what is taylor series? and how it is applied in finding the numerical value of trigonometric functions?
140439
In general, a Taylor Series expansion for a function is:\n\nf(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! +f'''(a)(x-a)^3/3! +...\n\nWhen a = 0, this is known as a Maclauren series.\n\nApplying this to trig functions, you get\n\nsin(x) = x - x^3/3! + x^5/5! + x^7/7!...\nand\ncos(x) = 1 - x^2/2! + x^4/4! + ...
729562
1/sin(x) - cos(x) * cos(x)/sin(x)\n1/sin(x) - cos^2(x)/sin(x)\n(1-cos^2(x))/sin(x)\nsin^2(x)/sin(x)\nsin(x)
908517
On the calc. you'll see SIN, COS, and TAN\n\nSin- Sine\nCos- Cosine\nTan- Tangent\n\n\nAll used for trigonometry.
433978
Simply equate the functions as it asks you to;\nso you get 1-sinx = 2sin2x\nThen use the trig identity sin2x=2sinxcosx\nso you get 1-sinx=4sinxcosx\n\nCan you solve it from there?\nI'd next use cosx = sqrt (1-(sinx)^2)\nThen square the equation and solve it for x as you've probably done a million times before.\n\nYou'll probably get a quadratic equation in sinx. Use the quadratic formula to get some values out, and remember that you have to give answers between -360 and 360 degrees.
419071
You can cheat and use a calculator to get the values, but I don't think this is what your teacher is looking for. \n\nFirst, use the relationship sin^2(x)+cos^2(x)=1 to find |sin(a)| and |cos(b)|. Pay close attention to the ranges of a and b to get the sign right-- sin(a) and cos(b) should both be negative. Doing this you should find:\n\nsin(a) = -12/13 \ncos(b) = (-2 sqrt(2))/3\n\nHere are two useful identities:\n\ncos(a-b) = cos(a)cos(b) + sin(a)sin(b)\ncos(2b) = 1 - 2sin^2(b)\n\nSince you now know cos(a),cos(b),sin(a) and sin(b), you can find that:\n\ncos(a-b)=(12-10sqrt(2))/39\ncos(2b)=7/9
626687
r(t) = 1+cos[t]\n\nPolar Conversion:\n x= cos[t]\n y= sin[t]\n\nm = weight of the cardioid\n\nxg = (1/m)∫x*1ds\nyg = (1/m)∫y*1ds\n\nds = sqrt[(r')^2 + (r^2)*((t')^2)]dt\n\n(ds)^2 = [(r')^2 + (r^2)*((t')^2)](dt^2)\n\n(ds)^2 = (dr(t))^2 + ((r(t))^2)*(dt^2)\n\n \n(ds)^2 = [(Sin[t])^2 + 1 + (Cos[t])^2](dt)^2\n\n(ds)^2 = 4*[(Cos[t/2])^2]*(dt)^2\n\nds = 2*Cos[t/2]*dt\n\nor m = ∫ds = ∫2*Cos[t/2]*dt = 2*sqrt[2]. {0<t<(pi/2)}\n\n\nxg = [1/(2*sqrt[2])]∫xds \n\nxg = [1/(2*sqrt[2])]∫(1 + Cos[t])Cos[t](2Cos[t/2])dt  {0<t<(pi/2)}\n\nxg = 6/5  ->  1.2\n\n\nsame for yg, but:\n\nyg = [1/(2*sqrt[2])]∫(1 + Cos[t])Sin[t](2Cos[t/2])dt  {0<t<(pi/2)} \n\nyg = (-1 + 4*sqrt[2])/5  ->  0.93
611207
The only way to do this is by using the reduction formula for int(sin(x)^n), which is valid for n>0.  Unfortuantely, it reduces to another integral for sin(x)^(n-2), for which the reduction formula no longer is valid.  Hence, the only way to do this integral is by using numerical integration techniques.\n\nProbably the best way to do this problem is to convert sin(x)^(1/2) to a taylor series.  \n\nI did this and got:\nx^0.5 - x^2.5/12 + x^4.5/1440 - x^6.5/24192 - 67/29030400*x^8.5 .....\n\nIntegrate:\n\n(2/3)*x^(3/2)-(1/42)*x^(7/2)+(1/7920)*x^(11/2)-(1/181440)*x^(15/2)-(67/275788800)*x^(19/2)......\n\nNeedless to say, after the first 4 terms, any error from a direct integration of sqrt(sin(x) is less than 10^-6, which I would say is pretty damn close.  \n\nThere ya go...
394432
sin(x) - cos(x) - tan(x) = -1\nsin(x) - cos(x) - ((sin(x))/(cos(x))) = -1\n(sin(x)cos(x)) - cos(x)^2 - sin(x) = -cos(x)\n(sin(x)cos(x)) - sin(x) = cos(x)^2 - cos(x)\n((sin(x))(cos(x) - 1) = (cos(x))(cox(x) - 1)\n\ndivide both sides by (cos(x) - 1)\n\nsin(x) = cos(x)\nor\nsin(x) = 1cos(x)\n\ndivide both sides by cos(x)\n\n(sin(x))/(cos(x)) = 1\n\ntan(x) = 1\n\ninverse tan both sides\n\nx = 45° or 225°\n\nAlso i found out that "x" also equals 0\n\nIf you go to www.quickmath.com, it will give you answers of (pi/4) which equals 45°, (-3pi/4) which equals 225°, and 0.
