Problem 33 - Precalculus : Building Concepts and Connections

Question

Find the exact values of all the trigonometric functions for the giocn calues of \(t .\) If a certain value is undefined, state sa Do not use a calculator. $$I=\frac{13 \pi}{6}$$

Step-by-Step Solution

Answer

The exact values of all trigonometric functions for \(t=\frac{13\pi}{6}\) are: \(\sin(t) = \frac{1}{2}\), \(\cos(t) = \frac{\sqrt{3}}{2}\), \(\tan(t) = \frac{1}{\sqrt{3}}\), \(\csc(t) = 2\), \(\sec(t) = \frac{2}{\sqrt{3}}\), and \(\cot(t) = \sqrt{3}\).

Step 1: Angle Reduction

First, reduce the angle to fall within one complete revolution [0, \(2\pi\)). We can subtract \(2\pi\) (which is equivalent to a full revolution in radians) from \(t\) until it falls within this range. So, \(t = \frac{13\pi}{6}-2\pi = \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6}\).

Step 2: Calculate Sine, Cosine and Tangent

Now, calculate the sine, cosine and tangent of the new angle using the unit circle. From the unit circle, we know that: \(\sin(\frac{\pi}{6}) = \frac{1}{2}\), \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\), and \(\tan(\frac{\pi}{6}) = \frac{\sin(\frac{\pi}{6}) }{ \cos(\frac{\pi}{6})} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\).

Step 3: Calculate Cosecant, Secant and Cotangent

Lastly, calculate the reciprocal trigonometric functions using the results from step 2: \(\csc(\frac{\pi}{6}) = \frac{1}{\sin(\frac{\pi}{6})} = 2\), \(\sec(\frac{\pi}{6}) = \frac{1}{\cos(\frac{\pi}{6})} = \frac{2}{\sqrt{3}}\), and \(\cot(\frac{\pi}{6}) = \frac{1}{\tan(\frac{\pi}{6})} = \sqrt{3}\).

Key Concepts

Angle Reduction

Angle reduction is an essential part of trigonometry, as it helps us simplify complex problems by bringing high-angle values into a familiar range, usually between 0 and \(2\pi\). This process involves adjusting an angle so it fits within a single full revolution of the circle.

To do this, we subtract integer multiples of \(2\pi\) (a circle's full rotation equivalent in radians) from the angle. For example, with \(t = \frac{13\pi}{6}\), we need to reduce this by a full revolution, so:

• Subtract \(2\pi\), which can be written as \(\frac{12\pi}{6}\) to match the denominator.
• The reduced angle becomes \(\frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6}\).

This calculation places the angle within the standard interval, making it easier to find the trigonometric functions' exact values.

Unit Circle

The unit circle is a fundamental concept in trigonometry. It is a circle with a radius of one unit centered at the origin, on a coordinate system, helping us to visually understand the sine, cosine, and tangent of angles. Here's how it works:

• Any point on the unit circle can be represented as \((\cos \theta, \sin \theta)\), where \(\theta\) is the angle from the positive x-axis.
• For example, at \(\theta = \frac{\pi}{6}\), the coordinates are \((\frac{\sqrt{3}}{2}, \frac{1}{2})\).

These coordinates directly give you the cosine and sine of the angle, providing an intuitive way of finding exact trigonometric values. Furthermore, using these coordinates, the tangent of \(\theta\) can be calculated using the formula: \( \tan \theta = \frac{\sin \theta}{\cos \theta}\). This method eliminates the need for a calculator and emphasizes the visual and logical understanding of angles and their functions.

Exact Values

Exact trigonometric values are critical in mathematics, enabling precise calculations without approximation. They are derived from known triangles or the unit circle and are especially useful for certain standard angles like \(\frac{\pi}{6}, \frac{\pi}{4},\) and \(\frac{\pi}{3}\).

• For \(\frac{\pi}{6}\), the exact values are: \(\sin(\frac{\pi}{6}) = \frac{1}{2}\) and \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\).
• The tangent is calculated with \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \).

With these values, you can also find the reciprocal functions: the cosecant, secant, and cotangent. They are computed as follows:

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

Understanding how to derive these exact values can greatly enhance problem-solving skills in trigonometry, steering clear from estimations and maintaining precision in calculations.