(a) in the 17th century, Machin obtained the formula: (pi/4)=4arctan(1/5)-arctan(1/239). Use a calculator to check this formula.\n\n(b) Use the Taylor polynomial approximation of degree 5 to the arctangent function to approximate the value of pi.  (Note: In 1873 William Shanks used this approach to calculate pi to 707 decimal places.  Unfortunately, in 1946 it was found that he made an error in the 528th place.  Currently, several billion decimal places are known.)\n\n(c) Why do the two series for arctangent converge so rapidly here?\n\n(d) Now we prove Machin's formula using the tangent addition formula\ntan(A+B) =(tanA+tanB)/(1-tanAtanB).\n(i) Let A=arctan(120/119) and B=(-arctan(1/239)) and show that arctan(120/119)-arctan(1/239)=arctan1.\n(ii) Let A=B=arctan(1/5) and show that 2arctan(1/5)=arctan(5/12).\nUse a similar method to show that 4arctan(1/5)=arctan(120/119).\n(iii) Derive Machin's formula.
132068
milan.milanovic.org/math/english/pi/machin.html \n\nCheck this site
56975
LHS:sin(-π/4) = -sin(π/4)  since sin(-x)=-sin(x)  (odd function)\n\nRHS:sin(5π+4) = sin(π +π/4) = - sin(π/4) since sin(π+x) = -sin(x) (the sin is -ve in the 3rd quardon)\n\nLHS=RHS\n\nsin(5π/4)=sin(-π/4)\n\nUsing the unit circle 5π/4 is a angle on the 3rd quadron having the same y component to this one of -π/4 on the 4th quadron\nThe symmetrie  proove the both 5π/4 and -π/4 have the same  -ve sine\n\nAs 3π/4 in the 2nd quadron and π/4 on the 1st quadron having the same +ve sine
518767
Let the first term of the series be a.Therefore,the second term would be ar,r being the common ratio.and the fifth term is ar^4\nDividing ,the fifth term  ar^4 by the second term ar,we get.\nar^4/ar=(81/64)/3  or,r^3=27/64.Therefore r=3/4.\nPutting the valu of r in the secon term i.e. ar=3,we get a=3/r=3/(3/4)=4.\nTherefore,the series is 4,3......81/64.......\nit is apparent that the trms of the series would have the values increasing and it would not come to an end because it has attained the value of  more than an unity.\nThe sum of the series would be infinity.\nAs the sum of the series is infinity,the sum of firs 10 trms of sum of any number of trms for that matter,and the sum of the complete sries would be infinity as any number subtratd from infinity gives  result of infinity.
635893
3 is a very, very rough estimate. It's calculated using a Taylor series. The best info I've found is here\nhttp://www.jimloy.com/geometry/pi.htm
546493
lim ((arctan(x))-(pi/2))*(x) = (lim(arctan(x)-lim(pi/2))*(x)\nx->+infinity                            is this step ok?\n\nanyway: looks like it's 0*x, so 0
147876
The formula is as follows:\n\nC\n- = pi\nd\n\nC=the circumference\nd=the diameter\nand pi=3.14\n\nThe next step is to put all the known information into the equation:\n\nC\n- = 3.14\n14\n\nOnce you have that, you can multiply 14 by both sides to get C alone. Like this:\n\n14( C/14) = (3.14)14\n\n14 is canceled on the side with C, and the you take 3.14 and multiply it by 14 to get 43.96. So now the equation looks like this:\n\nC = 43.96\n\nAnd there you have your answer. The circumference of the circle with a diameter of 14 is 43.96\n\nHope this helped.
561823
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.
358723
(x/2)*(100-x^2)^(1/2)+\n50*arcsin(x/10)\n\nSorry it's on two lines, yahoo cuts it off if I leave it on one. arcsin is the inverse sine function.
