2. Let [tex]$f(x)=3^x$[/tex], then show that [tex]$\frac{f(x+2)-f(x)}{2}=4\left(3^x\right)$[/tex].

3. Given [tex][tex]$\tan \theta=\frac{1}{2}$[/tex][/tex], and [tex]$\sin \theta\ \textless \ 0$[/tex], then find [tex]$\cos \theta$[/tex].



Answer :

Let's solve both parts of the question in detail.

### Part 2:
Given the function [tex]\( f(x) = 3^x \)[/tex], we are to show that:
[tex]\[ \frac{f(x+2) - f(2)}{2} = 4 \cdot 3^x \][/tex]

Step-by-Step Solution:

1. Calculate [tex]\( f(x+2) \)[/tex]:
[tex]\[ f(x+2) = 3^{x+2} \][/tex]

2. Calculate [tex]\( f(2) \)[/tex]:
[tex]\[ f(2) = 3^2 = 9 \][/tex]

3. Substitute [tex]\( f(x+2) \)[/tex] and [tex]\( f(2) \)[/tex] into the expression:
[tex]\[ \frac{f(x+2) - f(2)}{2} = \frac{3^{x+2} - 9}{2} \][/tex]

4. Express [tex]\( 3^{x+2} \)[/tex] in terms of [tex]\( 3^x \)[/tex]:
[tex]\[ 3^{x+2} = 3^x \cdot 3^2 = 9 \cdot 3^x \][/tex]

5. Substitute [tex]\( 3^{x+2} \)[/tex] back into the expression:
[tex]\[ \frac{9 \cdot 3^x - 9}{2} \][/tex]

6. Factor out the common term [tex]\( 9 \)[/tex]:
[tex]\[ \frac{9 (3^x - 1)}{2} \][/tex]

7. Simplify the expression:
[tex]\[ \frac{9 \cdot 3^x - 9}{2} = \frac{9}{2} \cdot (3^x - 1) = 4.5 \cdot 3^x - 4.5 \][/tex]

8. Notice that this doesn't seem to directly simplify to [tex]\( 4 \cdot 3^x \)[/tex] due to the presence of [tex]\(-4.5\)[/tex], indicating there might have been an error somewhere in the initial question setup or additional context needed (validates towards assertion step differences).

### Part 3:
Given [tex]\(\tan \theta = \frac{1}{2}\)[/tex], and [tex]\(\sin \theta < 0\)[/tex], find [tex]\(\cos \theta \cdot 2\)[/tex].

Step-by-Step Solution:

1. Since [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex], write the given tangent value as:
[tex]\[ \tan \theta = \frac{1}{2} \][/tex]

2. Using the identity for tangent:
[tex]\[ \tan^2 \theta = \frac{1}{4} \][/tex]

3. Using the Pythagorean identity for tangents:
[tex]\[ 1 + \tan^2 \theta = \sec^2 \theta \][/tex]
[tex]\[ 1 + \frac{1}{4} = \frac{5}{4} \][/tex]

4. Recall the relationship between secant and cosine:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
[tex]\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{5}{4} \][/tex]

5. Solve for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[ \cos^2 \theta = \frac{4}{5} \][/tex]

6. Since [tex]\(\sin \theta < 0\)[/tex], [tex]\(\theta\)[/tex] must lie in the third or fourth quadrant where cosine is positive in the fourth quadrant:
[tex]\[ \cos \theta = -\sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \][/tex]

7. Calculate [tex]\(2 \cos \theta\)[/tex]:
[tex]\[ 2 \cos \theta = 2 \cdot \left( \frac{2}{\sqrt{5}} \right) \][/tex]
[tex]\[ 2 \cos \theta = \frac{4}{\sqrt{5}} \][/tex]

So, the answer to [tex]\(\cos \theta \cdot 2\)[/tex] is [tex]\(\boxed{\frac{4}{\sqrt{5}}}\)[/tex].