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 BBS: The Abacus * HST/DS * Potterville, MI
Date: 03-12-93 (22:41)             Number: 48
From: MICHEL BERTLER               Refer#: 40
  To: CARLTON HOUSTON               Recvd: NO  
Subj: ArcFunctions                   Conf: (35) Quick Basi
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 CH> Does Anyone Out There Have A Good Algorythm For Finding The Angle From
 CH> The sin or cos?

Mathematical functions not intrinsic to GW-Basic can be calculated as
follows:

Function             GW-Basic equivalent
--------             -------------------
Secant               SEC(X)=1/COS(X)

Cosecant             CSC(X)=1/SIN(X)

Cotangent            COT(X)=1/TAN(X)

Inverse Sine         ARCSIN(X)=ATN(X/SQR(-X*X+1))

Inverse Cosine       ARCCOS(X)=ATN(X/SQR(-X*X+1))+
/2

Inverse Secant       ARCSEC(X)=ATN(X/SQR(X*X-1))+SGN(SGN(X)-1)*
/2

Inverse Cosecant     ARCCSC(X)=ATN(X/SQR(X*X-1))+SGN(X)-1)*
/2

Inverse Cotangent    ARCCOT(X)=ATN(X)+
/2

Hyperbolic Sine      SINH(X)=(EXP(X)-EXP(-X))/2

Hyperbolic Cosine    COSH(X)=(EXP(X)+EXP(-X))/2

Hyperbolic Tangent   TANH(X)=(EXP(X)-EXP(-X))/(EXP(X)+EXP(-X))

Hyperbolic Secant    SECH(X)=2/(EXP(X)+EXP(-X))

Hyperbolic Cosecant  CSCH(X)=2/(EXP(X)-EXP(-X))

Hyperbolic Cotangent COTH(X)=EXP(-X)/(EXP(X)-EXP(-X))*2+1

Inverse
Hyperbolic Sine      ARCSINH(X)=LOG(X/SQR(X*X+1))

Inverse
Hyperbolic Cosine    ARCCOSH(X)=LOG(X+SQR(X*X-1))

Inverse
Hyperbolic Tangent   ARCTANH(X)=LOG((1+X)/(1-X))/2

Inverse
Hyperbolic Cosecant  ARCCSCH(X)=LOG(SGN(X)*SQR(X*X+1)+1)/X

Inverse
Hyperbolic Secant    ARCSECH(X)=LOG(SQR(-X*X+1)+1)/X

Inverse
Hyperbolic Cotangent ARCCOTH(X)=LOG((X+1)/(X-1))/2


Michel

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