i used taylor series to calc sine inverse but it is giving big error for higher values. devise other formula?
490983
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> ? :-)
911476
5sin² x +13sin x - 6 = 0 This is a quadratic in sinx. Solve by factoring (if possible)\n\n(5sinx - 2)(sin x + 3) = 0\n\nOnly possibility is sin x = 0.4 as |sinx| ≤ 1 in real numbers. So sin x = -3 is disallowed.\n\nThus x ≈ 0.4115, π - 0.4115\n≈ 0.4115, 2.730
335034
Divide the power series. You can use the fact that sin^2z=(1/2)(1-cos(2z)) to get the series expansion easier.\n\nModified:\nDo long division. The initial z^2 divides into 1 to give an initial 1/z^2. After subtracting, the linear term of exp(z) will give a (4/3)z. Then the z^2 goes into that (4/3)z^(-1) times. So your residue is 4/3.\n\nBTW, use z not x for that denominator.
907694
I'm not sure if I get your question right, but I think you're referring to lookup tables (or LUTs).  I think the easiest way to explain the use of these, is through an example from the old days.  Twenty years ago (or so), it took a long time to calculate the sine of an angle.  So if you needed to make a lot of calculations with sines in them, what you did was: calculate the sines of each of the 360 degrees (or more if you needed a better precision) and store the results in an array.  If you then needed the sine of n, all you needed to do is look up the value of the n'th entry in the array --which was a lot faster and therefore efficient than calculating the value.\n\nOf course, this only helped if you needed to calculate a sine more than 360 times, otherwise you wouldn't win back the overhead of pre-calculating all those sines.\n\nHTH.\n\nPS: a conclusion of what I wrote should be that using arrays with[in] loops does NOT *always* improve efficiency.
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
839872
Okay lets try this one ;\n\n2sin²  – 3sinx  + 1 →\n( sin x – 1)(2sin x – 1) = 0\nSin x – 1 = 0  →  sinx = 1 → x =∏ /2 \nX = 2∏ +∏ /2 , …., x = 2k∏ +∏/2 , ( k Є Z ) \n2sin x – 1 = 0  → sinx = 1/2 → x = ∏/6\nAs sin (∏ -∏/6) = sin ∏/6 = 1/2 \nSo ; \nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\nGood Luck dear
765649
switch the variables.\n\nStart with\n\ny = sin(0.5x-3)\n\nand solve for x.\n\nto get rid of  the sin, we apply arcsin or sin^-1 (on calculators)\n\narcsin(y) = arcsin(sin(.5x - 3))\n\nthis gives\n\narcsin(y) = .5x - 3\n\nx = 2*[arcsin(y) + 3]
880226
it is not what you specified\n\nIt cannot be solved without identities\n\nsin ^5 t = sin^4 t sin t\n= (1-cos^2 t)^2 sin t\n\nnow let cost = x\nso diffentiate to get - sint dt = dx\n\nso we need to integrate\n\n-(1-x^2)^2 dx\n\n= -dx + 2 x^2 dx - x^4 dx\n\nintegrating we get\n\n-x + 2x^3/3 - x^5/5 + C\n= - cos t + 2 cos^3 t/3 - cos^5 t /5 + C\n\nwhere C is constant of integration
369033
Most handheld calculators just use Taylor series expansions to approximate the value.  For example, if the angle x is in radians, then\n\ncos(x) = 1 - (1/2) x^2 + (1/24) x^4 - (1/720) x^6 + ...\nsin(x) = x - (1/6) x^3 + (1/120) x^5 - (1/5040) x^7 + ...\ntan(x) = x + (1/3) x^3 + (2/15) x^5 + (17/315) x^7 + ...\n\nYour calculator has a polynomial (maybe of degree 50?) written into its memory which can calculate sine and cosine to a remarkable accuracy -- but it's only an approximation in the end.\n\nOther more powerful programs, such as Mathematica, MATLAB, or even Maple use slightly more sophisticated ideas.  They have a built in database which contain the *exact* value of the trigonometric functions for certain angles x.  For instance, they would know that cos(pi/3) = 1/2 -- something which is not obvious from the Taylor series expansion above!
