What is arcsin and how do I find it on a TI-83 graphing calculator?
602788
2nd, sin^-1 is arcsin.
168710
http://www.graphcalc.com/
429405
go to your graphing calculator and hit the Y= button, then input your problem, it looks like an arch, that goes extremly high.. it starts on -3
986397
dear u are integrating a derivative. so the answer is\ne^(arctan x) 0 through 1\n= e^(π/4) - e^0\n= e^(π/4) - 1
70027
1)-2+3i in trig form we would write it as rcost+rsint;\nr=sqrt[x^2+y^2]=sqrt (4+9)=sqrt(13). This square root is irrational, so that is the most accurate form of writing down, hence we would simply leave it as sqrt(13).\nNow, the angle. Tan t = sint/cost= y/x = 3/-2 = -1.5\nDo arctan on your calculator and obtain the angle. The numbers are not "very nice", so the angle isn't as well.\nWhen you got the angle t, your answer will be, \nrcost + irsin t, where r is the radius sqrt(13) and the angle is the "ugly t" you found by applying arctan to -1.5\n\n2) Present it in the trig form as well, as \n2sqrt(2)[cos3PI/4+isin3PI/4].\nApply De Moivre formula to elevate to the 8 power:\nR(cost+isint)^n= R^n(cos nt+isin nt)\n\n4096 (cos 6PI + isin 6PI)= 4096(1+0)= 4096\n\nAlternatively, elevate first to the second power.\n\n(-2+2i)^2=4 - 8i + 4i^2= 4 - 8i - 4= -8i\n\nNow elevate to the remaining fourth power:\n\n(-2+2i)^8 =  [(-2+2i)^2]^4 = (-8i)^4 = (-8)^4*(i)^4 =\n\n=(-8)^4* 1= (-8)^4 = 4096.\n\nThus, that way or another, the answer to your second question is 4096.\n\nHope that helps
443117
well ive got a TI-83, but i don't think there'e much differnce b/w the two\nI think you go to "mode" and find "Pol" click enter and uve got it.
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
887482
y=arccos(x - 5) \nDomain is 0 ≤ (x - 5) ≤ π      ie 5 ≤ x ≤ π + 5\nRange -1 ≤ y ≤ 1\n\nSo inverse function is\n\nx = arccos(y - 5)\n\nRange is 5 ≤ y ≤ π + 5\nDomain is -1 ≤ x ≤ 1\n\nie y - 5 = cosx \nSo y = cosx + 5\n\nDomain is -1 ≤ x ≤ 1\nRange is 5 ≤ y ≤ π + 5
838358
i can't give you the answer but i used my ti-83 and i can tell you the points of the graph...\n(1,0) (2,4) (3,8) (4,0)\n\nHope this helps
801229
90° < beta < 180° means 2nd quadrant.\nsin beta = 12/13 \nbeta=arcsine(12/13)=67.38° in the 1st quadrant\nor 180°-67.38°=112.62° in the 2nd.\n\ntan beta = 3/4 , sin beta < 0\nsin beta<0 means 3rd or 4th quadrant.\ntan beta>0 means 1st or 3rs quadrant so we are in the 3rd quadrant.\nbeta=arctan 3/4=36.87° in the 1st quadrant. or\n36.87°+180°=216.87° in the 3rd.
509283
In statistics we talked about residuals. a Residual is anestimate on how well your regression line fits your data. The residual is often notes as r^2. I'm not sure on exactly how to find it, but on the TI 83 plus, the residual is a flat line and the data is scattered around it. The data should be spread equally on both sides of the line. the more equal the data is spread, the better your residual line. When you find the "VarsStat1" on the TI 83, the residual value will be there noted as r^2.
880047
I can find this on a TI-84+, is that how you want to get it or do you need an algebraic solution?
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.
726975
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...
918506
Here is your answer - the normalcdf function is used:\nhttp://webserver3.lanecc.edu/math/staff/ShellabD/calculatorCentralFolder/TI-83NormCDF&Invnorm.html
931057
f(x) = sin½x + a \nSo x = sin(½f^(-1)( x)) + a\nx - a = sin(½f^(-1)( x))\nso ½f^(-1)( x) = arcsin(x - a)\nSo f^(-1)( x) = 2 arcsin(x - a) \nDomain: a -1 ≤ x ≤ a + 1\nRange: - π ≤ f^(-1)( x) ≤ π
930538
in the calculator the symbol for arctan is tan^-1 ( tan to the minus 1) '