Problem 62 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the exact values of \(\sin (u / 2), \cos (u / 2),\) and \(\tan (u / 2)\) using the half-angle formulas. $$\tan u=-\frac{5}{12}, \quad 3 \pi / 2

Step-by-Step Solution

Answer

The exact values of \(\sin(u/2) = \frac{\sqrt{51}}{10}\), \(\cos(u/2) = -\frac{1}{10}\), and \(\tan(u/2) = -\sqrt{51}\)

Step 1: Identify the Sine and Cosine of Angle \(u\)

The given \(\tan u = -\frac{5}{12}\) is negative which indicates that the angle is in the 4th quadrant, where sine is negative and cosine is positive. As we know that \(\tan u = \frac{\sin u}{\cos u}\), and taking Pythagorean identity \(\sin^2u + \cos^2u = 1\), we can find \(\sin u\) and \(\cos u\). \n Given, \(\tan u = -\frac{5}{12}\), squaring both sides, we get \(\tan^2u = \frac{25}{144}\). So, using the identity \(\tan^2u = \frac{\sin^2u}{\cos^2u}\), we get \(\sin^2u = 25\cos^2u\). Now using Pythagorean identity \(\sin^2u + \cos^2u = 1\), replace \(\sin^2u\) with \(25\cos^2u\), we get \(25\cos^2u + \cos^2u = 1\), giving \(\cos^2u = \frac{1}{26}\) and therefore \(\cos u = \frac{1}{\sqrt{26}} = \frac{\sqrt{26}}{26}\) as cosine is positive in the 4th quadrant. We can find \(\sin u = -\sqrt{1 - \cos^2u} = -\sqrt{1 - (\frac{1}{26})} = -\frac{5}{\sqrt{26}}\) as sine is negative in the 4th quadrant.

Step 2: Apply Half-Angle Formulas

Half angle formulas are - \(\sin(\frac{u}{2}) = \pm \sqrt{\frac{1-\cos u}{2}}\), \(\cos(\frac{u}{2}) = \pm \sqrt{\frac{1+\cos u}{2}}\), and \(\tan(\frac{u}{2}) = \pm \sqrt{\frac{1-\cos u}{1+ \cos u}}\). For \(\sin(\frac{u}{2})\) and \(\cos(\frac{u}{2})\), we choose the signs based on the quadrant where \(\frac{u}{2}\) lies. Given, \(3 \pi / 2 < u < 2 \pi\), \(\pi / 4 < \frac{u}{2} < \pi\) i.e., \(\frac{u}{2}\) lies in the 2nd quadrant, where sine is positive and cosine is negative. Plug in the value of \(\cos u = \frac{\sqrt{26}}{26}\) into the half-angle formulas, we get \(\sin(\frac{u}{2}) = \sqrt{\frac{1-\frac{\sqrt{26}}{26}}{2}} = \frac{\sqrt{51}}{10}\) and \(\cos(\frac{u}{2}) = -\sqrt{\frac{1+\frac{\sqrt{26}}{26}}{2}} = -\frac{\sqrt{1}}{10}\). For \(\tan(\frac{u}{2})\), the positive or negative sign should be chosen such that \(\tan(\frac{u}{2}) = \frac{\sin(\frac{u}{2})}{\cos(\frac{u}{2})}\). Here, \(\tan(\frac{u}{2}) = -\frac{\sqrt{51}}{1}\).

Step 3: Conclusion

The exact values of \(\sin(u/2)\), \(\cos(u/2)\), and \(\tan(u/2)\) using the half-angle formulas are \(\frac{\sqrt{51}}{10}\), \(-\frac{1}{10}\), and \(-\sqrt{51}\) respectively.

Key Concepts

Half-Angle Formulas

Half-angle formulas are essential tools in trigonometry used to find the values of sine, cosine, and tangent for half of a given angle. These formulas express trigonometric functions of half an angle in terms of the trigonometric functions of the original angle. They are given by:

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

These formulas can be particularly useful when direct computation of trigonometric functions at a certain angle is complex. A critical step in using these formulas is determining the correct sign for the result. This choice depends on the quadrant where the half angle lies. Quadrant properties are crucial in choosing the signs for this reason. For the problem at hand, since the angle \(u/2\) was found to be in the 2nd quadrant, where sine is positive and cosine is negative, the signs for the half-angle formulas were chosen accordingly.

Sine and Cosine of an Angle

In trigonometry, understanding the sine and cosine of an angle is key to solving many problems, such as the given exercise. The relationship between sine, cosine, and tangent for an angle is fundamental. If you know how to work with one of these values, you can find the others using known identities. A key identity for angle \(u\) is the Pythagorean identity: \(\sin^2 u + \cos^2 u = 1\).This formula allows you to find one trigonometric function given the other. For example, if \(\tan u\) is given, it expresses as a ratio of sine to cosine: \(\tan u = \frac{\sin u}{\cos u}\).Using this identity, both \(\sin u\) and \(\cos u\) can be derived. The relationship for tangent provided here, \(\tan u = -\frac{5}{12}\), was used to calculate both \(\sin u\) and \(\cos u\), factoring in their signs based on the quadrant provided in the exercise.

Quadrant Properties

Understanding how trigonometric functions behave in different quadrants is crucial for solving problems like the one presented. The unit circle is divided into four quadrants, each with unique properties impacting trigonometric signs:

• In the 1st quadrant, all trigonometric functions are positive.
• In the 2nd quadrant, sine is positive, but cosine and tangent are negative.
• In the 3rd quadrant, tangent is positive, while sine and cosine are negative.
• In the 4th quadrant, cosine is positive, but sine and tangent are negative.

In the exercise, it was important to realize that angle \(u\) is located in the 4th quadrant, which affected the signs of its sine and cosine. This awareness is equally critical when determining the signs for the half of the given angle \(u/2\), which was positioned in the 2nd quadrant. Hence, these quadrant properties directly impacted the calculation of the trigonometric functions using half-angle formulas.