi}{6}) = 2\) allowed us to conclude that \(\sec(\frac{2\pi}{3}) = -2\) accurately. Mastering these exact values is essential to solving more complex trigonometric problems quickly and effectively.