how can solve this " sin(X)+sin(2x)+sin(3x)+....+sin(nx) " ?
959491
sin(X)+sin(2x)+sin(3x)+....+sin(nx)\n= img[exp(ix)+exp(i2x)+exp(i3x)+...+exp(inx)]\n=img[(exp(i(n+1)x) -1) / (exp(ix) - 1)]\n=img[(exp(i(n+1)(x/2)) -1) /exp(i(n+1)(x/2)) * (exp(i(n+1)(x/2)) -exp(-i(n+1)(x/2))) / (exp(i(x/2)) - exp(-i(x/2)))\n=[sin((1/2)(n+1)x)/sin((1/2)x)]* img(exp(inx/2))\n=[sin((1/2)nx)/sin((1/2)x)]*[sin((1/2)(n+1)x)/sin((1/2)x)]/[sin((1/2)x]\n\n[exp is exponential, img is imaginary part, 'i' is usual imaginary i in complex numbers] I also suggest you write down this so that you will be more clear with the equations. Sorry but without equation editorits bit difficult to type math.\n\nIf you want an infinite sum, please take the limit as x-->infinity of the last term obtained.\n\nI hope this answeres your question. All the sest.
542056
Start by factoring:\n\n(cos^2(x))(sin^2(x))-(cos^2(x)) = (cos^2(x))(sin^2(x) - 1)\n= -(cos^2(x))(1 - sin^2(x))\n= -(cos^2(x))(cos^2(x))\n= -cos^4(x)
200647
sin^2(x)=0.5*(1-cos(2x)).this can be integrated
848587
sin^2x=(1-cos2x)/2\nantiderivative of (1-cos2x)/2\n=1/2x-(1/4)sin2x +c
546258
cos2x=cos(x+x)=cos^2x-sin^2x=1-sin^2x-sin^2x=1-2sin^2x\ncos2x=1-2sin^2x  \ncos2x+2sin^2x-sinx=1-2sin^2x+2sin^2x-sinx=1-sinx=0\nsinx=1\nx=2kπ+π/2  ,   k=...,-1,0,1,...
776925
-2 sin^2 x-3 sin x-1=0\n-(2 sin^2 x+3 sin x+1)=0\n2 sin^2 x+3 sin x+1=0\n(2 sin x+1)(sin x+1)=0\nsin x=-1/2 or -1\nx=7pi/6, 11pi/6, and 3pi/2\nI hope that helps!
952344
(1+tanx)/(1+cotx)\n=(1+sinx/cosx)/(1+cosx/sinx)\n=[(cosx+sinx)/cosx]/(sinx+cosx)/sinx]\n=(cosx+sinx)(sinx)/(cosx+sinx)cosx)\n=sinx/cosx\n=tanx\nhence proved
817619
you should use formula for\n\nsin 4x = sin 2(2x)\n\nassumin 2x = y only for understanding purpose\n\nsin 4x =sin 2y\n           = 2 sin y cos y\n           = 2 sin 2x cos 2x\nnow apply the formula for 2x
971617
y = sin(3x) + cos(4x)\ny = sin(2x + x) + cos(2x + 2x)\n\nsin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)\ncos(2x + 2x) = cos(2x)cos(2x) - sin(2x)sin(2x)\n\nsin(2x) = 2sinxcosx\ncos(2x) = cos(x)^2 - sin(x)^2 = 2cos(x)^2 - 1 = 1 - 2sin(x)^2\n\n(sin(2x)cos(x) + sin(x)cos(2x)) + (cos(2x)^2 - sin(2x)^2)\n\n(2sin(x)cos(x)cos(x) + sin(x)(2cos(x)^2 - 1)) + ((2cos(x)^2 - 1)^2 - (2sin(x)cos(x))^2)\n\n(2sin(x)cos(x)^2 + 2sin(x)cos(x)^2 - sin(x)) + (((2cos(x)^2 - 1)(2cos(x)^2 - 1)) - 4sin(x)^2cos(x)^2)\n\n(4sin(x)cos(x)^2 - sin(x)) + ((4cos(x)^4 - 2cos(x)^2 - 2cos(x)^2 + 1) - 4sin(x)^2cos(x)^2)\n\n4sin(x)cos(x)^2 - sin(x) + (4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2)\n\n4sin(x)cos(x)^2 - sin(x) + 