Programa para Hallar el valor de cos(nΘ)

Posted on by Rudeus Greyrat

Dado un valor de cos(Θ) y una variable  n . La tarea es encontrar el valor de cos(nΘ) usando la propiedad de las funciones trigonométricas.
Nota: n <= 15.
Ejemplos : 
 

Input : cos(Θ) = 0.5, n = 10
Output : -0.5

Input :cos(Θ) = 0.5, n = 3
Output : -0.995523

El problema se puede resolver usando el teorema de De moivre y el teorema del binomio como se describe a continuación:
Usando el teorema de De-Moivre, tenemos : 

\begin{align*} \cos(n\theta)+\iota \sin(n\theta)&=(\cos \theta+\iota \sin \theta)^n\\ &=\cos^n \theta +\binom{n}{1} \cos^{n-1} \theta (\iota \sin \theta)+\binom{n}{2} \cos^{n-2} \theta (\iota \sin \theta)^2+\\ & \binom{n}{3} \cos^{n-3} \theta (\iota \sin \theta)^3+ \cdots \\ \end{align*} Equating the result to real part to get the value of $\cos n\theta$ finally, we have\\ $\cos (n \theta)=\binom{n}{0}\cos^{n}\theta \sin^0 \theta-\binom{n}{2}\cos^{n-2}\theta \sin^2 \theta+\binom{n}{4}\cos^{n-4}\theta \sin^4 \theta- \cdots$ \\ \\ As we have value of $\cos \theta$, \\ we can find value of $\sin \theta $\\ $$\sin \theta=\sqrt{1-\cos^2 \theta}$$

Ahora, como se conoce el valor tanto de sin(Θ) como de cos(Θ). Ponga los valores en la ecuación anterior para obtener la respuesta.
A continuación se muestra la implementación de la idea anterior: 
 

C++

// CPP program to find the value of cos(n-theta)
 
#include <bits/stdc++.h>
 
#define MAX 16
 
using namespace std;
 
int nCr[MAX][MAX] = { 0 };
 
// Function to calculate the binomial
// coefficient upto 15
void binomial()
{
    // use simple DP to find coefficient
    for (int i = 0; i < MAX; i++) {
        for (int j = 0; j <= i; j++) {
            if (j == 0 || j == i)
                nCr[i][j] = 1;
            else
                nCr[i][j] = nCr[i - 1][j] + nCr[i - 1][j - 1];
        }
    }
}
 
// Function to find the value of cos(n-theta)
double findCosnTheta(double cosTheta, int n)
{
    // find sinTheta from cosTheta
    double sinTheta = sqrt(1 - cosTheta * cosTheta);
 
    // to store required answer
    double ans = 0;
 
    // use to toggle sign in sequence.
    int toggle = 1;
 
    for (int i = 0; i <= n; i += 2) {
        ans = ans + nCr[n][i] * pow(cosTheta, n - i) *
                              pow(sinTheta, i) * toggle;
        toggle = toggle * -1;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    binomial();
 
    double cosTheta = 0.5;
 
    int n = 10;
 
    cout << findCosnTheta(cosTheta, n) << endl;
 
    return 0;
}

Java

// Java program to find the value of cos(n-theta)
 
class GFG
{
    static int MAX=16;
    static int[][] nCr=new int[MAX][MAX];
 
// Function to calculate the binomial
// coefficient upto 15
static void binomial()
{
    // use simple DP to find coefficient
    for (int i = 0; i < MAX; i++) {
        for (int j = 0; j <= i; j++) {
            if (j == 0 || j == i)
                nCr[i][j] = 1;
            else
                nCr[i][j] = nCr[i - 1][j] + nCr[i - 1][j - 1];
        }
    }
}
 
// Function to find the value of cos(n-theta)
static double findCosnTheta(double cosTheta, int n)
{
    // find sinTheta from cosTheta
    double sinTheta = Math.sqrt(1 - cosTheta * cosTheta);
 
    // to store required answer
    double ans = 0;
 
    // use to toggle sign in sequence.
    int toggle = 1;
 
    for (int i = 0; i <= n; i += 2) {
        ans = ans + nCr[n][i] * Math.pow(cosTheta, n - i) *
                            Math.pow(sinTheta, i) * toggle;
        toggle = toggle * -1;
    }
 