417601
This is a tricky one. You have to use a couple of steps to get this one. First you use a u-substituion to "simplify" it and then use a trig substitution.\n\nIf you let u=cos(x) then du=-sin(x) then the integral becomes\n&#92;int -1/(1+u^2) du.\n\nNow let u=tan(s) and du=sec^2(s) and the integral is\n&#92;int -sec^2(s)/(1+tan^2(s)) ds = &#92;int -sec^2(s)/sec^2(s) ds = &#92;int -1 ds.\n\nThe integral is -s + C where C is any constant. We now have to undo the steps to get it back in terms of x. First, we undo the trig substituion. Since u=tan(s) then the integral is\n\n-arctan(u)+C\n\nSince u=cos(x) we get\n\n-arctan(cos(x))+C
509021
a) y= 10^(tan (πθ))\n= e^(ln10*tan(πθ))\n= e^u where u = ln10*tan(πθ\ndy/dθ =dy/du * du/dθ\n= e^u * ln10*πsec²(πθ)\n= ln10*πsec²(πθ)*e^(ln10*tan(πθ))\n= ln10.πsec²(πθ)*10^(tan (πθ))\n\nb) y= √[tln(t^4)]\n= √[4t*ln(t)]\n= 2√[t*ln(t)]\n= 2u^½ where u = t*ln(t)\ndy/dt = dy/du * du/dt\n= u^-½ * (ln(t) + 1)\n= (ln(t) + 1)/√[tln(t)]\n\nc) y = arctan(arcsin (x^½))\n= arctan u where u = arcsin v where v = x^½\ndy/dx = dy /du * du/dv * dv/dx\n= 1/(1+ u²) * 1/√[1 - v²] * ½x^(-½)\n= 1/[2√(1 - x)*(1 + (arcsin (x^½))²)]\n\nd)xe^y = y - 1\ne^y + xe^y * dy/dx = dy/dx\nSo dy/dy - xe^y * dy/dx = e^y\nie dy/dx (1 - xe^y) = e^y\ndy/dx = e^y/ (1 - xe^y)\n= 1/(e^(-y) - x)
224237
You would need more and more terms of the series to get an accurate result for the larger values of the argument.  Sum the positive and negative ones separately, from least to greatest absolute value, before adding those two subtotals together.\n\nHow does the Pade approximation do?  For example see\n\nhttp://math.fullerton.edu/mathews/n2003/pade/PadeApproximationMod/Links/PadeApproximationMod_lnk_15.html\n\nP subscript{5, 5} (x) for arcsin(x) = the fraction given by:\n\nnumerator: (922320 x - 717780 x^3 + 69049 x^5)\ndenominator: (15 (61488 - 58100 x^2 + 9675 x^4))\n\nIf you are using C, why not just call the\n\ndouble asin(double x);\n\nfunction in <math.h> ? :-)
524837
A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
491137
This type of problem will eventually seem EASY for you, so hang in there.  Let's begin by stressing that "a circle with radius 5" is simple to draw -- I recommend marking the 4 points that are "5 units" above, below, right and left of the center point.\n\nThat is, mark (-6,7) on your graph paper, then sketch four more points at (-1,7), (-11,7), (-6, 2) and (-6,12).  "Connect the dots" by drawing all 4 quarter-circle arcs between the points.\n\nThe equation for any circle -- in "rectangular coordinates" (which means (x,y)) -- is normally:  (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.\n\nBut what you're hoping to find is the equation of a circle in "polar coordinates" (which means (r,theta)).\n\nThe reason a sample triangle was drawn inside the given circle (I'm assuming you're looking at a textbook) is to illustrate the relationships among x, y, theta and r.  Let's inspect, using rules from trignometry:\n\nx = r cos (theta) + a\ny = r sin (theta) + b\nr = sqrt ((x-a)^2 + (y-b)^2)\ntheta = Arctan ((y-b)/(x-a))\n\nwhere (x,y) is any point on the circle, (a,b) is the circle's center, and r is the radius.\n\nThe formula for theta that I just provided may be precisely what you are looking for.  It works simply by first finding the "slope" of the line connection (x,y) and (a,b), then taking the inverse tangent of that slope, which 'converts' it to an "angle."\n\nHope this helps!
