calculus anyone? (or trig?!)

Alex

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Oct 26, 1999
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wtf?!?!?!? :| :|

this is seriously frustrating me... i REALLY DONT GET the whole quadrant thingy for like evaluating inverse trig expressions... anyone think they can explain it to me easily?

thx!

-frangstrated (hehe)

EDIT:
evaluate:
1) arctan(tan 3pi /4)
2) cos (arcsin 1/2)
3) cos (arctan 2 + arctan 3)

if anyone has any clues?
 

Fausto

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Nov 29, 2000
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I'd love to help, but I long ago used a butter knife to excise the part of my brain that knew calculus.

The memories were just too painful. :p
 

Alex

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Oct 26, 1999
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Originally posted by: Fausto1
I'd love to help, but I long ago used a butter knife to excise the part of my brain that knew calculus.

The memories were just too painful. :p

hehe ya only a few more years and i get to do that myself.. :p

but for now....BUMP! :D
 

Fausto

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Nov 29, 2000
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Originally posted by: franguinho
Originally posted by: Fausto1
I'd love to help, but I long ago used a butter knife to excise the part of my brain that knew calculus.

The memories were just too painful. :p

hehe ya only a few more years and i get to do that myself.. :p

but for now....BUMP! :D
I had a year of calc plus a year of Physical Chemistry (which involves a ton of very complex calc). Nearly killed me.

 

Hammer

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Oct 19, 2001
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Originally posted by: Fausto1
I'd love to help, but I long ago used a butter knife to excise the part of my brain that knew calculus.

The memories were just too painful. :p

lol my memory of it didn't last past the following semester. :p

course i was drinking a tad back then. ;)
 

Kev

Lifer
Dec 17, 2001
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what exactly dont you understand? maybe i could dust off my brain...
 

Alex

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Oct 26, 1999
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yeah after im done i dare not pursue further enlightenment in calculus lest it make my quietus... :p
 

Fausto

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Nov 29, 2000
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Originally posted by: Hammer
Originally posted by: Fausto1
I'd love to help, but I long ago used a butter knife to excise the part of my brain that knew calculus.

The memories were just too painful. :p

lol my memory of it didn't last past the following semester. :p

course i was drinking a tad back then. ;)
You? Drink a tad? :Q

/passes out from shock

 

Alex

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Oct 26, 1999
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Originally posted by: maladroit
what exactly dont you understand? maybe i could dust off my brain...

evauluate:
1) arctan(tan 3pi /4)
2) cos (arcsin 1/2)
3) cos (arctan 2 + arctan 3)

any clues?
 

Turin39789

Lifer
Nov 21, 2000
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that looks more like trig to me than calc. Calc is fun. Trig is the great darkness from the north that threatens the world of men.
 

Fausto

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Originally posted by: franguinho
hehe well either way im pretty screwed then is the bottom line :(
When the going gets tough, the tough start drinking.

It is Friday after all. :p

 

vtqanh

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Jan 4, 2001
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arctan(tan 3pi/4) = arctan(tan -pi/4) = -pi/4
cos (arcsin 1/2) = cos (pi/6) = sqrt(3)/2
I haven't figured out number 3 yet
 

vtqanh

Diamond Member
Jan 4, 2001
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ok, number 3:
arctan x + arctan y = arctan ((x+y) /(1-xy))
thus, arctan 2 + arctan 3 = arctan (5/(1-6)) = arctan (-1) = -pi/4
cos (-pi/4) = cos (pi/4) = sqrt(2)/2
 

Kyteland

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Dec 30, 2002
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Originally posted by: franguinho
Originally posted by: maladroit
what exactly dont you understand? maybe i could dust off my brain...

evauluate:
1) arctan(tan 3pi /4)
2) cos (arcsin 1/2)
3) cos (arctan 2 + arctan 3)

any clues?

Ok, I'll try to help with the first one.

Start off by claculating the equation inside the parenthesis. tan 3pi /4. You should have this value memorized from trig tables and it is -1.

you know tan(x) = sin(x)/cos(x) so tan(3pi /4) = sin(3pi /4)/cos(3pi /4)
you know sin(3pi /4) = sqrt(2)/2 <--- trig tables anyone?
you know cos(3pi /4) = -sqrt(2)/2 <---- tables again
therefore tan(3pi /4) = -1

so now you need to finish off with the last part of the equation.
You are left with arctan(-1) = ??

arctan is the inverse tangent function. If x=arctan(y) then tan(x)=y by definition.

you have arctan(-1)=??? so by the above tan(???)=-1. Oh, but wait!!! we just calculated above that tan(3pi /4)=-1. Problem solved!!! Oops, but that values isn't defined for the range of the acrtan() function. arctan() is only defined in the range of -pi/2<arctan(x)<pi/2. Since arctan has a period of pi you subtract that from your answer above.

3pi /4 - pi = -pi/4 and that is in the proper range so it is you answer.

You can use a similar method to solve all of the rest of the questions.