怎样计算任意角度的正弦值阿？

C_Stranger 2008-09-07 03:26:55

库函数太大了，不适合在小系统里面跑！
精度要求不高，能精确到小数点后面4为就够了！哪位大虾用帮忙写段c语言实现的代码！先谢谢了！

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xiao_dang 2008-09-07

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效率最高的莫过于用硬件来实现，我有个朋友就研究这个，哈哈

baihacker 2008-09-07

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1.查表后插值.
2.taylor展开.
3.其它级数展开.

C_Stranger 2008-09-07

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谢谢ls！
另外，有效率高一点的代码吗？

mLee79 2008-09-07

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查表加插值, 这样的精度应该够了, 改用 float 应该6位精度没问题...



#include <stdio.h>

#include <math.h>

#include <float.h>



double XSin( double x )

{

	static const double XSinTbl[] = {

		0.00000000000000000  , 0.015707317311820675 , 0.031410759078128292 , 0.047106450709642665 , 0.062790519529313374 ,

		0.078459095727844944 , 0.094108313318514325 , 0.10973431109104528  , 0.12533323356430426  , 0.14090123193758267  ,

		0.15643446504023087  , 0.17192910027940955  , 0.18738131458572463  , 0.20278729535651249  , 0.21814324139654256  ,

		0.23344536385590542  , 0.24868988716485479  , 0.26387304996537292  , 0.27899110603922928  , 0.29404032523230400  ,

		0.30901699437494740  , 0.32391741819814940  , 0.33873792024529142  , 0.35347484377925714  , 0.36812455268467797  ,

		0.38268343236508978  , 0.39714789063478062  , 0.41151435860510882  , 0.42577929156507272  , 0.43993916985591514  ,

		0.45399049973954680  , 0.46792981426057340  , 0.48175367410171532  , 0.49545866843240760  , 0.50904141575037132  ,

		0.52249856471594880  , 0.53582679497899666  , 0.54902281799813180  , 0.56208337785213058  , 0.57500525204327857  ,

		0.58778525229247314  , 0.60042022532588402  , 0.61290705365297649  , 0.62524265633570519  , 0.63742398974868975  ,

		0.64944804833018377  , 0.66131186532365183  , 0.67301251350977331  , 0.68454710592868873  , 0.69591279659231442  ,

		0.70710678118654757  , 0.71812629776318881  , 0.72896862742141155  , 0.73963109497860968  , 0.75011106963045959  ,

		0.76040596560003104  , 0.77051324277578925  , 0.78043040733832969  , 0.79015501237569041  , 0.79968465848709058  ,

		0.80901699437494745  , 0.81814971742502351  , 0.82708057427456183  , 0.83580736136827027  , 0.84432792550201508  ,

		0.85264016435409218  , 0.86074202700394364  , 0.86863151443819120  , 0.87630668004386369  , 0.88376563008869347  ,

		0.89100652418836779  , 0.89802757576061565  , 0.90482705246601958  , 0.91140327663544529  , 0.91775462568398114  ,

		0.92387953251128674  , 0.92977648588825146  , 0.93544403082986738  , 0.94088076895422557  , 0.94608535882754530  ,

		0.95105651629515353  , 0.95579301479833012  , 0.96029368567694307  , 0.96455741845779808  , 0.96858316112863108  ,

		0.97236992039767667  , 0.97591676193874743  , 0.97922281062176575  , 0.98228725072868872  , 0.98510932615477398  ,

		0.98768834059513777  , 0.99002365771655754  , 0.99211470131447788  , 0.99396095545517971  , 0.99556196460308000  ,

		0.99691733373312796  , 0.99802672842827156  , 0.99888987496197001  , 0.99950656036573160  , 0.99987663248166059  ,

		1.00000000000000000  };

#define XPI			(3.1415926535897932384626433832795)

#define XENTRY		(100)

#define XINCL		(XPI/2/XENTRY)

	int s = 0 , n;

	double dx , sx , cx;

	if( x < 0 )

		s = 1 , x = -x;

	x = fmod( x , 2 * XPI );

	if( x > XPI )

		s = !s , x -= XPI;

	if( x > XPI / 2 )

		x = XPI - x;

	n = (int)( x / XINCL );

	dx = x - n * XINCL;

	if( dx > XINCL / 2 )

		++n , dx -= XINCL;

	sx = XSinTbl[n];

	cx = XSinTbl[XENTRY-n];

	x = sx + dx*cx - (dx*dx)*sx/2

		- (dx*dx*dx)*cx/6 

		+ (dx*dx*dx*dx)*sx/24

		;

	

	return s ? -x : x;

}



double XCos( double x )

{

	return XSin( x + XPI/2 );

}



#define TESTM		(100.)

#define TESTN		(123456)

#define TESTINCL	(TESTM/TESTN)



#define TESTE1		(1E-6)

#define TESTE2		(1E-6)



int main()

{

	double diff  , diffs , m;

	double x , r1 , r2;

	int    i;	



	for( diff = diffs = 0 , i = 0 , x=0; i <= TESTN+10; ++i , x += TESTINCL )

	{

		r1 = sin(x);

		r2 = XSin(x);



		diffs += fabs(r1-r2);



		if( fabs(r1-r2) > diff )

		{

			diff = fabs(r1-r2);

			m = x;

		}		

	}

	printf( "%.18f %.18e %.18e %.18f %.18f\n" , m , diffs/TESTN  , diff , sin(m) , XSin(m) );

	

	for( diff = diffs = 0 , i = 0 , x=0; i <= TESTN+10; ++i , x += TESTINCL )

	{

		r1 = cos(x);

		r2 = XCos(x);

		

		diffs += fabs(r1-r2);

		

		if( fabs(r1-r2) > diff )

		{

			diff = fabs(r1-r2);

			m = x;

		}		

	}

	printf( "%.18f %.18e %.18e %.18f %.18f\n" , m , diffs/TESTN  , diff , cos(m) , XCos(m) );

return 0;

}

oo_v_oo 2008-09-07

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就用插值
LZ都已经学会有问题上CSDN，想是搞一个插值算法应该不在话下
哈哈

_石头_ 2008-09-07

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写了一个，不过没考虑负数！



float user_sin (float angle)

{

	float x, t,temp;

	int i;

	i=(int)angle/360;

	temp=(float)(angle-360*i);

	if(temp<=90)

		angle=temp;

	else if(90<temp<=270)

		angle=180-temp;

	else

		angle=temp-360;

	x=(float)(angle*3.141592/180);

	t=(float)(x-x*x*x/6+x*x*x*x*x/120);

	return(t);

}