Robinson Crusoe decides to explore his isle. On a sheet of paper he plans the following process.
His hut has coordinates `origin = [0, 0]`. From that origin he walks a given distance `d` on a line
that has a given angle `ang` with the x-axis. He gets to a point A.
(Angles are measured with respect to the x-axis) 
From that point A he walks the distance `d` multiplied by a constant `distmult` on a line that 
has the angle `ang` multiplied by a constant `angmult` and so on and on.
We have `d0 = d`, `ang0 = ang`; then `d1 = d * distmult`, `ang1 = ang * angmult` etc ...
Let us suppose he follows this process n times.
What are the coordinates `lastx, lasty` of the last point?
The function `crusoe` has parameters;
- n        : numbers of steps in the process
- d        : initial chosen distance
- ang      : initial chosen angle in degrees
- distmult : constant multiplier of the previous distance
- angmult  : constant multiplier of the previous angle
`crusoe(n, d, ang, distmult, angmult)` should return
`lastx, lasty` as an array or a tuple depending on the language.
### Example:
`crusoe(5, 0.2, 30, 1.02, 1.1)` ->
The successive `x` are : `0.0, 0.173205, 0.344294, 0.511991, 0.674744, 0.830674` (approximately)
The successive `y` are : `0.0, 0.1, 0.211106, 0.334292, 0.47052, 0.620695` (approximately)
and
```
lastx: 0.8306737544381833
lasty: 0.620694691344071
```
### A drawing:
![alternative text](https://i.imgur.com/SGHbb6Vl.png)
Successive points:
- x: `0.0, 0.9659..., 1.8319..., 2.3319..., 1.8319...`
- y: `0.0, 0.2588..., 0.7588..., 1.6248..., 2.4908...`
### Note
Please could you ask before translating: some translations are already written and published when/if the kata is approved.
from math import sin,cos,pi
def crusoe(n, d, ang, dis_tmult, ang_mult):
    x = 0
    y = 0
    for i in range(n):

        x += d * cos((ang/180)*pi)
        y += d * sin((ang/180)*pi)
        d = d * dis_tmult
        ang = ang * ang_mult
    
    return (x,y)

from math import sin,cos,pi
def crusoe(n, d, ang, dis_tmult, ang_mult):
    x = 0
    y = 0
    for i in range(n):

        x += d * cos((ang/180)*pi)
        y += d * sin((ang/180)*pi)
        d = d * dis_tmult
        ang = ang * ang_mult
    
    return (x,y)
