/*
Aim: Program to Find Value of sin(x) using Expansion Series Given Below:
sin(x) = x - x3/3! + x5/5! - x7/7!........
Developer: Devharsh Trivedi
Web: www.knowcrazy.com
*/
#include <stdio.h>
#include <math.h>
main()
{
float base, pwr, sum, c=1, m=2, i=3, g, h;
printf("\nEnter the base value: ");
scanf("%f", &base);
printf("\nEnter the power value: ");
scanf("%f", &pwr);
sum = base;
ab:
m = m * i;
h = pow(-1, c);
g = pow(base, i);
sum = sum + (h * g) / m;
i = i + 2;
c++;
m = m * (i - 1);
if (i <= pwr)
goto ab;
printf("\n\nSum = %f", sum);
getch();
}
the Taylor series of sinx
#include double x, y;y = sin (x);
What is a sinusoidal wave? This is a wave that appears to have curves. AC current/voltage. If you see a wave on a ossiloscope of what our AC (Alternating current) mains voltage that will be the answer to the question. DC (direct current) does not appear to have the same qualitys
The limit as x approaches zero of sin(x) over x can be determined using the squeeze theorem.For 0 < x < pi/2, sin(x) < x < tan(x)Divide by sin(x), and you get 1 < x/sin(x) < tan(x)/sin(x)That is the same as 1 < x/sin(x) < 1/cos(x)But the limit as x approaches zero of 1/cos(x) is 1,so 1 < x/sin(x) < 1which means that the limit as x approaches zero of x over sin(x) is 1, and that also means the inverse; the limit as x approaches zero of sin(x) over x is 1.You can also solve this using deriviatives...The deriviative d/dxx is 1, at all points. The deriviative d/dxsin(x) at x=0 is also 1.This means you have the division of two functions, sin(x) and x, at a point where their slope is the same, so the limit reduces to 1 over 1, which is 1.
the Taylor series of sinx
The word sine, not sinx is the trigonometric function of an angle. The answer to the math question what is the four series for x sine from -pi to pi, the answer is 24.3621.
2
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.
It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)
From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.
It seems that you can't express that integral in terms of a finite number of commonly used functions. In the Wolfram Alpha site (input: "integral cos sin x"), you can find the first few terms of an infinite series expansion.
integration of (sinx)^1/2 is not possible.so integration of root sinx is impossible
If you mean 1 - sinx = 0 then sinx = 1 (sin-1) x = 90
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.