Problem 50 - Precalculus with Calculus Previews

Question

Use the given information to find (a) \(\cos (x / 2)\), (b) \(\sin (x / 2)\), and (c) \(\tan (x / 2)\). $$ \cot x=-\frac{1}{4}, 90^{\circ}

Step-by-Step Solution

Answer

(a) \(\cos(x/2) = \sqrt{\frac{1 - \frac{1}{\sqrt{17}}}{2}}\), (b) \(\sin(x/2) = \sqrt{\frac{1 + \frac{1}{\sqrt{17}}}{2}}\), (c) \(\tan(x/2) = \sqrt{\frac{1 + \frac{1}{\sqrt{17}}}{1 - \frac{1}{\sqrt{17}}}}\).

Step 1: Identify the quadrant

Given that \(90^{\circ} < x < 180^{\circ}\), angle \(x\) lies in the second quadrant. In this quadrant, the sine of an angle is positive while both cosine and tangent are negative.

Step 2: Use cotangent to find tangent

The cotangent of angle \(x\) is given as \(\cot x = -\frac{1}{4}\). Therefore, the tangent can be found as \(\tan x = -4\) because \(\tan x = \frac{1}{\cot x}\).

Step 3: Use identity to find sine and cosine

Using the identity \(\tan x = \frac{\sin x}{\cos x}\), and knowing \(\tan x = -4\), we can set up the equation \(\sin x = -4 \cos x\). Since \(\sin^2 x + \cos^2 x = 1\), substitute \(-4 \cos x\) for \(\sin x\) to get: \((-4 \cos x)^2 + \cos^2 x = 1\).

Step 4: Simplify the trigonometric equation

Solve \(16 \cos^2 x + \cos^2 x = 1\) which simplifies to \(17 \cos^2 x = 1\). Solving for \(\cos^2 x\), we obtain \(\cos^2 x = \frac{1}{17}\). Hence, \(\cos x = -\frac{1}{\sqrt{17}}\) since \(\cos x\) is negative in the second quadrant.

Step 5: Find sine

With \(\cos x = -\frac{1}{\sqrt{17}}\), use \(\sin x = -4 \cos x\) to find \(\sin x = -4 \left(-\frac{1}{\sqrt{17}}\right) = \frac{4}{\sqrt{17}}\).

Step 6: Use half-angle formulas for cosine, sine, and tangent

For \(\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos x}{2}}\), substitute \(\cos x = -\frac{1}{\sqrt{17}}\) to find \(\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \frac{1}{\sqrt{17}}}{2}}\). Since \(45^{\circ} < \frac{x}{2} < 90^{\circ}\), \(\cos \left(\frac{x}{2}\right)\) is positive.

Step 7: Solve for sine using the half-angle formula

For \(\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{2}}\), substitute \(\cos x = -\frac{1}{\sqrt{17}}\), yielding \(\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \frac{1}{\sqrt{17}}}{2}}\). As \(\sin \left(\frac{x}{2}\right)\) is positive in the first quadrant, choose the positive value.

Step 8: Calculate tangent using its half-angle formula

Use \(\tan \left(\frac{x}{2}\right) = \frac{\sin x}{1 + \cos x}\) or \(\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}\). Substitute for the expressions derived above to find \(\tan \left(\frac{x}{2}\right)\).

Key Concepts

Half-Angle Formulas

Understanding half-angle formulas is crucial for finding the sine, cosine, and tangent of half an angle. These formulas provide a method to express trigonometric functions of half angles in terms of the full angle. Here are the key half-angle formulas:

• Cosine: \( \cos \left( \frac{x}{2} \right) = \pm \sqrt{\frac{1 + \cos x}{2}} \)
• Sine: \( \sin \left( \frac{x}{2} \right) = \pm \sqrt{\frac{1 - \cos x}{2}} \)
• Tangent: \( \tan \left( \frac{x}{2} \right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} \)

When using these formulas, the sign (positive/negative) is determined by the standard position of the resulting angle. This is often guided by which quadrant the half-angle lies within. For example, given that \(90^{\circ} < x < 180^{\circ}\) implies that \( \frac{x}{2} \) resides in the range \( 45^{\circ} < \frac{x}{2} < 90^{\circ} \), thus being in the first quadrant, where both sine and cosine are positive. It's this concept that ensures a correct evaluation of trigonometric functions.

Second Quadrant

The second quadrant in the unit circle spans from \(90^{\circ}\) to \(180^{\circ}\). In this quadrant, some unique characteristics of trigonometric functions are noted:

• Sine is positive: \( \sin \theta > 0 \)
• Cosine is negative: \( \cos \theta < 0 \)
• Tangent is negative: \( \tan \theta < 0 \)

These sign rules are pivotal when solving problems involving the second quadrant. The exercise you are working on falls into this territory, with \(x\) being in the second quadrant. This affects how you approach finding the values of other trigonometric functions. For example, since cosine is negative, when calculating \(\cos x\), it is crucial to reflect this sign in your calculations. This understanding helps in solving equations accurately, including those that involve half-angle formulas.

Trigonometric Functions

Trigonometric functions include sine, cosine, tangent, cotangent, secant, and cosecant and are foundational in trigonometry. Here is a brief overview:

• **Sine (\( \sin \))**: Relates to the ratio of the opposite side to the hypotenuse in a right triangle.
• **Cosine (\( \cos \))**: Represents the ratio of the adjacent side to the hypotenuse.
• **Tangent (\( \tan \))**: The ratio of the opposite side to the adjacent side. It can also be expressed as \( \frac{\sin x}{\cos x} \).
• **Cotangent (\( \cot \))**: The reciprocal of tangent, \( \cot x = \frac{1}{\tan x} \).
• **Secant (\( \sec \))**: The reciprocal of cosine, \( \sec x = \frac{1}{\cos x} \).
• **Cosecant (\( \csc \))**: The reciprocal of sine, \( \csc x = \frac{1}{\sin x} \).

These functions enable us to relate angles and side lengths in circles and right triangles. In problems involving radians and degrees, knowing these functions and their properties, such as signs in different quadrants, helps solve complex trigonometric equations. Always remember, mastering the trigonometric functions allows for deeper understanding when manipulating and solving the associated formulas.