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    binomial();
 
    double cosTheta = 0.5;
 
    int n = 10;
 
    System.out.println(String.format("%.5f",findCosnTheta(cosTheta, n)));
}
}
// This code is contributed by mits

Python3

# Python3 program to find the value of cos(n-theta)
 
import math
MAX=16
nCr=[[0 for i in range(MAX)] for i in range(MAX)]
 
# Function to calculate the binomial
# coefficient upto 15
def binomial():
     
    # use simple DP to find coefficient
    for i in range(MAX):
        for j in range(0,i+1):
            if j == 0 or j == i:
                nCr[i][j] = 1
            else:
                nCr[i][j] = nCr[i - 1][j] + nCr[i - 1][j - 1]
 
# Function to find the value of cos(n-theta)
def findCosnTheta(cosTheta,n):
     
    # find sinTheta from cosTheta
    sinTheta = math.sqrt(1 - cosTheta * cosTheta)
     
    # to store the required answer
    ans = 0
     
    # use to toggle sign in sequence.
    toggle = 1
    for i in range(0,n+1,2):
        ans = ans + nCr[n][i]*(cosTheta**(n - i)) *(sinTheta**i) * toggle
        toggle = toggle * -1
    return ans
     
# Driver code
if __name__=='__main__':
    binomial()
    cosTheta = 0.5
    n = 10
    print(findCosnTheta(cosTheta, n))
     
# this code is contributed by sahilshelangia

C#

// C# program to find the value of cos(n-theta)
using System;
public class GFG{
    static int MAX=16;
    static int [,]nCr=new int[MAX,MAX];
  
    // Function to calculate the binomial
    // coefficient upto 15
    static void binomial()
    {
        // use simple DP to find coefficient
        for (int i = 0; i < MAX; i++) {
            for (int j = 0; j <= i; j++) {
                if (j == 0 || j == i)
                    nCr[i,j] = 1;
                else
                    nCr[i,j] = nCr[i - 1,j] + nCr[i - 1,j - 1];
            }
        }
    }
 
    // Function to find the value of cos(n-theta)
    static double findCosnTheta(double cosTheta, int n)
    {
        // find sinTheta from cosTheta
        double sinTheta = Math.Sqrt(1 - cosTheta * cosTheta);
 
        // to store required answer
        double ans = 0;
 
        // use to toggle sign in sequence.
        int toggle = 1;
 
        for (int i = 0; i <= n; i += 2) {
            ans = ans + nCr[n,i] * Math.Pow(cosTheta, n - i) *
                                Math.Pow(sinTheta, i) * toggle;
            toggle = toggle * -1;
        }
 
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        binomial();
 
        double cosTheta = 0.5;
        int n = 10;
        Console.WriteLine(findCosnTheta(cosTheta, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript program to find the value of cos(n-theta)
 
MAX = 16
 
var nCr = Array.from(Array(MAX), () => new Array(MAX));
 
// Function to calculate the binomial
// coefficient upto 15
function binomial()
{
    // use simple DP to find coefficient
    for (var i = 0; i < MAX; i++) {
        for (var j = 0; j <= i; j++) {
            if (j == 0 || j == i)
                nCr[i][j] = 1;
            else
                nCr[i][j] = nCr[i - 1][j] + nCr[i - 1][j - 1];
        }
    }
}
 
// Function to find the value of cos(n-theta)
function findCosnTheta(cosTheta, n)
{
    // find sinTheta from cosTheta
    var sinTheta = Math.sqrt(1 - cosTheta * cosTheta);
 
    // to store required answer
    var ans = 0;
 
    // use to toggle sign in sequence.
    var toggle = 1;
 
    for (var i = 0; i <= n; i += 2) {
        ans = ans + nCr[n][i] * Math.pow(cosTheta, n - i) *
                            Math.pow(sinTheta, i) * toggle;
        toggle = toggle * -1;
    }
 
    return ans.toFixed(1);
}
 
// Driver code
binomial();
var cosTheta = 0.5;
var n = 10;
document.write( findCosnTheta(cosTheta, n));
 
</script>
Producción: 

-0.5

 

Publicación traducida automáticamente

Artículo escrito por sahilshelangia y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

Deja una respuesta

Tu dirección de correo electrónico no será publicada. Los campos obligatorios están marcados con *