How do I write the code to solve y=(2xcosx)^1/2 / (sinx)^1/2 ?

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Using math library, what is y=(2xcosx)^1/2 / (sinx)^1/2 ?

Posted 10-Oct-23 9:36am

Marvin Ochieng

Updated 12-Oct-23 9:42am

Deeksha Shenoy

v2

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jeron1 10-Oct-23 15:54pm

Maybe take a look at functionality included with this header.
https://cplusplus.com/reference/cmath/

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Rick York · Answer 1 · 2023-10-10T14:33:00

I recommend that you start with a tutorial like this one : cplusplus.com tutorial[^].

The part that is "^1/2" is probably the square root. You can obtain that by calling the function sqrt[^]. You will also need these two functions : cos[^] and sin[^].

By reading the tutorial you should learn out to call functions and utilize the computation operators for multiplication and division.

CPallini · Answer 2 · 2023-10-10T20:20:00

Solution 2

You can rewrite your expression as

y(x) = ((2 x cosx)/sinx)^(1/2)

Then, in C++

C++

y = sqrt( 2 * x * cos(x) / sin(x))

Try

C++

#include <iostream>
#include <cmath>

int main()
{
	double x = M_PI/3;
	
	double y = sqrt( 2 * x * cos(x) / sin(x));
	
	std::cout << "x = " << x << ", y(x) = " << y << std::endl;
	
}

Posted 10-Oct-23 20:20pm

CPallini

Updated 10-Oct-23 20:21pm

v2

Comments

Kenneth Haugland 12-Oct-23 8:09am

You could also replace cos(x)/sin(x) with either cot(x) or 1/tan(x), but I dont know if that will be better.

merano99 12-Oct-23 11:03am

You wanted to say cos(x)/sin(x) = tan(x)?

Richard Deeming 12-Oct-23 11:14am

That's not right - cos/sin = 1/tan, as Kenneth said.

Going back to high-school geometry:
* cos = A/H
* sin = O/H
* tan = O/A

So: cos/sin === (A/H)/(O/H) === A/O === 1/tan

merano99 12-Oct-23 11:41am

yes, that's right, I read it wrong.

merano99 · Answer 3 · 2023-10-11T09:50:00

Changing the formula and taking the square root only once has already been mentioned as a tip. In addition, you should take care that no division by zero occurs.

C++

double mycalc(double x) 
{
    double sin_x = std::sin(x);

    if (sin_x == 0.0) {
        return std::numeric_limits<double>::quiet_NaN();
    }

    double temp = 2.0 * x * std::cos(x) / sin_x;

    if (temp < 0.0) {
        return std::numeric_limits<double>::quiet_NaN();
    }

    return std::sqrt(temp);
}

Perhaps you can write it with tan(x) = sin(x)/cos(x) more performant.

How do I write the code to solve y=(2xcosx)^1/2 / (sinx)^1/2 ?

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