IDNLearn.com: Where your questions meet expert answers and community support. Ask your questions and receive comprehensive, trustworthy responses from our dedicated team of experts.
Sagot :
Let's tackle each problem step by step.
### Problem 5
Given that [tex]\(\tan \theta = \frac{1}{\sqrt{3}}\)[/tex], we need to find the value of:
[tex]\[ \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
1. Substitute [tex]\(\tan \theta\)[/tex] with [tex]\(\frac{1}{\sqrt{3}}\)[/tex]:
[tex]\[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} \][/tex]
2. Simplify the expression:
[tex]\[ = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} \][/tex]
Therefore, the value is [tex]\(\sqrt{3}\)[/tex].
### Problem 6
Given that [tex]\(\sin \theta = \frac{4}{5}\)[/tex], we need to find the value of:
[tex]\[ 3 \sec \theta - 5 \cos \theta \][/tex]
1. First, determine [tex]\(\cos \theta\)[/tex] using the Pythagorean identity:
[tex]\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
2. Find [tex]\(\sec \theta\)[/tex] which is the reciprocal of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \][/tex]
3. Substitute into the given expression:
[tex]\[ 3 \sec \theta - 5 \cos \theta = 3 \times \frac{5}{3} - 5 \times \frac{3}{5} = 5 - 3 = 2 \][/tex]
Therefore, the value is 2.
### Problem 7
Given that [tex]\(4 \tan \theta = 3\)[/tex], show that:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11} \][/tex]
1. First, solve for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{3}{4} \][/tex]
2. Use the identity [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex], we can let:
[tex]\[ \sin \theta = 3k \quad \text{and} \quad \cos \theta = 4k \][/tex]
3. Normalize using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ (3k)^2 + (4k)^2 = 1 \quad \Rightarrow \quad 9k^2 + 16k^2 = 1 \quad \Rightarrow \quad 25k^2 = 1 \quad \Rightarrow \quad k = \frac{1}{5} \][/tex]
So, [tex]\(\sin \theta = \frac{3}{5}\)[/tex] and [tex]\(\cos \theta = \frac{4}{5}\)[/tex].
4. Substitute these values into the given expression:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{4 \cdot \frac{3}{5} + 3 \cdot \frac{4}{5}}{8 \cdot \frac{3}{5} + 5 \cdot \frac{4}{5}} = \frac{\frac{12}{5} + \frac{12}{5}}{\frac{24}{5} + \frac{20}{5}} = \frac{\frac{24}{5}}{\frac{44}{5}} = \frac{24}{44} = \frac{6}{11} \][/tex]
Hence, the expression simplifies to [tex]\(\frac{6}{11}\)[/tex], proving the equality.
### Problem 8
Given that [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex], show that:
[tex]\[ 4 \cos^3 \theta - 3 \cos \theta = 0 \][/tex]
1. Substitute [tex]\(\cos \theta\)[/tex] into the expression:
[tex]\[ 4 \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \left(\frac{\sqrt{3}}{2}\right) \][/tex]
2. Calculate each term individually:
[tex]\[ \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3\sqrt{3}}{8} \][/tex]
3. Now plug it back into the original equation:
[tex]\[ 4 \cdot \frac{3\sqrt{3}}{8} - 3 \cdot \frac{\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{12\sqrt{3}}{8} = 0 \][/tex]
Therefore, we have shown that [tex]\(4 \cos^3 \theta - 3 \cos \theta = 0\)[/tex].
### Problem 5
Given that [tex]\(\tan \theta = \frac{1}{\sqrt{3}}\)[/tex], we need to find the value of:
[tex]\[ \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
1. Substitute [tex]\(\tan \theta\)[/tex] with [tex]\(\frac{1}{\sqrt{3}}\)[/tex]:
[tex]\[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} \][/tex]
2. Simplify the expression:
[tex]\[ = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} \][/tex]
Therefore, the value is [tex]\(\sqrt{3}\)[/tex].
### Problem 6
Given that [tex]\(\sin \theta = \frac{4}{5}\)[/tex], we need to find the value of:
[tex]\[ 3 \sec \theta - 5 \cos \theta \][/tex]
1. First, determine [tex]\(\cos \theta\)[/tex] using the Pythagorean identity:
[tex]\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
2. Find [tex]\(\sec \theta\)[/tex] which is the reciprocal of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \][/tex]
3. Substitute into the given expression:
[tex]\[ 3 \sec \theta - 5 \cos \theta = 3 \times \frac{5}{3} - 5 \times \frac{3}{5} = 5 - 3 = 2 \][/tex]
Therefore, the value is 2.
### Problem 7
Given that [tex]\(4 \tan \theta = 3\)[/tex], show that:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11} \][/tex]
1. First, solve for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{3}{4} \][/tex]
2. Use the identity [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex], we can let:
[tex]\[ \sin \theta = 3k \quad \text{and} \quad \cos \theta = 4k \][/tex]
3. Normalize using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ (3k)^2 + (4k)^2 = 1 \quad \Rightarrow \quad 9k^2 + 16k^2 = 1 \quad \Rightarrow \quad 25k^2 = 1 \quad \Rightarrow \quad k = \frac{1}{5} \][/tex]
So, [tex]\(\sin \theta = \frac{3}{5}\)[/tex] and [tex]\(\cos \theta = \frac{4}{5}\)[/tex].
4. Substitute these values into the given expression:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{4 \cdot \frac{3}{5} + 3 \cdot \frac{4}{5}}{8 \cdot \frac{3}{5} + 5 \cdot \frac{4}{5}} = \frac{\frac{12}{5} + \frac{12}{5}}{\frac{24}{5} + \frac{20}{5}} = \frac{\frac{24}{5}}{\frac{44}{5}} = \frac{24}{44} = \frac{6}{11} \][/tex]
Hence, the expression simplifies to [tex]\(\frac{6}{11}\)[/tex], proving the equality.
### Problem 8
Given that [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex], show that:
[tex]\[ 4 \cos^3 \theta - 3 \cos \theta = 0 \][/tex]
1. Substitute [tex]\(\cos \theta\)[/tex] into the expression:
[tex]\[ 4 \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \left(\frac{\sqrt{3}}{2}\right) \][/tex]
2. Calculate each term individually:
[tex]\[ \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3\sqrt{3}}{8} \][/tex]
3. Now plug it back into the original equation:
[tex]\[ 4 \cdot \frac{3\sqrt{3}}{8} - 3 \cdot \frac{\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{12\sqrt{3}}{8} = 0 \][/tex]
Therefore, we have shown that [tex]\(4 \cos^3 \theta - 3 \cos \theta = 0\)[/tex].
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
