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Re: [CPROG] invers cosinus



----- Original Message -----
From: "Kasper Bonne" <sslug@sslug>
> ----- Original Message -----
> From: <sslug@sslug>
> > Jeg leder efter en invers cosinus funktion, der beregner antal grader
> > imellem 0 og 180.
> >
> > Funktionen double acos(double x) fra math.h c-standard-lib beregner
> > arc-cosine of x, er det det samme som invers cosinus? - Den giver output
i
> > radiantal, der åbenbart kun ligger imellem 0 og 90 grader.
>
> Problemet er at f.eks. både `cos( pi/4 )´ og `cos( 3*pi/4 )´ begge giver
det
> samme resultat nemlig `0.5´ og når du så tager `acos( 0.5 )´ får du kun
det
> mindste tal: `pi/4´ for at få det andet skal du sige `pi - acos( 0.5 )´.
>
> For så at få det i grader skal du gange med `180/pi´ ~ `57.2957795131´
>
> DVS at løsningen i grader på `acos( 0.5 )´ er lig med: `acos( 0.5 ) *
> 180/pi´ eller `(pi - acos( 0.5 )) * 180/pi´

UPS - jeg tror lige jeg fik en lille hjerneblødning der :)
Jeg prøver lige igen med de rigtige tal...

Problemet er at f.eks. både `cos( pi/4 )´ og `cos( 7*pi/4 )´ begge giver det
samme resultat nemlig `sqroot(2)/2´ og når du så tager `acos( sqroot(2)/2 )´
får du kun det mindste tal: `pi/4´ for at få det andet skal du sige `2*pi -
acos( sqroot(2)/2 )´.

For så at få det i grader skal du gange med `180/pi´ ~ `57.2957795131´

DVS at løsningen i grader på `acos( sqroot(2)/2 )´ er lig med: `acos(
sqroot(2)/2 ) * 180/pi´ eller `(2*pi - acos( 0.5 )) * 180/pi´

Det andet er for sinus -- bortset fra at der stadig ikke giver 0.5 -- jeg
ved ikke hvad jeg havde gang i, men dette her skulle være okay.

/Kasper

- - - - - -
Get PGP key from www.keyserver.net - Key ID: 0xDE21F438
Name = Kasper Bonne
Fingerprint = B583 0E69 166A 15EF B6F8 B248 7558 DB7C DE21 F438
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