Answer :
Sure, let's address each of these one by one:
### Problem 9
Given: [tex]\(\sin \theta = \frac{4}{5}\)[/tex]
1. Find [tex]\(\cos \theta\)[/tex]:
To find [tex]\(\cos \theta\)[/tex], we use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{4}{5}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{16}{25} + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos^2 \theta = 1 - \frac{16}{25} = \frac{9}{25} \][/tex]
[tex]\[ \cos \theta = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \][/tex]
Final results:
[tex]\[ \cos \theta = \frac{3}{5}, \quad \tan \theta = \frac{4}{3} \][/tex]
### Problem 10
Given: [tex]\(\cos \theta = \frac{5}{13}\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex]:
Using the Pythagorean identity again:
[tex]\[ \sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + \frac{25}{169} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169} \][/tex]
[tex]\[ \sin \theta = \sqrt{\frac{144}{169}} = \frac{12}{13} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{12}{13}, \quad \tan \theta = \frac{12}{5} \][/tex]
### Problem 11
Given: [tex]\(\tan \theta = 1\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = 1 \implies \sin \theta = \cos \theta \][/tex]
Using the identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ 2 \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = \frac{1}{2} \][/tex]
[tex]\[ \sin \theta = \cos \theta = \frac{\sqrt{2}}{2} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{\sqrt{2}}{2}, \quad \cos \theta = \frac{\sqrt{2}}{2} \][/tex]
### Problem 12
Given: [tex]\(\sin \theta = \frac{1}{2}\)[/tex]
1. Find [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{1}{4} + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos \theta = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \][/tex]
Final results:
[tex]\[ \cos \theta = \frac{\sqrt{3}}{2}, \quad \tan \theta = \frac{1}{\sqrt{3}} \][/tex]
### Problem 13
Given: [tex]\(\cos \theta = \frac{9}{41}\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex]:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{9}{41}\right)^2 + \sin^2 \theta = 1 \][/tex]
[tex]\[ \frac{81}{1681} + \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{81}{1681} = \frac{1600}{1681} \][/tex]
[tex]\[ \sin \theta = \sqrt{\frac{1600}{1681}} = \frac{40}{41} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{40}{41}, \quad \tan \theta = \frac{40}{9} \][/tex]
### Problem 14
Given: [tex]\(\tan \theta = \sqrt{3}\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \sqrt{3} \][/tex]
We recognize this situation from common trigonometric values where:
[tex]\[ \sin \theta = \frac{\sqrt{3}}{2}, \quad \cos \theta = \frac{1}{2} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{\sqrt{3}}{2}, \quad \cos \theta = \frac{1}{2} \][/tex]
### Problem 9
Given: [tex]\(\sin \theta = \frac{4}{5}\)[/tex]
1. Find [tex]\(\cos \theta\)[/tex]:
To find [tex]\(\cos \theta\)[/tex], we use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{4}{5}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{16}{25} + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos^2 \theta = 1 - \frac{16}{25} = \frac{9}{25} \][/tex]
[tex]\[ \cos \theta = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \][/tex]
Final results:
[tex]\[ \cos \theta = \frac{3}{5}, \quad \tan \theta = \frac{4}{3} \][/tex]
### Problem 10
Given: [tex]\(\cos \theta = \frac{5}{13}\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex]:
Using the Pythagorean identity again:
[tex]\[ \sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + \frac{25}{169} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169} \][/tex]
[tex]\[ \sin \theta = \sqrt{\frac{144}{169}} = \frac{12}{13} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{12}{13}, \quad \tan \theta = \frac{12}{5} \][/tex]
### Problem 11
Given: [tex]\(\tan \theta = 1\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = 1 \implies \sin \theta = \cos \theta \][/tex]
Using the identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ 2 \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = \frac{1}{2} \][/tex]
[tex]\[ \sin \theta = \cos \theta = \frac{\sqrt{2}}{2} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{\sqrt{2}}{2}, \quad \cos \theta = \frac{\sqrt{2}}{2} \][/tex]
### Problem 12
Given: [tex]\(\sin \theta = \frac{1}{2}\)[/tex]
1. Find [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{1}{4} + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos \theta = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \][/tex]
Final results:
[tex]\[ \cos \theta = \frac{\sqrt{3}}{2}, \quad \tan \theta = \frac{1}{\sqrt{3}} \][/tex]
### Problem 13
Given: [tex]\(\cos \theta = \frac{9}{41}\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex]:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{9}{41}\right)^2 + \sin^2 \theta = 1 \][/tex]
[tex]\[ \frac{81}{1681} + \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{81}{1681} = \frac{1600}{1681} \][/tex]
[tex]\[ \sin \theta = \sqrt{\frac{1600}{1681}} = \frac{40}{41} \][/tex]
2. Find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{40}{41}, \quad \tan \theta = \frac{40}{9} \][/tex]
### Problem 14
Given: [tex]\(\tan \theta = \sqrt{3}\)[/tex]
1. Find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \sqrt{3} \][/tex]
We recognize this situation from common trigonometric values where:
[tex]\[ \sin \theta = \frac{\sqrt{3}}{2}, \quad \cos \theta = \frac{1}{2} \][/tex]
Final results:
[tex]\[ \sin \theta = \frac{\sqrt{3}}{2}, \quad \cos \theta = \frac{1}{2} \][/tex]
