First of all you have to insert the following line in your code:
#include
sin is defined function which takes argument fo double type and returns double type:
sin(x)
If you write something like:
double x, y;
cout << "Enter value for argument for sin/n";
cin >> x;
cout << "/nResult is: " << y;
The same for sqrt which is square root of x (sqrt(x));
The function pow has two arguments (pow(x, y)) of doubletype and in returns the value of double type. For instance,
pow(2, 3) in mathematical language can be written as 23.
You don't. You can include math.h though, and use its functions eg. sin, cos, tg, sqrt, etc.
Math is a pre-defined class in the package java.lang. It contain various predefined functions in it that help the users to carry out various mathematical operations in their programs. Various Math Functions are:: Math.pow(x,y) => x^y Math.sqrt(x) => Square root of 'x' Math.tan(x),sin(x),cos(x) => tan, sin, cos of 'x' respectively there are also other functions like Math.ceil,Math.floor,Math.mod,etc.
It's called a sine wave because the waveform can be reproduced as a graph of the sine or cosine functions sin(x) or cos (x).
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.
#include- Standard Input /Output Functionsclearerr()fclose()feof()ferror()fflush()fgetc()fgetpos()fgets()fopen()fprintf()fputc()fputs()fread()freopen()fscanf()fseek()fsetpos()ftell()fwrite()getc()getchar()gets()perror()printf()putc()putchar()puts()remove()rename()rewind()scanf()setbuf()setvbuf()sprintf()sscanf()tmpfile()tmpnam()ugetc()vfprintf()vprintf()vsprintf()#include-Standard Mathematical Functionsabs()acos()asin()atan()atan2()ceil()cos()cosh()div()exp()fabs()floor()fmod()frexp()labs()ldexp()ldiv()log()log10()modf()modf()pow()sin()sinh()sqrt()tan()tanh()#include- Standard String Handling Functions-charactersatof()atoi()atol()isalnum()isalpha()iscntrl()isdigit()isgraph()islower()isprint()ispunct()isspace()isupper()isxdigit()memchr()memcmp()memcpy()memmove()memset()strcat()strchr()strcmp()strcoll()strcpy()strcspn()strerror()strlen()strncat()strncmp()strncpy()strpbrk()strrchr()strspn()strstr()strtod()strtok()strtol()strtoul()strxfrm()tolower()toupper()#include-Date and Time Functionsasctime()clock()ctime()difftime()gmtime()localtime()mktime()strftime()time()#include- Memory Functionscalloc()free()malloc()realloc()#include-(Other Functions-exit())#include-(Other Functions-Standard Library Functions)C standard library
sin 105 = sin (60+45) = sin60cos45 + cos60sin45sin 105 = ((sqrt(3)/2)((sqrt(2)/2)) + ((1/2)((sqrt(2)/2)))sin 105 = (sqrt(6) + sqrt(2)) / 4
0.5*cos(x)/sqrt(sin(x))
sin(75) = sin(45 + 30) = sin(45)*cos(30) + cos(45)*sin(30) = [1/sqrt(2)]*[sqrt(3)/2] + [1/sqrt(2)]*[1/2] = 1/[2*sqrt(2)]*[sqrt(3) + 1] that is [sqrt(3) + 1] / [2*sqrt(2)]
You don't. You can include math.h though, and use its functions eg. sin, cos, tg, sqrt, etc.
cos(195) = cos(180 + 15) = cos(180)*cos(15) - sin(180)*sin(15) = -1*cos(15) - 0*sim(15) = -cos(15) = -cos(60 - 45) = -[cos(60)*cos(45) + sin(60)*sin(45)] = -(1/2)*sqrt(2)/2 - sqrt(3)/2*sqrt(2)/2 = - 1/4*sqrt(2)*(1 + sqrt3) or -1/4*[sqrt(2) + sqrt(6)]
Because it contains definitions of mathematical functions such as cos, sin, tan, power (pow) and so on. Thus if you want to use mathematical functions in your programs you have to include math.h
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
cos = sqrt(1 - sin^2)
11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
sqrt(2)
1 cot(theta)=cos(theta)/sin(theta) cos(45 degrees)=sqrt(2)/2 AND sin(45 degrees)=sqrt(2)/2 cot(45 deg)=cos(45 deg)/sin(deg)=(sqrt(2)/2)/(sqrt(2)/2)=1
63.3333333333You can search for any mathematical expression, using functions such as: sin, cos, sqrt, etc. You can find a complete list of functions here.RadDegx!Invsinlnπcoslogetan√AnsEXPxy()%AC789÷456×123−0.=+