Problem 39 - Algebra and Trigonometry

Question

35-40 Find \(\sin \frac{x}{2}, \cos \frac{x}{2},\) and \(\tan \frac{x}{2}\) from the given information. \(\sec x=\frac{3}{2}, \quad 270^{\circ} < x < 360^{\circ}\)

Step-by-Step Solution

Answer

\( \sin \frac{x}{2} = \frac{\sqrt{6}}{6}, \cos \frac{x}{2} = -\frac{\sqrt{30}}{6}, \tan \frac{x}{2} = -\frac{\sqrt{5}}{5} \).

Step 1: Identify Secant and Cosine

Given that \( \sec x = \frac{3}{2} \), recall that \( \sec x = \frac{1}{\cos x} \). Therefore, we find \( \cos x = \frac{2}{3} \).

Step 2: Determine the Sign of Cosine

Since the angle \( x \) is in the fourth quadrant \( (270^{\circ} < x < 360^{\circ}) \), in this quadrant, cosine is positive. So, \( \cos x = \frac{2}{3} \).

Step 3: Use the Pythagorean Identity to Find Sine

Using \( \sin^2 x + \cos^2 x = 1 \), substitute \( \cos x = \frac{2}{3} \): \( \sin^2 x + \left(\frac{2}{3}\right)^2 = 1 \). Simplifying gives \( \sin^2 x + \frac{4}{9} = 1 \). \( \sin^2 x = \frac{5}{9} \), so \( \sin x = -\frac{\sqrt{5}}{3} \) since sine is negative in the fourth quadrant.

Step 4: Use Half-Angle Formula for Sine

The half-angle formula for sine is \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \). Substitute \( \cos x = \frac{2}{3} \): \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \frac{2}{3}}{2}} = \pm \sqrt{\frac{1}{6}} = \pm \frac{\sqrt{6}}{6} \).

Step 5: Use Half-Angle Formula for Cosine

The half-angle formula for cosine is \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \). Substitute \( \cos x = \frac{2}{3} \): \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \frac{2}{3}}{2}} = \pm \sqrt{\frac{5}{6}} = \pm \frac{\sqrt{30}}{6} \).

Step 6: Use Half-Angle Formula for Tangent

The half-angle formula for tangent is \( \tan \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} \). Substitute \( \cos x = \frac{2}{3} \): \( \tan \frac{x}{2} = \pm \sqrt{\frac{1 - \frac{2}{3}}{1 + \frac{2}{3}}} = \pm \sqrt{\frac{1}{5}} = \pm \frac{1}{\sqrt{5}} \). Simplify to \( \pm \frac{\sqrt{5}}{5} \).

Step 7: Determine Signs for Half-Angle Values

Since \( x \) is in the fourth quadrant, \( \frac{x}{2} \) would be in the second quadrant \( (135^{\circ} < \frac{x}{2} < 180^{\circ}) \) where sine is positive, cosine is negative, and tangent is negative. Thus, \( \sin \frac{x}{2} = \frac{\sqrt{6}}{6} \), \( \cos \frac{x}{2} = -\frac{\sqrt{30}}{6} \), and \( \tan \frac{x}{2} = -\frac{\sqrt{5}}{5} \).

Key Concepts

Secant Function

The secant function is an essential part of trigonometry. It is directly related to the cosine function through the equation \( \sec x = \frac{1}{\cos x} \). This means, if you know the value of one, you can easily determine the other. For instance, if \( \sec x = \frac{3}{2} \), you can find \( \cos x = \frac{2}{3} \) by taking the reciprocal.
The secant function is important because it helps to evaluate trigonometric values when direct measurement may not be possible. Remember, the secant function is defined wherever the cosine function is not zero, which excludes odd multiples of \( 90^{\circ} \).
Understanding secant helps in solving various trigonometric problems efficiently, especially when combined with identities and other functions.

Half-Angle Formulas

Half-angle formulas are a powerful tool in trigonometry, especially useful for solving problems involving angles that are half of a known angle. These formulas allow us to find the sine, cosine, and tangent of half an angle. For sine, the half-angle formula is \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \). For cosine, it's \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \), and for tangent, it's \( \tan \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} \).
Half-angle formulas are extremely helpful when working with trigonometric functions in non-standard positions or when addressing component angles in larger geometric figures. They open pathways to easily calculate values that might have been complexly hidden behind other trigonometric layers. Always remember that the selection of signs (positive or negative) in these formulas relies heavily on the quadrant of the resultant angle. This is a key aspect to consider in ensuring correct results.

Quadrants in Trigonometry

In trigonometry, the coordinate plane is divided into four quadrants, each affecting the sign of the trigonometric functions. As you move counterclockwise from the positive x-axis:

• The first quadrant (0° to 90°): all trigonometric functions (sine, cosine, and tangent) are positive.
• The second quadrant (90° to 180°): sine is positive, while cosine and tangent are negative.
• The third quadrant (180° to 270°): tangent is positive, while sine and cosine are negative.
• The fourth quadrant (270° to 360°): cosine is positive while sine and tangent are negative.

Knowing which quadrant an angle lies in is critical because it helps determine the sign of trigonometric values.
For instance, with a given angle in the fourth quadrant, as in our example with \( \sec x \), cosine is positive, which is why \( \cos x = \frac{2}{3} \). When using half-angle formulas, the angle's quadrant determines whether the half-angle formula should yield a positive or negative result. Recognizing these core aspects of quadrants aids in effectively solving trigonometric problems.