# include <conio.h>
# include <stdio.h>
unsigned int factorial(int);
void main()
{
unsigned int f;
int a,i,g=0,c=1,j=-1;
clrscr();
scanf("%d",&a);
printf("sine series up to %d terms",a);
for(i=1;i<a;i++)
{
f=factorial(c);
j=j*(-1);
printf(" [(%d(x^%d))/(%d)] ",j,g,f);
if(i!=a)
printf("+");
c=c+2;
g=g+2;
}
getch();
}
unsigned int factorial(int x)
{
int fact=1,i;
for(i=1;i<=x;i++)
fact=fact*i;
return fact;
}
C-language was derived from B-language.
C Language is First Step of Programming Language, Help for C Language you are show the correct answer
in case of the c languages we are very flexible to to solve the problems in the step by step order (because of using the c-functions,other variables,other operators), debugging also so very easy because trace the exact steps which are given in the problem domains.
C is a pop language. C is a case sensetive language. C is motherof all language. C is block structure language. C is a high level language. C is advace of B language. C developed by D.richties in 1972 at AT & T Bell lab in USA. Sachin Bhardwaj 986854722 skbmca@gmail.com
it is not regular language .it is high level language
find the program in c-pgms.blogspot.com
java language moreover solve the problems witch is encounter in c and c++ that s why we use the java language...
Please do.
You will need to use the distributive law to solve discrete series by grouping. The distributive law is a(b + c) = ab + ac. You will be removing the common factors as you go.
Consider any triangle ABC, and let AD be the altitude from A on to BC. Then sin(B) = AD/AB so that AD = AB*sin(B) and sin(C) = AD/AC so that AD = AC*sin(C) Therefore AB*sin(B) = AC*sin(C) or c*sin(B) = b*sin(C) where the lower case letter represents the side opposite the angle with the upper case name. Divide both sides by bc to give sin(B)/b = sin(C)/c. Similarly, using the altitude from B you can show that sin(A)/a = sin(C)/c. Combining with the previous result, sin(A)/a = sin(B)/b = sin(C)/c.
Yes, but you would need to know a degree measure too. [Sin(A)/a] = [Sin(B)/b] = [Sin(C)/c] [a/Sin(A)] = [b/Sin(B)] = [c/Sin(C)]
Just as you would do it manually, I mean there is no predefined 'solve_equation' function in C.
You need to use the sine rule. If the three angles are A, B and C and the sides opposite them are named a, b and c then, by the sine rule, a/sin(A) = b/sin(b) = c/sin(C) Therefore b = a*sin(B)/sin(A) = a*y where y = sin(B)/sin(A) can be calculated and c = a*sin(C)/sin(A) = a*z where z = sin(C)/sin(A) can be calculated. then perimeter = p = a + b + c = a + ay + az = a*(1 + y + z) therefore a = p/(1 + y + z) or a = p/[1 + sin(B)/sin(A) + sin(C)/sin(A)]. Everything on the right hand side is known and so a can be calculated. Once that has been done, b = a*y and c = a*z.
28 The Law of Sines: a/sin A = b/sin B = c/sin C 24/sin 42˚ = c/sin (180˚ - 42˚ - 87˚) since there are 180˚ in a triangle. 24/sin 42˚ = c/sin 51˚ c = 24(sin 51˚)/sin 42˚ ≈ 28
In a triangle ABC, with side a opposite angle A, side b opposite angle B and side c opposite angle C, the sine rule is: sin(A)/a = sin(B)/b = sin(C)/c The cosine rule is: a2 = b2 + c2 - 2bc*cos(A) and, by symmetry, b2 = c2 + a2 - 2ca*cos(B) c2 = a2 + b2 - 2ab*cos(C)
The Law of sines: a/sin A = b/sin B = c/sin CIf the angle C in the triangle ABC is 90 degrees, then the triangle ABC is a right triangle, where c is the measure of the hypotenuse, a is the measure of the leg opposite the angle A, and b is the measure of the leg opposite the angle B.Let us observe what happens when sin C = sin 90 degrees = 1.c/sin C = a/sin A cross multiply;c sin A = a sin C divide by c both sides;(c sin A)/c = (a sin C)/c simplify c on the left hand side;sin A = (a sin C)/c = [(a)(1)]/c = a/csin A = (measure of leg opposite the angle A)/(measure of hypotenuse)From the Law of Cosine we know that cos A= (b^2 + c^2 - a^2)/(2bc). If we substitute a^2 + b^2 for c^2, we have:cos A = (b^2 + (a^2+ b^2) - a^2 )/(2ab)cos A = 2b^2 /2ab simplify;cos A = b/c = (measure of leg adjacent the angle A)/(measure of hypotenuse) Therefore tan A = sin A/cos A =(a/c)/(b/c) = (a/c)(c/b) = a/b = (measure of leg opposite the angle A)/(measure of leg adjacent to angle A). And cot A = cos A/sin A = (b/c)/(a/c) = (b/c)(c/a) = b/a = (measure of leg adjacent to angle A)/(measure of leg opposite the angle A).
To find unknown sides or angles of a triangle. For triangle ABC, if C is a right angle and you are using angle A the side a is the opposite of A, side b is the adjacent side of angle A and c is the hypotenuse. Ex: Sin A = a/c, if you know any 2 you can solve for the 3rd. Cos A = b/c, if you know any 2 you can solve for the 3rd. Tan A = a/b, and again, if you know any 2 you can solve for the 3rd.