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Re: [CPROG] invers cosinus



ligger imellem 0 og 90 grader.
Yes, og tusind tak for det pæne svar.
God weekend ;-) Marie Louise

> 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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