Problem 41 - Precalculus Mathematics for Calculus

Question

Find \(\sin \frac{x}{2}, \cos \frac{x}{2},\) and \(\tan \frac{x}{2}\) from the given information. $$\text { sec } x=\frac{3}{2}, \quad 270^{\circ}

Step-by-Step Solution

Answer

\(\sin \frac{x}{2} = \frac{1}{\sqrt{6}}, \cos \frac{x}{2} = -\frac{\sqrt{5}}{\sqrt{6}}, \tan \frac{x}{2} = -\frac{1}{\sqrt{5}}.\)

Step 1: Determine the quadrant

Since \(270^\circ < x < 360^\circ\), angle \(x\) lies in the fourth quadrant. In this quadrant, the sine function is negative, the cosine function is positive, and the tangent function is negative.

Step 2: Convert Secant to Cosine

The given information is \(\sec x = \frac{3}{2}\). Recall that \(\sec x = \frac{1}{\cos x}\). Therefore, \(\cos x = \frac{2}{3}\). Since \(x\) is in the fourth quadrant, \(\cos x\) is indeed positive, matching the value found.

Step 3: Use Pythagorean Identity to Find Sine

Use the identity \(\sin^2 x + \cos^2 x = 1\) to find \(\sin x\). Substitute \(\cos x = \frac{2}{3}\): \[\sin^2 x + \left(\frac{2}{3}\right)^2 = 1 \\sin^2 x + \frac{4}{9} = 1 \\sin^2 x = 1 - \frac{4}{9} = \frac{5}{9} \\sin x = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}\]The sine is negative because \(x\) is in the fourth quadrant.

Step 4: Use Half Angle Formulas

To find \(\sin \frac{x}{2}\), \(\cos \frac{x}{2}\), and \(\tan \frac{x}{2}\), we use the half-angle identities:For \(\sin \frac{x}{2}\):\[\sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} = \pm \sqrt{\frac{1 - \frac{2}{3}}{2}} = \pm \sqrt{\frac{1}{6}} = \pm \frac{1}{\sqrt{6}}\]For \(\cos \frac{x}{2}\):\[\cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} = \pm \sqrt{\frac{1 + \frac{2}{3}}{2}} = \pm \sqrt{\frac{5}{6}} = \pm \frac{\sqrt{5}}{\sqrt{6}}\]For \(\tan \frac{x}{2}\):\[\tan \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \pm \sqrt{\frac{1 - \frac{2}{3}}{1 + \frac{2}{3}}} = \pm \sqrt{\frac{1}{5}} = \pm \frac{1}{\sqrt{5}}\]Since \(x\) is in the fourth quadrant, \(x/2\) is in the second quadrant, where sine is positive, cosine is negative, and tangent is negative. Apply appropriate signs.

Step 5: Assign Final Values

\(\sin \frac{x}{2} = \frac{1}{\sqrt{6}}, \cos \frac{x}{2} = -\frac{\sqrt{5}}{\sqrt{6}}, \tan \frac{x}{2} = -\frac{1}{\sqrt{5}}.\) These are the final values, with signs determined based on the quadrant of \(\frac{x}{2}\).

Key Concepts

Half-Angle Formulas

The half-angle formulas are fundamental in trigonometry for finding the sine, cosine, and tangent of half of a given angle. These formulas are particularly useful when you lack direct access to values for these functions at non-standard angles. Here’s a quick dive into these formulas:

• For determining \(\sin \frac{x}{2}\), use the identity: \[ \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \]
• For \(\cos \frac{x}{2}\), the formula becomes: \[ \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \]
• And for \(\tan \frac{x}{2}\), we use: \[ \tan \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} \]

All these formulas involve the cosine of the original angle, \(x\). This makes it essential to first determine \(\cos x\) accurately. The choice of sign (positive or negative) is equally critical. This choice depends on the quadrant that \(\frac{x}{2}\) lies in. As \(x\) from the exercise was between \(270^\circ < x < 360^\circ\), \(\frac{x}{2}\) falls in the second quadrant. Hence, we assign positive signs for sine, and negative signs for cosine and tangent in such scenarios.

Pythagorean Identity

The Pythagorean identity is one of the core trigonometric identities, expressing the fundamental relationship between the sine and cosine of an angle. It's expressed as:
\[ \sin^2 x + \cos^2 x = 1 \]
In practical terms, this identity allows you to find one trigonometric function value if you have the other. For example, in our exercise, you begin by knowing \(\cos x = \frac{2}{3}\). Applying the Pythagorean identity helps to find \(\sin x\) via:
\[ \sin^2 x = 1 - \cos^2 x = 1 - \left(\frac{2}{3}\right)^2 = \frac{5}{9} \]
This gives us \(\sin x = -\frac{\sqrt{5}}{3}\). Note the negative sign, which is the result of determining that \(x\) is located in the fourth quadrant, where sine is negative. Thus, understanding and applying the Pythagorean identity in this context allows you to complete the information needed to use half-angle formulas later.

Quadrant Analysis

In trigonometry, analyzing which quadrant an angle lies in is pivotal for determining the signs of trigonometric function values. The four quadrants of a circle are divided, each affecting the signs differently.

• First Quadrant: All trigonometric functions are positive.
• Second Quadrant: Sine is positive; cosine and tangent are negative.
• Third Quadrant: Tangent is positive; sine and cosine are negative.
• Fourth Quadrant: Cosine is positive; sine and tangent are negative.

For our exercise, determining that \(x\) is in the fourth quadrant (\(270^\circ < x < 360^\circ\)) affected the sign of \(\sin x\) and verified the sign of \(\cos x\). Later, assessing that \(\frac{x}{2}\) is in the second quadrant allowed us to choose the correct signs for the half-angle values. Understanding which quadrant you're working with is crucial for avoiding mistakes in sign assignment, ensuring accurate calculations throughout trigonometric evaluations.