Cosine or Sine function for a math 'sunspot' problem?
138327
I derive essentially the same function as you do:\nf(x) = 70+50*cos(((2*pi)/11)*(x-1))\n\nThis function is the actual function of sunspot activity with the first maximum occuring in year 1 and maximums occuring every 11 years thereafter including 1750 and 1948.\n\nThe reason your calculator is giving you wrong results is that it is in "degree" mode rather than in "radian" mode. (Don't feel too bad, it took me almost an hour to figure out I had my calculator in the wrong mode too.)
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
765813
There's a pretty good online grapher at http://id.mind.net/~zona/ezGraph/ezGraph.html that you can use.  It works basically like a graphing calculator.\n\nIt appears this function should be an upside-down sign curve with an amplitude of 10 and a vertical shift of 9, which means the center of the graph will be at 9, it will go up to 19 and down to -1.  It's not 100% clear what you mean by what you have written after sin.  If you mean the sine of 2pi/15 times x (since the minus zero is irrelevent), you would have a period of 15, so you'd complete 1 complete cycle at 15 radians; the halfway point would be 7.5 radians.
923039
email me, calcu_lust@yahoo.com, i'll write out a nice solution.\n\nEdit: Go to my website, and aforementioned nice solution is typed out as a .pdf for you to download, with all the correct symbols and such:\nhttp://geocities.com/calcu_lust/yahooans.pdf
938105
learn your identities well and as far as possible reduce all ratios to basic ratios of sine and cosine
748672
using Maclaurin series you find the series for sin and cos and then use long division for the Maclaurin series of tan
566422
cos150=\n\ncos(180-30)=\n\ncos(p-30)=\n\n-cos30=\n\n-sin60
459338
Go to http://mathworld.wolfram.com/Cosine.html
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
407911
So you are doing some numerical analysis... if you have a spreadsheet program you can use that. Use the trapezoid rule or Simpson's rule for some small increment in x.  Your spreadsheet program is able to evaluate arctan.  The second question can be evaluated with Newton's method.
924923
Ah Hi again dear \n\nwell \nSin(∏ - α) = -Sinα \nSo we have ; \nsin 210   =  sin( 180 + 30 ) = - sin30 = -1/2\ncos210   = cos( 180 + 30 ) = - cos120 =  - √3/2\ntan 210 = tan( 180 + 30 ) =√ 3/3\ncot 210  = cot ( 180 + 30 ) = √3 \n\nGood Luck dear
731974
You should be doing your own homework.\n\nYou need to know the diameter of Jupiter in meters.\nConvert that number of meters to AU.\nUse the definition of "sin" (or "tan").\nOne second of arc is 1/3600 of a degree.
983185
Take the first few derivatives of sin -- it is cyclical and you will see a pattern.  After you see the pattern, you will be able to say exactly what the 53rd derivative is, without having to actually take the derivative 53 times!\n\n~ ♥ ~
997471
IF you want to try and get it in that EXACT form dicatated by Mathematica, I suggest you start with u=1+sin(x) and then solve for x in terms of u and do some resubstitution.  You'll get the arcsin function, which becomes a polynomial in terms of another dummy variable when you start the first u*dv substitution.  Notice I said "FIRST", because to get it in the form that you want, you'll have to do it about 6-10 times.  Also, you'll have to do some "right triangle simplification" throughout the problem as well.  Of course, then you get to factor it.  Have at it.  I did this very thing, stopping at the 3rd u*dv substitution because it is just ridiculous.  \nI did, however, get several of the terms.  I think if you decide to do this, you're going to need lots of paper.  I was doing it symbollically in a CAD program to keep things organized and such, and it was already 5 pages.  Oh, and not to mention that this "answer" given by Mathematica isn't even a closed form "answer".  It left it in terms of a polylogarithm, which is still an integral hidden in notation.  \n\nThe point is that the fuction (among others) is best integrated numerically because there is no useful closed form for this function.  This is what I did and it took me 2 minutes.  \n\nx^2/(1+sin(x))=x^2 - x^3 + x^4 - (5/6)x^5 + (2/3)x^6 - (61/120)x^7 + (17/45)x^8 - (277/1008)x^9 .................\n\nIntegrate termwise, which is fricking easy.  \n\n(1/3)x^3 - (1/4)x^4 + (1/5)x^5 - (5/36)x^6 + (2/21)x^7......\n\nNotice it is not undefined (like the original function) and the error term goes to 10^-7 after the first 5 terms.  I'd say that's as accurate as anyone would ever really need to be, and if not, they can always expand this function further.\n\nTell your uncle he's a bastard just for me.  He's teasing you.
276671
(e^(cose^e+sin^e)/2)^pi\nhahaha ..just kidding
1047641
since 1= |sin z|^2+|cos z|^2\nthen |cos Z|=sqrt(1-(sin x cosh y +i cos x sinh y)^2)\nAnyway if you were looking for a more elaborated one using the "e" definitions for the hyperbolics function I could do it but naah
461039
sin2A = 2sinA.cosA\nSo,\nsin2A = 2*12/13*5*13\n      = 120/169\n\ngood luck!!