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Trigonometric Functions


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#1 alisson64

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Posted 19 February 2006 - 07:31 PM

Hi all !

I am a newbie with some quez .. Please help



I have a rather odd problem with my cp 300. I dont know how to type a function as the following : f(x)= 1+arctan(x)-arccos (x)

The problem is that I cant find how to write the -arccos,arcsin,arcsec,arccsc in the manual..I would be greatful if you can help with that .. :rant:


I wonder if their is a direct way to write Arctan,SEC,cosec,arcsin,arccos ..etc without rewriting.The hp49g has a special func called trig for those.
Is their something similar to that ??? :hammer:

I mean even hyperbolic trig functions..
ThanX


#2 The_AFX_Master

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Posted 19 February 2006 - 11:20 PM

on the virtual keypad, tap on "TRIG", now you see

SIN SIN^-1
COS COS^-1
TAN TAN^-1

in bold letter, arcsin, arccos,arctan if you want sec, cosec, cotan... you know that

sec = 1/cos x = (cos (x))^-1 (not confuse with cos^-1(x) that are arccos
cosec = 1/sin x =(sin (x))^-1 (not confuse with sin^-1(x) that are arcsin

the same with cotangent

#3 Kilburn

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Posted 20 February 2006 - 08:19 AM

Define sec(x)=1/cos(x)
Define csc(x)=1/sin(x)
Define cot(x)=1/tan(x)

Define sec^-1(x)=cos^-1(1/x)
Define csc^-1(x)=sin^-1(1/x)
Define cot^-1(x)=tan^-1(1/x)

(Replace ^-1 by the -1 superscript:go in abc, click on MATH and select the -1 symbol)

OK? ;)

#4 alisson64

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Posted 22 February 2006 - 10:40 PM

on the virtual keypad, tap on "TRIG", now you see

SIN SIN^-1
COS COS^-1
TAN TAN^-1

in bold letter, arcsin, arccos,arctan if you want sec, cosec, cotan... you know that

sec = 1/cos x = (cos (x))^-1 (not confuse with cos^-1(x) that are arccos
cosec = 1/sin x =(sin (x))^-1 (not confuse with sin^-1(x) that are arcsin

the same with cotangent


ThanX Master very much ...but I want adirect solution without rewriting Please :D

Define sec(x)=1/cos(x)
Define csc(x)=1/sin(x)
Define cot(x)=1/tan(x)

Define sec^-1(x)=cos^-1(1/x)
Define csc^-1(x)=sin^-1(1/x)
Define cot^-1(x)=tan^-1(1/x)

(Replace ^-1 by the -1 superscript:go in abc, click on MATH and select the -1 symbol)

OK? ;)


A Killer solution Kilburn !! Can you elaborate more :D I am Zero in Basic yet :banghead:

It is a program which I should write and put in memo to be recalled from user defined funX later !! Isnt It?

Please elaborate more :banghead:

#5 Lovecasio

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Posted 23 February 2006 - 01:36 AM

Hi. :)
You don't need to access the Program application. You can do it in Main. IN order for the function to appear in the CATALOG tab, you should create the function in "library" folder.
In Main application input the following:
Definde library\sec(x)=1/cos(x)
...
Then to recall this function, in Main or other app, simply type sec(x) <span class=EXE' /> or find it in CATALOG tab (virtual keyboard)/User

#6 Kilburn

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Posted 23 February 2006 - 09:45 AM

;) But a program can also be useful sometimes.

DelVar library\sec,library\csc,library\cot,library\sec^-1,library\csc^-1,library\cot^-1
Define library\sec(x)=1/cos(x)
Define library\csc(x)=1/sin(x)
Define library\cot(x)=1/tan(x)
Define library\sec^-1(x)=cos^-1(1/x)
Define library\csc^-1(x)=sin^-1(1/x)
Define library\cot^-1(x)=tan^-1(1/x)

It should be OK. I added the line DelVar.... to delete all the variables who have these name, because Define hates name conflicts! :roflol: As Lovecasio says, your functions must be defined in the library folder to be usable in the soft keypad. (in 'cat' select 'User' in the combo-box)

Of course, you can also define sech, csch, coth, sech^-1, csch^-1, coth^-1 B)
But for the hyperbolic functions, I'm not sure... Is is the same formula?

sech(x)=1/cosh(x)
csch(x)=1/sinh(x)
coth(x)=1/tanh(x)?



EDIT:
OK, I wrote the program. I could not be hard to implement sech, csch, coth, etc... Just don't forget to add library\sech... to the DelVar line. ;)

http://kilburn.free....igo_install.mcs

Enjoy! ^_^


#7 alisson64

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Posted 23 February 2006 - 07:13 PM

;) But a program can also be useful sometimes.

