Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Given that \(\tan \alpha = \frac{15}{8}\) and \(0^\circ < \alpha < 90^\circ\), let's find the exact values of \(\sin \frac{\alpha}{2}\), \(\cos \frac{\alpha}{2}\), and \(\tan \frac{\alpha}{2}\).
### Step-by-Step Solution
Step 1: Calculate \(\sec \alpha\)
From the identity \(1 + \tan^2 \alpha = \sec^2 \alpha\), we have:
[tex]\[ \sec \alpha = \sqrt{1 + \tan^2 \alpha} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \left(\frac{15}{8}\right)^2} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \frac{225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{64 + 225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{289}{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{\sqrt{289}}{\sqrt{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{17}{8} \][/tex]
Step 2: Calculate \(\cos \alpha\)
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), we have:
[tex]\[ \cos \alpha = \frac{1}{\sec \alpha} \][/tex]
[tex]\[ \cos \alpha = \frac{1}{\frac{17}{8}} \][/tex]
[tex]\[ \cos \alpha = \frac{8}{17} \][/tex]
Step 3: Calculate \(\sin \alpha\)
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we have:
[tex]\[ \sin^2 \alpha = 1 - \cos^2 \alpha \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \left(\frac{8}{17}\right)^2 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{289 - 64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{225}{289} \][/tex]
[tex]\[ \sin \alpha = \sqrt{\frac{225}{289}} \][/tex]
[tex]\[ \sin \alpha = \frac{15}{17} \][/tex]
Step 4: Calculate \(\sin \frac{\alpha}{2}\)
Using the half-angle formula for sine, \(\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\):
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{17 - 8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{9}{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}} \][/tex]
Step 5: Calculate \(\cos \frac{\alpha}{2}\)
Using the half-angle formula for cosine, \(\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\):
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{17 + 8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{25}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{25}{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{\sqrt{25}}{\sqrt{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}} \][/tex]
Step 6: Calculate \(\tan \frac{\alpha}{2}\)
Using the half-angle formula for tangent, \(\tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}\):
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{1 + \frac{8}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{9}{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{3}{5} \][/tex]
### Answers
a. \(\sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}}\)
b. \(\cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}}\)
c. [tex]\(\tan \frac{\alpha}{2} = \frac{3}{5}\)[/tex]
### Step-by-Step Solution
Step 1: Calculate \(\sec \alpha\)
From the identity \(1 + \tan^2 \alpha = \sec^2 \alpha\), we have:
[tex]\[ \sec \alpha = \sqrt{1 + \tan^2 \alpha} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \left(\frac{15}{8}\right)^2} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \frac{225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{64 + 225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{289}{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{\sqrt{289}}{\sqrt{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{17}{8} \][/tex]
Step 2: Calculate \(\cos \alpha\)
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), we have:
[tex]\[ \cos \alpha = \frac{1}{\sec \alpha} \][/tex]
[tex]\[ \cos \alpha = \frac{1}{\frac{17}{8}} \][/tex]
[tex]\[ \cos \alpha = \frac{8}{17} \][/tex]
Step 3: Calculate \(\sin \alpha\)
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we have:
[tex]\[ \sin^2 \alpha = 1 - \cos^2 \alpha \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \left(\frac{8}{17}\right)^2 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{289 - 64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{225}{289} \][/tex]
[tex]\[ \sin \alpha = \sqrt{\frac{225}{289}} \][/tex]
[tex]\[ \sin \alpha = \frac{15}{17} \][/tex]
Step 4: Calculate \(\sin \frac{\alpha}{2}\)
Using the half-angle formula for sine, \(\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\):
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{17 - 8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{9}{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}} \][/tex]
Step 5: Calculate \(\cos \frac{\alpha}{2}\)
Using the half-angle formula for cosine, \(\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\):
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{17 + 8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{25}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{25}{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{\sqrt{25}}{\sqrt{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}} \][/tex]
Step 6: Calculate \(\tan \frac{\alpha}{2}\)
Using the half-angle formula for tangent, \(\tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}\):
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{1 + \frac{8}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{9}{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{3}{5} \][/tex]
### Answers
a. \(\sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}}\)
b. \(\cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}}\)
c. [tex]\(\tan \frac{\alpha}{2} = \frac{3}{5}\)[/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
