Let `f: N -> R and g : N -> R` be two functions and `f(1)=08, g(1)=0.6`, `f(n+1)=f(n)cos(g(n))-g(n)sin(g(n)) and g (n+1)=f(n) sin(g(n))+g(n) cos(g(n))` for `n>=1`. `lim_(n->oo) f(n)` is equal to

Options

A. (a)`-1`
B. (b)`0`
C. (c)`1`
D. (d)does not exist

Answer

To solve the problem step by step, we will analyze the functions \( f(n) \) and \( g(n) \) defined recursively and find the limit of \( f(n) \) as \( n \) approaches infinity.

### Step 1: Initial Values
We are given:
- \( f(1) = 0.8 \)
- \( g(1) = 0.6 \)

### Step 2: Recursive Definitions
The functions are defined recursively as follows:
- \( f(n+1) = f(n) \cos(g(n)) - g(n) \sin(g(n)) \)
- \( g(n+1) = f(n) \sin(g(n)) + g(n) \cos(g(n)) \)

### Step 3: Calculate \( f(2) \) and \( g(2) \)
Using the initial values:
- \( f(2) = f(1) \cos(g(1)) - g(1) \sin(g(1)) \)
- \( g(2) = f(1) \sin(g(1)) + g(1) \cos(g(1)) \)

Calculating \( f(2) \):
\[ f(2) = 0.8 \cos(0.6) - 0.6 \sin(0.6) \]
Calculating \( g(2) \):
\[ g(2) = 0.8 \sin(0.6) + 0.6 \cos(0.6) \]

### Step 4: General Pattern
We observe that \( f(n) \) and \( g(n) \) can be interpreted as the coordinates of a point on the unit circle, where:
- \( f(n) = r \cos(\theta_n) \)
- \( g(n) = r \sin(\theta_n) \)

where \( r \) is the magnitude and \( \theta_n \) is the angle.

### Step 5: Magnitude Calculation
The magnitude \( r \) can be calculated as:
\[ r^2 = f(n)^2 + g(n)^2 \]
From the initial values:
\[ r^2 = 0.8^2 + 0.6^2 = 0.64 + 0.36 = 1 \]
Thus, \( r = 1 \).

### Step 6: Angle Behavior
The angle \( \theta_n \) evolves as:
\[ \theta_{n+1} = \theta_n + g(n) \]
Since \( g(n) \) is bounded and oscillatory, the angle \( \theta_n \) will eventually cover all angles in the interval \( [0, 2\pi] \).

### Step 7: Limit of \( f(n) \)
As \( n \to \infty \), the values of \( f(n) \) will oscillate between -1 and 1. However, since the angle \( \theta_n \) will become uniformly distributed, the limit of \( f(n) \) will converge to the average value of \( \cos(\theta) \) over a full cycle, which is 0.

### Conclusion
Thus, we conclude that:
\[ \lim_{n \to \infty} f(n) = 0 \]