842198
I think you mean\n\n(sin^4 t - cos^4 t )/ sin 2t where t is for O\n= (sin ^2 t+ cos^2t)(sin^2 - cos ^ 2 t) as (a+b)(a-b) = a^2 - b^2\n= 1. (- cos 2t)/sin 2t  using sin^2t + cos^2 t = 1 and cos^2 t- sin^2 t = cos 2t\n= - cot 2t
970213
Use the Chain Rule repeatedly:\n\nOriginal function: y = (tan(sin x))^2\n\ndy/dx = 2 tan(sin x) (d/dx) tan(sin x)\n\n= 2 tan(sin x) (sec(sin x))^2 (d/dx) sin x\n\n= 2 tan(sin x) (sec(sin x))^2 cos x, QED
1079454
The previous answers are all WRONG. It is not easy to give a general formula, but here is the definition and the proceedure:\n\nFirst of all, remember that the power series expansion of sin x is\nsin x=x -(1/6) x^3 +(1/120)x^5-+...\n\nIf you plug in At for x, you get\nsin(At)=At-(t^3/6)A^3 +(t^5/120)A^5-+...\n\nThe powers here are obtained by matrix multiplication. For a *diagonal* matrix, you can simply take the sine of the diagonal entries (as the previous posters suggested), but that doesn't work for non-diagonal matrices. It should be pointed out that exp(At) or cos(At) would be defined by power series in a similar way.\n\nIn most applications, the matrix A is symmetric, so can be diagonalized (by finding eigenvalues, etc). In this case, there is\na diagonal matric D and another matrix P so that\nA=PDP^{-1}.\nThe matrix D has the eigenvalues of A along the diagonal and the matrix P has the corresponding eigenvectors down the columns.\nAnyway, for such matrices,\nsin(At)=P{sin(Dt)]P^{-1}.\nSince sin(Dt) is easy, this allows the computation of sin(At).\n\nFor non-diagonalizable matrices, the computation is harder.
407985
just expand using logarithmic series ....or use taylor series expansion f(X+h)=f(x)+hf'(x)+h^2/2! f''(x)+...........here u can express f (x)=ln(x+2) as f(x+2-2)=ln(x+2)take h=-2 and X=x+2 and as given integral of 1/(x+2)=ln(x+2) that means f'(x)=1/(x+2) ..differentiate it to get f''(x) and put it in the above series ...this one may be used if some x is given and u need 2 find the approx value
1050391
EDIT: I was looking at the answer above mine, and the cos(0) and sin(0) values are reversed, so it is not x=0.\n\nOK!! No problem. First of all, let me just say I teach math, but there is no way I would want to memorize all of the trig identities, so open up your precalc book, and if it is a good book, the major identities will be located on the front or back cover. If not, then go to the applicable page.\n\nStart with what is given:\n(sq root:3)sinx + cosx = 1\n\nAll I did was go through the identity lists and see what seemed to match closest to what you have there, and I noticed that:\n\ntan(x/2) = (1 - cosx) / sinx\n\nThen I visualized doing a little bit of algebra with your problem, and saw that if you subtracted cosx from both sides you would have:\n\nsqrt(3) sinx = 1 - cosx\n\nWhich is kind of similar to tan(x/2), except we need a sinx in the denominator. And if you divide both sides by sinx, that is what you get! So now we have:\n\nsqrt(3) = (1 - cosx) / sinx\n\nBut we already know that this is the same as tan(x/2)\n\nSo, we have:\n\nsqrt(3) = tan(x/2)\n\nFrom here, since you want to find the actual value for x which will be an angle measure, you go to inverse tangent:\n\ntan(-1) sqrt(3) = (x/2)\n\nI also have a handy table of all the major trig values open (it is in the back cover of my precalc book) and notice that tan(60) = sqrt(3)\n\nWe are almost done:\n\n60 degrees = x / 2\n\nMultiply by 2,\n\nx = 120 degrees.\n\nOf course, you should check that answer and make sure it is correct!!
347908
I got the same answer as math girl, but did it SLIGHTLY different.  I recognized that the form of the integral was one of the sine inverse functions.  That is, the derivative of Sin^-1 (x) = 1/(1-x^2)^.5  \n\nIf you recognize this relationship, the problem is an easy u substitution away.\nNamely, factor out 25 from the denominator in the square root, leaving you with \ndx/(25(1-(4x^2)/25))^.5\n=dx/(5(1-((2/5)x)^2)^.5)\nThen make a the u substitution of u = (2/5)x, giving dx = (5/2)du\n\nSubstituting back gives (1/5)(5/2) outside of the integral and inside the integral du/(1-u^2)^.5\n\nThe integral is (1/2)Sin^-1 (u) + C\nSubstitute back and get (1/2) Sin^-1 ((2/5)x) + C\n\nWriting it down with symbols on paper makes this method very easy, even if following it in this medium difficult. \n\nThe key being knowing your inverse trig derivatives.  If you don't know them, you are better off using math girl's method.
1148883
I assume what looks 'funny' to you is the wild oscillations near x=0.  The reason is that for x near zero, the quantity 1/x gets really large, and as x gets closer to zero the value of 1/x changes a LOT for a small change in x.  What this means for sin(1/x) is that the sine function goes through many oscillations over just a small range of x (near zero) because its argument changes so much.  Does that make sense?\n\nSee if you can make an argument for what behavior to expect for sin (1/x) when x is far from zero instead (positive or negative).
748672
using Maclaurin series you find the series for sin and cos and then use long division for the Maclaurin series of tan