912572
sin^3 β + cos^3 β  = (sinβ + cosβ)(sin^2β + cos^2β - sinβcosβ)\n= (sinβ + cosβ)(1 - sinβcosβ)\n\nsin^3 β + cos^3 β / sin β + cos β\n = (sinβ + cosβ)(1 - sinβcosβ)/( sin β + cos β)\n= 1 - sinβcosβ
844133
This is an interesting question:\n\n1)  First, make a plot of y = -sinx\n2) You'll see an infinite amount of lines that are passing through the origin and tangent to the function.\n3)  The line with the largest slope is either closest to x = 0 or the one  where x => infinity.\n4)  By making lines with a ruler that are tangent to the function, you'll see that the slope gets smaller as x => infinity.\n5)  Therefore, the line is the positive slope closest to zero at pi/2 < x < pi*3/2.\n6)  Since the slope is the derivative of the function, slope = -cos x\n7)  Using this info, you know that the tangent line passes through the origin, so slope = y/x = -cosx\n8)  Now, you have 2 equations:\ny = -sinx\ny/x = -cosx\n9)  Combine two equations to eliminate y:\ny/(y/x) = -sinx/-cosx\nx = tan(x)\n10)  You should use newton's method to solve this equation between pi/2 < x < pi*3/2.  I just plotted it on Excel and get a value of x = 4.49.\n\nBTW, can you tell me what level math class this is?  I'm amazed at how early students study advanced math nowadays.
1122586
The "Maclaurin series" is a restricted version of the Taylor\nseries [1] which leads to the approximation for sin(x) of:\n\nsin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + (x^9/9!) ...\n\nThere are several sites on the web which discuss these\napproximations -- many point out that you should use\n"Horner's Rule" to minimize the computation involved\nin evaluating them.[2]\n\nIf you're trying to implement these "professionally",\nyou should have argument reduction (ie. to reduce the\ninput argument down to a small range like 0 to pi/4).\nFor an example of source from GNU, see [3] and [4].\n\n\nBut just to see the computation in action, you could\nexplicitly code something like:\n\n#define F1  (1)\n#define F3  (F1*2*3)\n#define F5  (F3*4*5)\n#define F7  (F5*6*7)\n#define F9  (F7*8*9)\n#define F11 (F7*10*11)\n\ndouble\nsin(x)\n{\n  double x2  = x*x;\n  double x3  = x2*x;\n  double x5  = x3*x2;\n  double x7  = x5*x2;\n  double x9  = x7*x2;\n  double x11 = x9*x2;\n\n  return x - (x3/F3) + x5/F5 - x7/F7 + x9/F9 - x11/F11;\n}\n\nJust for fun I coded the evaluation as a\nloop (with NO consideration for efficiency\nor accuracy):\n\n\n#include <stdio.h>\n#include <math.h>\n\ndouble\nsinx(double x)\n{\n\ndouble factorial = 1.0;\ndouble x2 = x*x;\ndouble sign = 1;\n\ndouble term = x;\ndouble value = 0;\n\nfor (int i=1; i<11; i+=2) {\n  value += (sign*term/factorial);\n  term = term * x2;\n  sign *= -1;\n  factorial = factorial * (i+1) * (i+2);\n}\n\nreturn value;\n}\n\nvoid\ntest(double v)\n{\nprintf("sin(%f) = %f, "\n  "approx sin(%f) = %f&#92;n",\n  v, sin(v), v, sinx(v));\n}\n\nint\nmain(int argc, char*argv[])\n{\ntest(0.0);\ntest(0.2);\ntest(0.5);\ntest(1.0);\ntest(1.57079632679);\ntest(2.0);\ntest(3.0);\ntest(3.14159);\n\nreturn 0;\n}
681952
Here you go:\nhttp://en.wikipedia.org/wiki/Fibonacci_series
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
278456
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
417752
As others have pointed out, one use is to compute the values of functions--particularly transcendental functions like sin(x) or log(x) or whatever.\n\nOn top of that, there are differential equations that are most easily solved by generating a power-series solution.  Cumbersome, but it may be all you get.\n\nAlso realize that the power series isn't usually what is used to compute things on calculators--an approximation, computed FROM the power series is usually what gets used, giving the value to whatever number of digits precision is needed; mathematical tools to accelerate the convergence are applied (I used to know them, thirty years ago, but have forgotten 'em), allowing a MUCH quicker computation of values.  Being the basis for the accelerated estimations is perhaps their greatest usefulness.
890202
in trigonomertic form\n\nz = r cos t + ir sin t = 1 + i\n\nso r cos t = 1\n r sin t = 1\n\nsquare and add r^2 = 2\nso r = sqrt(2)\n\nnow cos t = 1/sqrt(2) and sint = 1/sqrt(2)\n\nso t = pi/4 (as cos pi/4 = 1/sqrt(2) and sin is also positive so 1st quadrant)\n\nso t = pi/4 and r = sqrt(2)
339966
The integral of sin(x)/x cannot be written in finite terms with the 'elementary functions', which consist of trig, inverse trig, exponential, logarithms, and algebraic functions. It can be written as a power series or in other ways, depending on what you want to do with it. Good amth programs will compute values for this integral (as approximations).\n\nAs for the integral of sin(x)/x from 0 to infinity, the result is pi/2, but that usually is done with complex analysis. It cannot be done easily with elementary methods.