Problem 47 - A Graphical Approach to Precalculus with Limits

Question

$$\tan \frac{x}{2}, \text { given } \tan x=-\frac{\sqrt{5}}{2} \text { and } \frac{\pi}{2}

Step-by-Step Solution

Answer

\( \tan \frac{x}{2} = \sqrt{3} \) and \( \sin \frac{x}{2} = \sqrt{\frac{\sqrt{5} - 1}{2\sqrt{5}}} \).

Step 1: Understanding the Problem for \( \tan \frac{x}{2} \)

We need to find \( \tan \frac{x}{2} \) given \( \tan x = -\frac{\sqrt{5}}{2} \). Since \( \frac{\pi}{2} < x < \pi \), \( x \) is in the second quadrant where tangent is negative, which is consistent with \( \tan x = -\frac{\sqrt{5}}{2} \).

Step 2: Use the Half-Angle Formula for Tangent

Use the half-angle formula for tangent: \( \tan \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} \). We need to find \( \cos x \) using \( \tan x = -\frac{\sqrt{5}}{2} \).

Step 3: Calculate \( \cos x \) and \( \sin x \)

Recall that \( \tan x = \frac{\sin x}{\cos x} = -\frac{\sqrt{5}}{2} \). Assign \( \sin x = \sqrt{5} \) and \( \cos x = -2 \) (since \( x \) is in the second quadrant where \( \cos x < 0 \)). Using \( \sin^2 x + \cos^2 x = 1 \), solve for \( \cos x = -\frac{2}{\sqrt{9}} = -\frac{2}{3} \).

Step 4: Solve for \( \tan \frac{x}{2} \)

With \( \cos x = -\frac{2}{3} \), plug into the half-angle identity for tangent: \( \tan \frac{x}{2} = \pm \sqrt{\frac{1 + \frac{2}{3}}{1 - \frac{2}{3}}} = \pm \sqrt{3} \). Since \( \frac{\pi}{2} < x < \pi \), \( \frac{x}{2} \) is between \( \frac{\pi}{4} \) and \( \frac{\pi}{2} \), meaning \( \tan \frac{x}{2} > 0 \). Thus, \( \tan \frac{x}{2} = \sqrt{3} \).

Step 5: Understanding the Problem for \( \sin \frac{x}{2} \)

We need to find \( \sin \frac{x}{2} \) given \( \tan x = 2 \) and \( 0 < x < \frac{\pi}{2} \). In the first quadrant, all trigonometric functions are positive.

Step 6: Use the Half-Angle Formula for Sine

The formula to find \( \sin \frac{x}{2} \) is \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \). We first find \( \cos x \) using \( \tan x = 2 \).

Step 7: Calculate \( \cos x \) and \( \sin x \) for \( \tan x = 2 \)

Given \( \tan x = 2 \), assign \( \sin x = 2k \) and \( \cos x = k \). Then \( (2k)^2 + k^2 = 1 \) leads to \( 5k^2 = 1 \) or \( k = \frac{1}{\sqrt{5}} \). Thus, \( \sin x = \frac{2}{\sqrt{5}} \) and \( \cos x = \frac{1}{\sqrt{5}} \).

Step 8: Solve for \( \sin \frac{x}{2} \)

Using the formula \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \frac{1}{\sqrt{5}}}{2}} \), simplify: \( \sin \frac{x}{2} = \sqrt{\frac{\sqrt{5} - 1}{2\sqrt{5}}} \). Since \( 0 < x < \frac{\pi}{2} \), \( \frac{x}{2} \) is also in the first quadrant, so we take the positive root. Thus, \( \sin \frac{x}{2} = \sqrt{\frac{\sqrt{5} - 1}{2\sqrt{5}}} \).

Key Concepts

Tangent

The trigonometric function known as tangent is often denoted by \( \tan \). It is one of the core functions used in trigonometry, and it provides the ratio between the opposite side and the adjacent side of a right triangle. Mathematically, \( \tan x = \frac{\sin x}{\cos x} \).

• The tangent function's value depends on the values of sine and cosine.
• In different quadrants, the tangent can be positive or negative.
• For example, in the second quadrant, the tangent is negative because the sine is positive while the cosine is negative.

Understanding the behavior of the tangent function across the quadrants is crucial for solving problems related to its half-angles.

Sine

Sine is another essential trigonometric function, defined as \( \sin x \), which is the ratio of the length of the opposite side of a right triangle to its hypotenuse. It's important to understand how sine behaves in different quadrants.

• In quadrants 1 and 2, sine is positive, while in quadrants 3 and 4, sine is negative.
• Sine, along with cosine, is fundamental in forming the equation \( \sin^2 x + \cos^2 x = 1 \).
• This equation helps find either sine or cosine if the other is known.

In the context of the half-angle formulas, sine values determine the final sign that should be used when solving equations involving half-angles.

Half-Angle Formulas

In trigonometry, half-angle formulas are useful identities for finding the sine, cosine, and tangent of half-angles, like \( \frac{x}{2} \). These formulas are derived from double angle formulas and can greatly simplify calculations.

• For tangent: \( \tan \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} \).
• For sine: \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \).
• The sign before the square root is determined by the angle's quadrant — where the half-angle lies.

These formulas are particularly helpful when solving for angles not present on common trigonometric tables, allowing deeper analysis and understanding of the behavior of trigonometric functions.

Quadrants

Angles in trigonometry are often measured in terms of quadrants, which are the four sections of a coordinate plane separated by the x and y axes.

• Quadrant 1 (0 to \( \frac{\pi}{2} \)): All trigonometric functions (sine, cosine, tangent) are positive.
• Quadrant 2 (\( \frac{\pi}{2} \) to \( \pi \)): Sine is positive, cosine and tangent are negative.
• Quadrant 3 (\( \pi \) to \( \frac{3\pi}{2} \)): Tangent is positive, sine and cosine are negative.
• Quadrant 4 (\( \frac{3\pi}{2} \) to 2\( \pi \)): Cosine is positive, sine and tangent are negative.

Understanding which quadrant an angle resides in helps determine the sign and value of trigonometric functions involved in calculations, like in half-angle problems.