Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. If \(\cos \alpha=\frac{4}{5}, 0<\alpha<\frac{\pi}{2},\) find the exact value of: (a) \(\cos \frac{\alpha}{2}\) (b) \(\sin \frac{\alpha}{2}\) (c) \(\tan \frac{\alpha}{2}\)

Short answer

(a) \ \frac{3 \sqrt{10}}{10} \: (b) \ \frac{\sqrt{10}}{10} \: (c) \ \frac{1}{3} \

Step-by-step solution

- Determine \(\cos \frac{\alpha}{2}\)

To find \(\cos \frac{\alpha}{2}\), use the half-angle formula: \[\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}}\]. Since \ \alpha \ is in the first quadrant, \ \frac{\alpha}{2} \ is also in the first quadrant, meaning cosine is positive. Plug in \ \cos \alpha = \frac{4}{5} \: \[\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{4}{5}}{2}} = \sqrt{\frac{\frac{9}{5}}{2}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} = \frac{3 \sqrt{10}}{10}\]

- Determine \(\sin \frac{\alpha}{2}\)

To find \(\sin \frac{\alpha}{2}\), use the half-angle formula: \[\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}\]. Since \ \alpha \ is in the first quadrant, \ \frac{\alpha}{2} \ is also in the first quadrant, meaning sine is positive. Plug in \ \cos \alpha = \frac{4}{5} \: \[\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{\frac{1}{5}}{2}} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}\]

- Determine \(\tan \frac{\alpha}{2}\)

To find \(\tan \frac{\alpha}{2}\), use the identity: \[\tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\]. Substitute the values found in the previous steps: \[\tan \frac{\alpha}{2} = \frac{\frac{\sqrt{10}}{10}}{\frac{3 \sqrt{10}}{10}} = \frac{\sqrt{10} / 10}{3 \sqrt{10} / 10} = \frac{1}{3}\]

Key concepts

Half-Angle Formulas

The half-angle formulas are a crucial part of trigonometry, especially when finding the exact values of trigonometric functions. These formulas allow us to calculate the sine, cosine, and tangent of half an angle when the full angle's trigonometric function values are known. For example, the half-angle formula for cosine is: \(\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}}\). Likewise, the formula for sine is: \(\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}\). Lastly, to find the tangent, you can use the identity \(\tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\).

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are fundamental in the study of angles and their relationships. They describe the ratios of the sides of a right triangle concerning its angles. For instance, \(\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}}\), and \(\tan \alpha = \frac{\text{opposite}}{\text{adjacent}}\). In the context of half-angle formulas:

• The cosine half-angle formula \(\cos \frac{\alpha}{2}\) allows us to find the cosine of half an angle without directly knowing the angle itself.
• The sine half-angle formula \(\sin \frac{\alpha}{2}\) provides an analogous method for the sine function.
• The tangent half-angle formula \(\tan \frac{\alpha}{2}\) is derived using the values of \(\sin \frac{\alpha}{2}\) and \(\cos \frac{\alpha}{2}\).

Exact Values

Finding the exact values of trigonometric functions is often necessary for a precise understanding of angles and their properties. In our example, where \(\cos \alpha = \frac{4}{5}\) and \(0 < \alpha < \frac{\pi}{2}\), we calculated the half-angle functions using exact values:

• First, using \(\cos \alpha = \frac{4}{5}\), we determined the exact value of \(\cos \frac{\alpha}{2} \): \(\cos \frac{\alpha}{2} = \frac{3 \sqrt{10}}{10}\).
• Next, we found \(\sin \frac{\alpha}{2}\) using the same \(\cos \alpha\): \(\sin \frac{\alpha}{2} = \frac{\sqrt{10}}{10}\).
• Lastly, using these results, we obtained \(\tan \frac{\alpha}{2} \): \(\tan \frac{\alpha}{2} = \frac{1}{3}\).

By understanding exact values and how to apply half-angle formulas, you can solve a wide range of trigonometric problems accurately.