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Hyperbolic trig


Crash

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Ok, so i just started hyperbolic trig at uni and although its sweet, i mean just the basics with a different name but im having trouble with a couple of q's

 

sinh(2x)= 2sinh(x)cosh(x), for all xE|R

 

d/dx(tanh^2(3x) = 6sech^2(3x)tanh(3x), xE|R

 

8sin^4(x)= 3-4cos(2x) + cos(4x), xE|R

 

i need the working, actually even just a method or explnation would be much appreciated,

 

Cheers

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for the first one start one the right with the definition of sinh and vosh in terms of exponetials, the answer is then one step away.

 

recall tanh = sinh/cosh and that you can differentiate sinh and cosh easily from the defintions as exponentials, so now apply the usual rules of differntiation.

 

is the third supposed to be sin and cos? again just start from the right and simplify cos(2x) in terms of sin

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So im guessing its

 

 

what is "it"?

 

=(exp(x) - exp(-x)/2)^2

=(exp(-2x) + exp(2x)/2)

 

no, how do you square things again? and why are you squaring something?

 

=2sinhcosh

Is that right?

 

no.

 

2sinh(x)cosh(x)=2(e^x-e^-x)(e^x+e^-x)/4 = (e^2x - e^ -2x)/2 = sinh(2x)

 

that is how it goes.

 

also what is the derivative of cosh^2(3x) and shine?

 

you can differentiate sinh and cosh right? look at their definitions

 

sinhx=(e^x-e^-x)/2

 

you can differentiate that.

 

given you can differentiate f(x) what is the derivative of f(3x)^2? nothing to do with hyperbolics, jsut your statndard calculus.

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i was squaring it because i started from the other side, is sinh(2x) not = (exp(x)-exp(-x)/2)^2 i was just thinking because sin(2x) can be img,(exp(ix))^2

 

sorry i meant 'so im guessing its like this', i was in a rush cause i had a lecture started in 5 mins.

 

Cheers for the help to dude, to tired to do it tonight. ill prob be back tommorrow though ;)

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ok, i got that down but im now stuck on damned complex roots,

z=-16+i16sqrt3

 

so i polarise, get 32cissqrt3

z=rexp(i*theta)

 

sqrt(3)/5 for the nth principle root and 2 for the radius but thats where my mind just dies on me, its too late to be doing this!!!!! Do maths degrees get easier or harder after second year?

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my second year was horrible lol. too many bad module choices. absolutely flopped on one of the stats module which brough my average way down. and then I seem to get good marks in stuff I have no interest in and which i have no plans pursuing further. sucks.

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I would say the second year combines difficult work with lack of freedom of choice. Though surely hyperbolic trig is taught in high school? OF course it may be I a confusing my further maths with single maths A level. In any case hyperbolic trig and identities will not be the most difficult material you meet by a very long way, but you will when you meet the more demanding material be mathematically more mature and it won't seem as hard to learn.

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I would say the second year combines difficult work with lack of freedom of choice. Though surely hyperbolic trig is taught in high school? OF course it may be I a confusing my further maths with single maths A level. In any case hyperbolic trig and identities will not be the most difficult material you meet by a very long way, but you will when you meet the more demanding material be mathematically more mature and it won't seem as hard to learn.

 

Nah, i think NZ schooling is gay for lack of a better explanatiojn. plus we just got a new qualification system (NCEA) which is total balls, it failed in several other countries. we got introduced to hyperbolic trig in first year courses, but that was all, im only a first year but i taking the advancing second year paper :( what level of mathamatics is best for physics major? i suppose the best level you can attain?

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You'll find the hardest thing is to figure out how to "do" questions mathematically, ie with rigour. It won't matter what the topic is. For instance, what is your answer to shwoing sinh(2x)=2sinh(x)cosh(x)? Mine would read:

 

We know that sinh(y):=(e^y - e^{-y})/2 and cosh(y:)=(e^y + e^{-y})/2, and thus it follows that

 

2sinh(x)cosh(x)=2(e^x-e^{-x})(e^x+e^{-x})/4 = (e^2x-e^{-2x})/2=sinh(2x)

 

I imagine you wouldn't write as many words, or any at all. But it'll be a good habit to start writing in words what you know, what the defintions are and what you want to show, then it often pops out how to do it. And if not anyone who comes to mark it will at least know what the problem is and where you need help.

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