Problem 56 - Algebra and Trigonometry

Question

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

a. \( \sin \frac{\alpha}{2} = 2 \) b. \( \cos \frac{\alpha}{2} = - \frac{\sqrt{2}}{2} \) c. \( \tan \frac{\alpha}{2} = -8 \)

Step 1: Evaluate \( \tan \alpha \)

From the exercise, \( \tan \alpha = \frac{8}{15} \). This implies that the side opposite to angle \( \alpha \) is 8 and the adjacent side is 15. Since \( 180^{\circ}<\alpha<270^{\circ} \), α is in the third quadrant where both sine and cosine are negative. Using Pythagoras theorem to find the hypotenuse \( r \), we get \( r = \sqrt{(8)^2 + (-15)^2} = 17 \)

Step 2: Evaluate \( \sin \alpha \) and \( \cos \alpha \)

We use the definition of sine and cosine in terms of the sides of a right triangle. Therefore, \( \sin \alpha = \frac{opposite}{hypotenuse} = \frac{8}{17} \) and \( \cos \alpha = \frac{adjacent}{hypotenuse} = \frac{-15}{17} \)

Step 3: Evaluate \( \sin \frac{\alpha}{2} \)

Using the half-angle identity for sine, we have \( \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha} {2}} = \sqrt{\frac{1 - (-15/17)} {2}} = \frac{\sqrt{16}}{2} = \frac{4}{2} = 2 \)

Step 4: Evaluate \( \cos \frac{\alpha}{2} \)

Using the half-angle identity for cosine, we have \( \cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha} {2}} = \sqrt{\frac{1 + (-15/17)} {2}} = - \frac{\sqrt{2}}{2} \) (negative because cosine has to be negative for \( \alpha \) in quadrant 2)

Step 5: Evaluate \( \tan \frac{\alpha}{2} \)

Using the formula \( \tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha} \) we get \( \tan \frac{\alpha}{2} = \frac{8/17}{1 - 15/17} = -8 \)

Key Concepts

Trigonometric Functions

Understanding trigonometric functions is essential for solving many problems in mathematics and physics. These functions relate the angles of a triangle to the ratios of its sides. We often deal with sine (sin), cosine (cos), and tangent (tan), each representing a specific relationship within a right triangle.

In the context of the given exercise, the tangent of angle \( \alpha \) is defined as the ratio of the side opposite to angle \( \alpha \) to the side adjacent to \( \alpha \) in a right-angled triangle. For our case, \( \tan \alpha = \frac{8}{15} \). Learning these functions and their relationships through the unit circle and their corresponding half-angle identities opens up a world of possibilities for solving equations and understanding motion and waves.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, this is written as \( a^2 + b^2 = c^2 \), with \(c\) being the hypotenuse.

In our exercise, we applied the Pythagorean theorem to find the hypotenuse \(r\) after knowing the lengths of the other two sides from the tangent ratio. This theorem is crucial not only in mathematics but also in various fields such as physics, engineering, and even in everyday problem-solving tasks. It is also closely linked with the concept of the unit circle in trigonometry.

Unit Circle

The unit circle is a powerful tool in trigonometry. It is a circle with a radius of one, centered at the origin of a coordinate plane. One of the unit circle's primary uses is to define the trigonometric functions for all angles, even those larger than \(90^\circ\), which is not possible in a right triangle.

Each point on the unit circle has coordinates \( (\cos \theta, \sin \theta) \) which correspond to the cosine and sine of the angle \( \theta \) formed by the radius with the positive x-axis. Using the unit circle, we can understand why, for example, the cosine function is negative in the second and third quadrants: the coordinates of points in those quadrants have a negative x-value. Our exercise takes advantage of this, determining the sign of trigonometric functions based on which quadrant angle \( \alpha \) lies in.