Problem 57 - 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. \(\tan \frac{\alpha}{2}\) $$\sec \alpha=-\frac{13}{5}, \frac{\pi}{2} < \alpha < \pi$$

Step-by-Step Solution

Answer

The exact values are: a) \(\sin \frac{\alpha}{2} = \frac{2\sqrt{13}}{13}\), b) \(\cos \frac{\alpha}{2} = \frac{3\sqrt{13}}{13}\), c) \(\tan \frac{\alpha}{2} = \frac{2}{3}\)

Step 1: Determine the cosine

Since the secant of an angle is 1 divided by the cosine of the same angle, the cosine of the angle \(\alpha\) can be found by calculating \(\cos \alpha=-\frac{1}{\sec \alpha}=-\frac{1}{-\frac{13}{5}}=\frac{5}{13}\). Remember that in the given interval \(\pi/2 < \alpha < \pi\), cosine is negative, but the secant is already negative.

Step 2: Find the sine

As cosine and sine are connected via the Pythagorean identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we can determine the sine as \(\sin \alpha = \sqrt{1 - (\cos \alpha)^2} = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}\). In the given interval \(\pi/2 < \alpha < \pi\), sine is positive.

Step 3: Calculate the half-angle formulas

Now, apply the half-angle formulas. \n\na. \(\sin \frac{\alpha}{2} = +\sqrt{\frac{1 - \cos \alpha}{2}} = \sqrt{\frac{1 - \frac{5}{13}}{2}} = \sqrt{\frac{8}{26}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}\) (We use the positive square root because \(\frac{\alpha}{2}\) falls in the first or second quadrant where sine is positive.)\n\nb. \(\cos \frac{\alpha}{2} = +\sqrt{\frac{1 + \cos \alpha}{2}} = \sqrt{\frac{1 + \frac{5}{13}}{2}} = \sqrt{\frac{18}{26}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}\) (We use the positive square root because \(\frac{\alpha}{2}\) falls in the first or second quadrant where cosine is positive.)\n\n c. \(\tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} = \frac{\frac{2\sqrt{13}}{13}}{\frac{3\sqrt{13}}{13}} = \frac{2}{3}\)

Key Concepts

Half-Angle Formulas

The half-angle formulas help us find the sine, cosine, and tangent of half of an angle. They are especially useful when dealing with angles in trigonometry that aren't directly measurable. For an angle \( \alpha \), the half-angle formulas are:

• \( \sin \frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos \alpha}{2}} \)
• \( \cos \frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos \alpha}{2}} \)
• \( \tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha} = \frac{1 - \cos \alpha}{\sin \alpha} \)

The sign of the square root depends on the quadrant where \( \frac{\alpha}{2} \) lies. In this exercise, since \( \frac{\alpha}{2} \) is in the first or second quadrant, both sine and cosine are positive. This understanding helps us choose the right sign for the solution.

Pythagorean Identity

The Pythagorean identity is a fundamental relationship in trigonometry that links sine and cosine together:\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]Using this identity, we can find one trigonometric function if we have the other. In this exercise, we were given \( \sec \alpha = -\frac{13}{5} \) and calculated \( \cos \alpha = \frac{5}{13} \). Then, we used the Pythagorean identity to find:\[ \sin \alpha = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13} \]This identity ensures that our trigonometric calculations remain consistent and accurate, forming the basis for further exploration into other trigonometric identities.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any angle. These identities are the tools that allow us to simplify and solve trigonometric equations. Some common identities include:

• Reciprocal identities, such as \( \sec \alpha = \frac{1}{\cos \alpha} \)
• Quotient identities, which involve \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \)

In this particular exercise, we used reciprocal identities to find the cosine from the secant. Recognizing and using these identities can help in solving complex trigonometric problems and finding exact values, especially when combined with specific conditions for the angle.

Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin on a coordinate plane. It's a fundamental tool for understanding trigonometric functions and provides a graphical representation of these identities.On the unit circle, every point \((x, y)\) corresponds to \( (\cos \theta, \sin \theta) \). This designation helps us easily find values of trigonometric functions based on the angle \( \theta \) formed with the positive x-axis.
In our exercise, the interval \( \frac{\pi}{2} < \alpha < \pi \), relates to angles in the second quadrant, where we know:

• Cosine is negative.
• Sine is positive.

The unit circle assists in visually and conceptually understanding why these relationships hold true, aiding in predicting the signs and behaviors of trigonometric functions across different quadrants.