255367
1. Make sure the angle of interest is in radians.  Radians are the "natural" (non)-units of angles, because they relate the angle measurement to the length of arc the angle cuts off on a unit circle.  So 2*Pi radians are equivalent to 360 degrees.  Basically, if you have that sucker in degrees, multiply it by 2 * Pi / 360.  Unit conversion is done. \n \n2. Make sure the angle is between -Pi and Pi.  Because sine is a periodic function, with period 2*Pi, just subtract or add 2*Pi enough times to get the angle between those two points.\n\nnow for the meat of the problem:\n\nsin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ... + \n+ (-1)^(n-1)*x^(2n-1)/(2n-1)! + ...\n\nExample: sin(15)\n\nStep 1: 15 degrees = 15*(2*Pi/360) radians = Pi/12 radians \n \nStep 2: I need Pi/12 in decimal (let's see, Pi=3.14159 (memorize that much, it's pretty easy... "Pi when calculating sine, 3 point 1 4 1 5 9!")) \nPi/12 approx = .261799 (we only had 6 sig figs, so we can't add any more than that) \n \nNow to the series: \n         first term:  x=.261799 \n         1st estimate: .261799 \n         second term: -x^3/3! = -.0029905  \n         2nd estimate: .258809 \n         third term: x^5/5! = .0000102     \n         3rd estimate: .258819 \n         fourth term: -x^7/7! = -.167*10^-7  \n         stop here - past precision\n\nThe answer is positive \nsin(15) = 0.258819\nsin(18) = 0.309017
101831
Bottom line: pi is the ratio between the circumference and the diameter of a circle.\n\nAn infinite sum is one way of approximating pi: just add up lots and lots of the elements in the sum, and you get there eventually.  The infinite sum itself, of course, is pi with utter precision.\n\nThere is a trick (read about it in Science News) where you can compute a specific hexidecimal digit of pi, say the 10948573929384756th digit, without computing any others.  Cute, not useful to me.\n\nThe trick that is used most was developed by Srinvasa Ramanujan, an incredibly brilliant, mostly self-taught mathematician from India.  He developed a recursive formula that converges to pi.  Each iteration roughly doubles the number of decimal digits precision of pi in the answer.  I am given to understand that this is the tool used to do many of the "record breaking length" calculations of pi.  A Google search should dig it up.  I'd dig it up, but I can't find my translation of his notebooks...\n\nYou might want to look up the Buffon's Needle problem in a good book on probability theory...  It is a wonderful, stochastic, computation of the value of pi.  It's also slower than slow to converge...  But FUN to a math freak!
384365
cosx=tanx\ncosx=sinx/cosx\ncos^2(x)=sinx\n1-sin^2(x)=sinx    {pythagoras}\nsin^2(x)+sinx-1=0\nuse quadratic formula\nsinx=(-1+or-sqrt(5))/2\n=0.6180339888\nor -1.6180339888\nsine has to lie between +/-1\nsinx= 0.6180339888\nx=arcsin(0.6180339888)\n  =38.17270763\n+ve value for sine lies in 1st and 2nd\nquadrants \ntherefore,x=38.17270763 degrees or\n141.8272924 degrees\n\nthese two values were checked in the \noriginal equation and were found \nto be correct\n\ni hope that this helps
439446
OK, first let's try to figure out the arctan part.  Arctan(-1) means the angle whose tangent is -1, right?  And the tangent is the sine divided by the cosine.  So you're looking for angles whose sines and cosines are equal, but with opposite signs.  They point out to the right and down, or out to the left and up, at 45-degree angles.  That means the angles are 135 degrees and 315 degrees, or 3pi/4 and 7pi/4 radians.\n\nNow, the question asked you to find the sines of these two angles.  The sine of 3pi/4 is sqrt(2)/2, and the sine of 7pi/4 is -sqrt(2)/2.  So your answer is +/- sqrt(2)/2.