📄 sinx.asm
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******************************************************
* 用泰勒级数开展开式计算一个角度的正弦值 *
*sin(x)=x(1-x*x/2*3(1-x*x/4*5(1-x*x/6*7(1-x*x/8*9))))*
******************************************************
.title "sinx.asm"
.mmregs
.def start
STACK: .usect "STACK",10
start: STM #STACK+10,SP
LD #d_x,DP
ST #6487H,d_x ;x-->d_x
CALL sin_start
end: B end
sin_start:
.def sin_start
d_coeff .usect "coeff",4
.data
table: .word 01C7H ;c1=1/(8*9)
.word 030BH ;c2=1/(6*7)
.word 0666H ;c3=1/(4*5)
.word 1556H ;c4=1/(2*3)
d_x .usect "sin_vars",1
d_squr_x .usect "sin_vars",1
d_temp .usect "sin_vars",1
d_sinx .usect "sin_vars",1
c_1 .usect "sin_vars",1
.text
SSBX FRCT
STM #d_coeff,AR5
RPT #3
MVPD #table,*AR5+
STM #d_coeff,AR3
STM #d_x,AR2
STM #c_1,AR4
ST #7FFFH,c_1
SQUR *AR2+,A ;A=x^2
ST A,*AR2 ;(AR2)=x^2
||LD *AR4,B ;B=1
MASR *AR2+,*AR3+,B,A ;A=1-x^2/72,T=x^2
MPYA A ;A=T*A=x^2(1-x^2/72)
STH A,*AR2 ;(d_temp)=x^2(1-x^2/72)
MASR *AR2-,*AR3+,B,A ;A=1-x^2/42(1-x^2/72),T=x^2(1-x^2/72)
MPYA *AR2+ ;B=x^2(1-x^2/42(1-x^2/72))
ST B,*AR2 ;(d_temp)=x^2(1-x^2/42(1-x^2/72))
||LD *AR4,B ;B=1
MASR *AR2-,*AR3+,B,A ;A=1-x^2/20(1-x^2/42(1-x^2/72))
MPYA *AR2+ ;B=x^2(1-x^2/20(1-x^2/42(1-x^2/72)))
ST B,*AR2 ;(d_temp)=B
||LD *AR4,B ;B=1
MASR *AR2-,*AR3+,B,A ;A=1-x^2/6(1-x^2/20(1-x^2/42(1-x^2/72)))
MPYA d_x ;B=x(1-x^2/6(1-x^2/20(1-x^2/42(1-x^2/72))))
STH B,d_sinx ;sin(theta)
RET
.end
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