Sure, let's go through these problems step by step.
### Problem 7:
Given: [tex]\(4 \tan \theta = 3\)[/tex]
First, let's find [tex]\(\tan \theta\)[/tex]:
[tex]\[
\tan \theta = \frac{3}{4}
\][/tex]
We know that:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
Let:
[tex]\[
\sin \theta = a \quad \text{and} \quad \cos \theta = b
\][/tex]
Given:
[tex]\[
\tan \theta = \frac{a}{b} = \frac{3}{4}
\][/tex]
This implies:
[tex]\[
a = 3k \quad \text{and} \quad b = 4k \quad \text{for some constant} \ k
\][/tex]
Let's substitute these expressions into the given fraction [tex]\(\frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta}\)[/tex]:
First, the numerator:
[tex]\[
4 \sin \theta + 3 \cos \theta = 4a + 3b = 4(3k) + 3(4k) = 12k + 12k = 24k
\][/tex]
Now, the denominator:
[tex]\[
8 \sin \theta + 5 \cos \theta = 8a + 5b = 8(3k) + 5(4k) = 24k + 20k = 44k
\][/tex]
Thus, the fraction becomes:
[tex]\[
\frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{24k}{44k} = \frac{24}{44} = \frac{6}{11}
\][/tex]
So we have shown that:
[tex]\[
\frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11}
\][/tex]
### Problem 8:
Given: [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex]
We need to show that:
[tex]\[
4 \cos^3 \theta - 3 \cos \theta = 0
\][/tex]
First, let's calculate [tex]\(\cos^3 \theta\)[/tex]:
[tex]\[
\cos^3 \theta = \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3 \sqrt{3}}{8}
\][/tex]
Now, substitute [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex] and [tex]\(\cos^3 \theta = \frac{3 \sqrt{3}}{8}\)[/tex] into the expression:
[tex]\[
4 \cos^3 \theta - 3 \cos \theta = 4 \left(\frac{3 \sqrt{3}}{8}\right) - 3 \left(\frac{\sqrt{3}}{2}\right)
\][/tex]
Simplify each term:
[tex]\[
4 \cos^3 \theta = 4 \left(\frac{3 \sqrt{3}}{8}\right) = \frac{12 \sqrt{3}}{8} = \frac{3 \sqrt{3}}{2}
\][/tex]
[tex]\[
3 \cos \theta = 3 \left(\frac{\sqrt{3}}{2}\right) = \frac{3 \sqrt{3}}{2}
\][/tex]
So, we can write:
[tex]\[
4 \cos^3 \theta - 3 \cos \theta = \frac{3 \sqrt{3}}{2} - \frac{3 \sqrt{3}}{2} = 0
\][/tex]
Thus, we have shown that:
[tex]\[
4 \cos^3 \theta - 3 \cos \theta = 0
\][/tex]
