# Find N values of X1, X2, … Xn such that X1 < X2 < … < XN and sin(X1) < sin(X2) < … < sin(XN)

• Difficulty Level : Medium
• Last Updated : 16 Nov, 2021

Given a number N, the task is to find the N integer values of Xi such that X1 < X2 < … < XN and sin(X1) < sin(X2) < … < sin(XN).
Examples:

Input: N = 5
Output:
X1 = 0 sin(X1) = 0.000000
X2 = 710 sin(X2) = 0.000060
X3 = 1420 sin(X3) = 0.000121
X4 = 2130 sin(X4) = 0.000181
X5 = 2840 sin(X5) = 0.000241
Input: N = 3
Output:
X1 = 0 sin(X1) = 0.000000
X2 = 710 sin(X2) = 0.000060
X3 = 1420 sin(X3) = 0.000121

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Approach: The idea is to use the fractional value of PI(&PI); i.e., Pi = 355/113 as it given best rational value of PI of accuracy being 0.000009%.

```As,
PI = 355/113
=> 113*PI = 355
=> 2*(113*PI) = 710

As sin() function has a period of 2*PI,
Therefore sin(2*k*PI + Y) = sin(Y);```

As per the above equation to get the value X1 < X2 < … < XN and sin(X1) < sin(X2) < … < sin(XN) we must find the value of sin(X) with an increment of 710.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to print all such Xi s.t.``// all Xi and sin(Xi) are strictly``// increasing``void` `printSinX(``int` `N)``{``    ``int` `Xi = 0;``    ``int` `num = 1;` `    ``// Till N becomes zero``    ``while` `(N--) {` `        ``cout << ``"X"` `<< num << ``" = "` `<< Xi;``        ``cout << ``" sin(X"` `<< num << ``") = "``             ``<< fixed;` `        ``// Find the value of sin() using``        ``// inbuilt function``        ``cout << setprecision(6)``             ``<< ``sin``(Xi) << endl;` `        ``num += 1;` `        ``// increment by 710``        ``Xi += 710;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `N = 5;` `    ``// Function Call``    ``printSinX(N);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to print all such Xi s.t.``// all Xi and sin(Xi) are strictly``// increasing``static` `void` `printSinX(``int` `N)``{``    ``int` `Xi = ``0``;``    ``int` `num = ``1``;` `    ``// Till N becomes zero``    ``while` `(N-- > ``0``)``    ``{` `        ``System.out.print(``"X"` `+ num + ``" = "` `+ Xi);``        ``System.out.print(``" sin(X"` `+ num + ``") = "``);` `        ``// Find the value of sin() using``        ``// inbuilt function``        ``System.out.printf(``"%.6f"``, Math.sin(Xi));``        ``System.out.println();``        ``num += ``1``;` `        ``// Increment by 710``        ``Xi += ``710``;``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``5``;` `    ``// Function Call``    ``printSinX(N);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program for the above approach``import` `math` `# Function to print all such Xi s.t.``# all Xi and sin(Xi) are strictly``# increasing``def` `printSinX(N):` `    ``Xi ``=` `0``;``    ``num ``=` `1``;` `    ``# Till N becomes zero``    ``while` `(N > ``0``):` `        ``print``(``"X"``, num, ``"="``, Xi, end ``=` `" "``);``        ``print``(``"sin(X"``, num, ``") ="``, end ``=` `" "``);` `        ``# Find the value of sin() using``        ``# inbuilt function``        ``print``(``"{:.6f}"``.``format``(math.sin(Xi)), ``"\n"``);` `        ``num ``+``=` `1``;` `        ``# increment by 710``        ``Xi ``+``=` `710``;``        ``N ``=` `N ``-` `1``;` `# Driver Code``N ``=` `5``;` `# Function Call``printSinX(N)` `# This code is contributed by Code_Mech`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to print all such Xi s.t.``// all Xi and sin(Xi) are strictly``// increasing``static` `void` `printSinX(``int` `N)``{``    ``int` `Xi = 0;``    ``int` `num = 1;` `    ``// Till N becomes zero``    ``while` `(N-- > 0)``    ``{``        ``Console.Write(``"X"` `+ num + ``" = "` `+ Xi);``        ``Console.Write(``" sin(X"` `+ num + ``") = "``);` `        ``// Find the value of sin() using``        ``// inbuilt function``        ``Console.Write(``"{0:F6}"``, Math.Sin(Xi));``        ``Console.WriteLine();``        ``num += 1;` `        ``// Increment by 710``        ``Xi += 710;``    ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 5;` `    ``// Function Call``    ``printSinX(N);``}``}` `// This code is contributed by SoumikMondal`

## Javascript

 ``
Output:
```X1 = 0 sin(X1) = 0.000000
X2 = 710 sin(X2) = 0.000060
X3 = 1420 sin(X3) = 0.000121
X4 = 2130 sin(X4) = 0.000181
X5 = 2840 sin(X5) = 0.000241```

Time Complexity: O(N)

Auxiliary Space: O(1)

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