Given a value of cos(?) and a variable 
. The task is to find the value of cos(n?) using property of trigonometric functions.
Note: n <= 15.
Examples:
Input : cos(?) = 0.5, n = 10 Output : -0.5 Input :cos(?) = 0.5, n = 3 Output : -0.995523
The problem can be solved using De moivre’s theorem and Binomial theorem as described below:
Using De-Moivre’s theorem, we have:
Now, as the value of both sin(?) and cos(?) is known. Put the values in above equation to get the answer.
Below is the implementation of the above idea:
C++
// CPP program to find the value of cos(n-theta)#include <bits/stdc++.h>#define MAX 16using namespace std;int nCr[MAX][MAX] = { 0 };// Function to calculate the binomial// coefficient upto 15void 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 codeint 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 15static 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 codepublic 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 mathMAX=16nCr=[[0 for i in range(MAX)] for i in range(MAX)]# Function to calculate the binomial# coefficient upto 15def 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 = 16var nCr = Array.from(Array(MAX), () => new Array(MAX));// Function to calculate the binomial// coefficient upto 15function 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 codebinomial();var cosTheta = 0.5;var n = 10;document.write( findCosnTheta(cosTheta, n));</script> |
Output:
-0.5
Time Complexity: O(MAX2)
Auxiliary Space: O(MAX2)
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