Problem 56 - Precalculus Essentials

Question

In Exercises \(55-58,\) use the given information to find the exact value of each of the following: a. \(\sin \frac{\alpha}{2}\) b. \(\cos \frac{\alpha}{2}\) c. \(\tan \frac{\alpha}{2}\) $$\tan \alpha=\frac{8}{15}, 180^{\circ} < \alpha < 270^{\circ}$$

Step-by-Step Solution

Answer

\(\sin \frac{\alpha}{2} = \frac{4}{\sqrt{15}}\), \(\cos \frac{\alpha}{2} = -\frac{2}{\sqrt{15}}\), \(\tan \frac{\alpha}{2} = 2\)

Step 1: Determine \(\cos \alpha\)

Using the Pythagorean identity, \(\cos \alpha = \frac{\pm\sqrt{1 - \tan^2 \alpha}}{1}\). Since \(180^{\circ} < \alpha < 270^{\circ}\), \(\cos\alpha\) is negative in the third quadrant. Hence, \(\cos \alpha = -\frac{\sqrt{1-\left(\frac{8}{15}\right)^2}}{1} = -\frac{7}{15}\)

Step 2: Find \(\sin \frac{\alpha}{2}\) and \(\cos \frac{\alpha}{2}\)

Using the half-angle formulas, \(\sin \frac{\alpha}{2} = \pm\sqrt{\frac{1 -\cos \alpha}{2}}\) and \(\cos \frac{\alpha}{2} = \pm\sqrt{\frac{1 +\cos \alpha}{2}}\). As the angle \(\alpha\) is in the third quadrant, the angle \(\frac{\alpha}{2}\) will be in the second quadrant where sine is positive and cosine is negative. So, \(\sin \frac{\alpha}{2} = \sqrt{\frac{1 -(-\frac{7}{15})}{2}} = \frac{4}{\sqrt{15}}\) after simplification. \(\cos \frac{\alpha}{2} = -\sqrt{\frac{1 +(-\frac{7}{15})}{2}} = -\frac{2}{\sqrt{15}}\) after simplification.

Step 3: Find \(\tan \frac{\alpha}{2}\)

\(\tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\). The ratio of \(\sin \frac{\alpha}{2}\) to \(\cos \frac{\alpha}{2}\) is \(-\frac{\frac{4}{\sqrt{15}}}{-\frac{2}{\sqrt{15}}} = 2\).

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is like having a Swiss Army knife for solving trigonometry problems—they are versatile tools that can be applied in various scenarios.

Essentially, trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. For instance, the half-angle formulas used in the exercise are part of this extensive set of identities. These particular identities enable us to express the sine, cosine, and tangent of half an angle in terms of the sine and cosine of the original angle.

There are numerous trigonometric identities, but some of the most commonly used include the reciprocal identities, Pythagorean identities, addition and subtraction formulas, double angle formulas, and of course, the half-angle formulas. Mastery of these identities allows you to simplify complex trigonometric expressions and solve trigonometric equations with greater ease.

Pythagorean Identity

At the heart of trigonometry lies the Pythagorean identity, an expression derived from the Pythagorean theorem that relates to the length of the sides of a right triangle. The most basic form of the Pythagorean identity is \[ \sin^2 x + \cos^2 x = 1 \].

This fundamental identity asserts that the square of the sine of an angle plus the square of the cosine of that same angle is always equal to one. In the step-by-step solution, using the given value of \( \tan \alpha \), the Pythagorean identity was employed to find \( \cos \alpha \) — a critical step for subsequently applying the half-angle formulas. By expanding our understanding of these identities, we can solve for unknown quantities in trigonometry problems and validate the relationships between different trigonometric functions.

Unit Circle

The unit circle is a fundamental concept in trigonometry that provides a geometric interpretation of trigonometric functions. It's a circle with a radius of one, centered at the origin of a coordinate plane.

Each point on the unit circle corresponds to an angle, with its coordinates representing the cosine and sine of that angle (\[ (\cos \theta, \sin \theta) \]). This relationship is crucial for visual learners as it allows for the visualization of how sine and cosine values are determined.

Moreover, the unit circle helps us understand why certain trigonometric functions are positive or negative in different quadrants. For example, in the exercise given, the angle \( \alpha \) lies in the third quadrant, where both sine and cosine values are negative which correlates with the negative result of \( \cos \alpha \) obtained through application of Pythagorean identity.