Problem 31 - Contemporary Precalculus

Question

Find \(\sin \frac{x}{2}, \cos \frac{x}{2},\) and \(\tan \frac{x}{2}\) under the given conditions. $$\tan x=\frac{1}{2} \quad\left(\pi

Step-by-Step Solution

Answer

Question: Given that \(\tan x = \frac{1}{2}\) and \(\pi < x < \frac{3\pi}{2}\), find the expressions for \(\sin\frac{x}{2}, \cos\frac{x}{2},\) and \(\tan\frac{x}{2}\). Answer: \(\sin\frac{x}{2} = \sqrt{\frac{7}{10}}, \cos\frac{x}{2} = -\sqrt{\frac{3}{10}},\) and \(\tan\frac{x}{2} = \sqrt{\frac{7}{3}}\).

Step 1: Determine the signs of \(\sin x\) and \(\cos x\)

Since we're given \(\pi < x < \frac{3\pi}{2}\), we know that \(x\) is in the third quadrant. In this quadrant, both \(\sin x\) and \(\cos x\) are negative.

Step 2: Use the given value of \(\tan x\) to find \(\sin x\) and \(\cos x\)

We know that \(\tan x =\frac{\sin x}{\cos x}=\frac{1}{2}\). Since both \(\sin x\) and \(\cos x\) are negative in the third quadrant, we can write \(\sin x = -k\) and \(\cos x = -2k\), for some positive value of \(k\). Thus, \(\tan x = \frac{-k}{-2k} = \frac{1}{2}\). Now we have \(\sin x = -k\) and \(\cos x = -2k\).

Step 3: Apply the Pythagorean trigonometric identity

We know that \(\sin^2 x + \cos^2 x = 1\). Using the values of \(\sin x\) and \(\cos x\) from the previous step, we have: \((-k)^2 + (-2k)^2 = 1 \Rightarrow k^2 + 4k^2 = 1 \Rightarrow 5k^2 = 1 \Rightarrow k^2 = \frac{1}{5}\). Therefore, \(k = \frac{\sqrt{5}}{5}\). So, \(\sin x = -\frac{\sqrt{5}}{5}\) and \(\cos x = -\frac{2\sqrt{5}}{5}\).

Step 4: Use half-angle identities to find \(\sin\frac{x}{2}, \cos\frac{x}{2},\) and \(\tan\frac{x}{2}\)

We will use the following half-angle identities: 1. \(\sin\frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}}\) 2. \(\cos\frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}}\) 3. \(\tan\frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{1+\cos x}}\) For \(\sin\frac{x}{2}\) and \(\cos\frac{x}{2}\), since \(\frac{x}{2}\) is in the second quadrant, \(\sin\frac{x}{2}\) is positive and \(\cos\frac{x}{2}\) is negative. Using the first half-angle identity: \(\sin\frac{x}{2} = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1-\left(-\frac{2\sqrt{5}}{5}\right)}{2}}= \sqrt{\frac{7}{10}}\) Using the second half-angle identity: \(\cos\frac{x}{2} = -\sqrt{\frac{1+\cos x}{2}} = - \sqrt{\frac{1-\left(-\frac{2\sqrt{5}}{5}\right)}{2}} = -\sqrt{\frac{3}{10}}\) Using the third half-angle identity: \(\tan\frac{x}{2} = \sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\frac{1-\left(-\frac{2\sqrt{5}}{5}\right)}{1+\left(-\frac{2\sqrt{5}}{5}\right)}} = \sqrt{\frac{7}{3}}\) So, we have: \(\sin\frac{x}{2} = \sqrt{\frac{7}{10}}, \cos\frac{x}{2} = -\sqrt{\frac{3}{10}},\) and \(\tan\frac{x}{2} = \sqrt{\frac{7}{3}}\).

Key Concepts

Half-Angle Identities

Half-angle identities are crucial tools in trigonometry, especially when you want to find the trigonometric functions of an angle that is half of another known angle. These identities simplify the process and help us easily determine the sine, cosine, and tangent for half-angle values. They are:

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

These formulas provide both positive and negative results. Choosing which sign to use depends on the quadrant in which the half-angle lies. In the given exercise, for example, when \( x \) is in the third quadrant, the half-angle, \( \frac{x}{2} \), would be in the second quadrant. This means \( \sin \frac{x}{2} \) is positive, while \( \cos \frac{x}{2} \) is negative. Thus, understanding the quadrants helps us select the appropriate sign.

Trigonometric Functions

Trigonometric functions include the basic sine, cosine, and tangent, which are vital for examining relationships in right-angled triangles. They also extend to the unit circle where angles can take any real number value.

• **Sine (\(\sin\))** - This function represents the vertical coordinate of a point on the unit circle. It is positive when above the x-axis and negative when below.
• **Cosine (\(\cos\))** - This measures the horizontal position of a point on the unit circle. Like sine, it is positive to the right of the y-axis and negative to the left.
• **Tangent (\(\tan\))** - This function is defined as the ratio of sine to cosine, \(\tan x = \frac{\sin x}{\cos x}\), and gives the slope of the line that the angle creates with the x-axis.

In any trigonometric problem, determine whether each function is positive or negative by identifying the quadrant of the angle. This fundamental understanding ensures correct computation and supports learning more complex identities and formulas.

Pythagorean Identities

The Pythagorean identities are derived directly from the Pythagorean theorem applied to a unit circle. In trigonometry, these identities are essential for simplifying expressions and solving trigonometric equations. They include:

• \( \sin^2 x + \cos^2 x = 1 \)
• \( 1 + \tan^2 x = \sec^2 x \)
• \( 1 + \cot^2 x = \csc^2 x \)

In the provided exercise, the first identity \( \sin^2 x + \cos^2 x = 1 \) is particularly critical. It allows us to solve for sine or cosine when one is already known. By rearranging and substituting known values, you can derive one function from the other, ensuring you have all necessary components for deeper calculations such as the half-angle identities. Recognizing these identities and their power in proof and problem-solving can significantly enhance your understanding and ability to work with trigonometric functions.