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  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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#2 Re: Help Me ! » cosine inverse of a number>1 » 2009-02-12 22:17:07

Ya i think I did this in a maths module last year..  No wonder my lecturer gave me fierce look wen I said I don't know..lol

Thank you for your help:)

#3 Help Me ! » cosine inverse of a number>1 » 2009-02-12 15:53:56

charismath
Replies: 2

I was asked abt the cosine inverse of a number greater than 1.. my reflex answer was obviously that the range of the cosine function is from -1 to +1.. but this was apparently not the expected answer!!:/

would the cosine inverse of a number greater than 1 be a a complex number?

#4 Re: Dark Discussions at Cafe Infinity » May I ask somthing? » 2009-02-12 05:15:23

I became a full member just few days ago...YUPIIIIIIIIIIII big_smile

Hope u become one too soon!!;)

#6 Re: Introductions » hi, just joined. » 2009-02-11 15:38:01

Hi Clint..
I have just one Astrophysics module and I must say that one is enough for me to sleep in class..lol.. examinable in May..
Glad to have u here..If I ever need your help, I'll ask!! wink

#9 Re: Jai Ganesh's Puzzles » English language puzzles » 2009-02-09 07:31:11

Romba nandri Ganesh... mikka santhosham..;) Joined Tamil chat room to get this..lol.. Thx a lot..

#10 Re: Jai Ganesh's Puzzles » English language puzzles » 2009-02-08 02:59:04

Wow JaneFairfax is so quick at replying.. Think i'll start a Jane fan club soon;)

#16 Re: Jai Ganesh's Puzzles » English language puzzles » 2009-02-05 05:01:49


oops.. yeah dat ws #496.. Fourier transforms are making me lose my head..lol

#18 Re: Help Me ! » Differientiation help » 2009-02-03 19:10:46

U got it right Helpmath!

Oh.. I'm glad I could help u.. am a newbie here like u..
Try more similar problems.. For practice..;)

#19 Re: Help Me ! » Differientiation help » 2009-02-02 20:12:59

ok lemme try to help u..But i'm in class at the moment so I dnt have time to go through the latex guide..so plz adjust..Copy on paper,ul understand better..

q = ax + by    (1)                   r = bx - ay   (2)
where a and b are constants

lets start with the 1st part, the easiest one., partial diferential (dq/x)y
Look at equ(1). assume that y is a constant. n u surely know that differential of a constant is 0.

Hence (dq/dx)y=a

Coming to the second part (dx/dq)r

you will notice that we now need to xpress x in terms of q and r
To do this take (1) and make y the subject of formula.
y = (q-ax) / b

Now replace in (2)

r = bx -a(q-ax) / b
Multiply the whole thing by b..    br=b^2.x - aq+a^2.x

a^2.x + b^2.x = br + aq

Make x the subjet of formula..
you'l obtain     x=br /(a^2+b^2) + aq/(a^2+b^2)

this equ..  Look at the firs term.. it consists of only constants a,b and variable r.. Since we r diferentiating with respect to q, we take r as constant.. hence differential of the first term willl be 0...

Now no spoon feeding for u.. differentiate yourself the second term with respet to q. you'l get the value for (dx/dq)r

and multiply with (dq/dx)y to get your final answer;)
Try it.. if u still cant do it, i'll give u the final answer..

#20 Re: Jai Ganesh's Puzzles » English language puzzles » 2009-02-02 01:27:07


hmm this one i knew but the others.. pfff.. so many new words am learning thanks to JaneFairfax and Ganesh:)  together with the definitions can u plz use them in sentences (if u have time), so that i know in which contexts they fit..

#22 Re: Jai Ganesh's Puzzles » English language puzzles » 2009-01-31 08:09:39

Wow.. thx.. this forum is amazing!!;) i edited it.. see!!!

#23 Re: Dark Discussions at Cafe Infinity » I saw it! » 2009-01-31 08:06:19

i gt this technical problm today.. abt 4h back.. i didnt knw the problm came from google.. my lil bro keeps downloading useless programs.. i thout he did stg to my mozilla.. i scolded him a lot..
I feel so dumb now..

#25 Re: This is Cool » Mathematical terms in other languages » 2009-01-30 19:04:51

Careless25 I'm a fan of Ganesh and JaneFairfax..

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