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calculus anyone? (or trig?!)

Alex

Diamond Member
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?
 
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. 😛
 
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. 😛

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

but for now....BUMP! 😀
 
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. 😛

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

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

 
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. 😛

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

course i was drinking a tad back then. 😉
 
yeah after im done i dare not pursue further enlightenment in calculus lest it make my quietus... 😛
 
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. 😛

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

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

/passes out from shock

 
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?
 
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.
 
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. 😛

 
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
 
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
 
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.
 
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