DelVar library\sec,library\csc,library\cot,library\sec^-1,library\csc^-1,library\cot^-1
Define library\sec(x)=1/cos(x)
Define library\csc(x)=1/sin(x)
Define library\cot(x)=1/tan(x)
Define library\sec^-1(x)=cos^-1(1/x)
Define library\csc^-1(x)=sin^-1(1/x)
Define library\cot^-1(x)=tan^-1(1/x)

It should be OK. I added the line DelVar.... to delete all the variables who have these name, because Define hates name conflicts! :roflol: As Lovecasio says, your functions must be defined in the library folder to be usable in the soft keypad. (in 'cat' select 'User' in the combo-box)

Of course, you can also define sech, csch, coth, sech^-1, csch^-1, coth^-1 B)
But for the hyperbolic functions, I'm not sure... Is is the same formula?

sech(x)=1/cosh(x)
csch(x)=1/sinh(x)
coth(x)=1/tanh(x)?

EDIT:
OK, I wrote the program. I could not be hard to implement sech, csch, coth, etc... Just don't forget to add library\sech... to the DelVar line. ;)

http://kilburn.free....igo_install.mcs

Enjoy! ^_^


ThanXXXXX :bow: :bow: :bow: ThanXXXXX Kilburn and Lovecasio :bow: :bow:

Merci Beaucoup :bow: :bow: :thumbsup:

#8 TacoFred

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Posted 01 March 2006 - 02:55 AM

sinh(x) = (e^x-e^-x)/2
cosh(x) = (e^x+e^-x)/2

tanh(x) = sinh(x)/cosh(x)

sech(x) = 1/cosh(x)
csch(x) = 1/sinh(x)
coth(x) = cosh(x)/sinh(x)

#9 Kilburn

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Posted 01 March 2006 - 10:30 AM

sinh(x) = (e^x-e^-x)/2
cosh(x) = (e^x+e^-x)/2

tanh(x) = sinh(x)/cosh(x)

sech(x) = 1/cosh(x)
csch(x) = 1/sinh(x)
coth(x) = cosh(x)/sinh(x)


Perfect! So juste add this ^_^

Define library\sech(x)=1/cosh(x)
Define library\csch(x)=1/sinh(x)
Define library\coth(x)=1/tanh(x)
Define library\sech^-1(x)=cosh^-1(1/x)
Define library\csch^-1(x)=sinh^-1(1/x)
Define library\coth^-1(x)=tanh^-1(1/x)


#10 alisson64

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Posted 01 March 2006 - 05:13 PM

Perfect! So juste add this ^_^

Define library\sech(x)=1/cosh(x)
Define library\csch(x)=1/sinh(x)
Define library\coth(x)=1/tanh(x)
Define library\sech^-1(x)=cosh^-1(1/x)
Define library\csch^-1(x)=sinh^-1(1/x)
Define library\coth^-1(x)=tanh^-1(1/x)




But the system does not recocnize those funX as indigenous or its own funX therefor you can not use them in cas sys calcukations :banghead: :banghead: their must be some way to make the sys to recocnize them as indigenous or internal .... Kilburn.. This must be a big programing miss from casio ???? :blink: :blink:

They must fix it in the comming OS 3 :rant: Nobody should resort to rewriting trig funX to suit cp300 need
it should be the opposite..All sort of trigFunX and operations like conversions should be seperatly incorporated in the new OS 3 like the trigfunction in hp49 and much more because cp300 is a math learning instrument.. Amath teacher - a friend of min - was horrifide by the bad trigfunX in cp300 which he suppose to evaluate and recommand as a mathlearning instrument to many schools had told me that he wont recommand it if casio does not make it better than hp49+ in the commimg OS3 :unsure: I have to admitt that he has a point their :blink: I realy thought it was missunderstanding but it seems to me a misshapp in programming :blink: What do you think ????ThanXXXX for all replays

#11 GameAddins

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Posted 01 March 2006 - 05:57 PM

on the virtual keypad, tap on "TRIG", now you see

SIN SIN^-1
COS COS^-1
TAN TAN^-1

in bold letter, arcsin, arccos,arctan if you want sec, cosec, cotan... you know that

sec = 1/cos x = (cos (x))^-1 (not confuse with cos^-1(x) that are arccos
cosec = 1/sin x =(sin (x))^-1 (not confuse with sin^-1(x) that are arcsin

the same with cotangent


on the virtual keypad now tap "hyp" (right above sin), for hyperbolic, and you see:

sinh sinh^-1
cosh cosh^-1
tanh tanh^-1

Did you guys miss that?

#12 alisson64

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Posted 01 March 2006 - 10:42 PM

on the virtual keypad now tap "hyp" (right above sin), for hyperbolic, and you see:

sinh sinh^-1
cosh cosh^-1
tanh tanh^-1

Did you guys miss that?



Hi GameAddin

No body missed anything !!!! :) try to read all the disscusion :D

can you solv f(x) = arcsec(x)\arccsc(x) +arccoth(X)-3coth(x) .....etc.. without rewriting??? :banghead: :banghead:
Their is no way for direct input for such trigfunX in cp300... and that is a big programming miss in the cas.

ThanX

Hi Kilburn !!
The following are the difinitions of the hyp.trig funX

coth(x)=1/tanh(x)
sech(x)=1/cosh(x)
csch(x)=1/sinh(x)
acoth(x)=atanh(1/x)
asech(x)=acosh(1/x)
acsch(x)=asinh(1/x)

They are the same as the normal trigfunX...
Best regards




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