4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2\n\ncos(x)^2 = 1 - sin(x)^2\n\n4sin(x)(1 - sin(x)^2) - sin(x) + 4(1 - sin(x)^2)^2 - 4(1 - sin(x)^2) + 1 - 4sin(x)^2(1 - sin(x)^2)\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4((1 - sin(x)^2)(1 - sin(x)^2)) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - sin(x)^2 - sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - 2sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4 - 8sin(x)^2 + 4sin(x)^4 - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n(4 + 4)sin(x)^4 - 4sin(x)^3 + (-8 + 4 - 4)sin(x)^2 + (4 - 1)sin(x) + (4 - 4 + 1)\n\nsin(3x) + cos(4x) = 8sin(x)^4 - 4sin(x)^3 - 8sin(x)^2 + 3sin(x) + 1\n\nfor a graph, go to http://www.calculator.com/calcs/GCalc.html\n\njust type in sin(3x) + cos(4x), click enter
851033
(cosx-sinx)^2=cos^2x+sin^2x-2sinxcosx so\nit can be 1-2sinxcosx+cosx+sinx
240201
u dont want to prove that...u want to solve for x.  you want to find the value of x so that sin(x)=cos(x).  a possible solution would be: root three over two.  aka square root of three divided by two.  This isnt a proof....cause to be a proof it would have to read sin(x)=cos(x) which is not always true so it isnt a proof.
972476
........._______________\nx - 5 ) x³ - 3x² - 13x + 15 ( x²\n......... x³ - 5x²\n......... -------- ....↓\n............... 2x² - 13x ....... ( 2x\n............... 2x² - 10x\n............... ----------- ....↓ \n............... ....... -3x + 15. (-3\n............... ....... -3x + 15\n............... ....... -----------\n............... ....... -----------\n\nSo P(x) = x³ - 3x² - 13x + 15\n= (x - 5)(x² + 2x - 3)
1028428
well..\n\nI'll give you one:\n\ntanx + cotx = 1/(sinx cosx)\n\ntanx = sinx/cosx\ncotx = cosx/sinx\n\nsubstituting:\n\nsinx/cosx + cosx/sinx\n\nmultiply each term to get a common denominator:\n\nsinx/cosx *(sinx/sinx) + cosx/sinx *(cosx/cosx)\n\nsin^2 x / (cosx*sinx) + cos^2x/(cosx*sinx)\n\nsince sin^2x + cos^2 x = 1, and you can add the two terms together because they have a common denominator, the entire expression equals:\n\n1/(sinxcosx)
564945
4*sin(x)cos^2(x) \n= 2*{2*sin(x)cos(x)}cos(x) \n= 2*sin(2x)cos(x)\n= sin(3x) + sin(x)\nTherefore, sin(x)cos^2(x) = [sin(3x) + sin(x)]/4\n\n4*sin^2(x)cos(x)\n= 2*{2*sin(x)cos(x)}sin(x)\n= 2*sin(2x)sin(x)\n= cos(x) - cos(3x)\nTherefore, sin^2(x)cos(x) = [cos(x) - cos(3x)]/4\n\nHence the expression inside the integral sign is\n[sin(3x) + sin(x) - cos(x) + cos(3x)]/4\n\nThus the integral becomes\n(1/4)*∫[sin(3x) + sin(x) - cos(x) + cos(3x)]dx\n= (1/4)*[∫sin(3x)dx + ∫sin(x)dx - ∫cos(x)dx + ∫cos(3x)dx]\n= (1/4)*[{-cos(3x)}/3 - cos(x) - sin(x) + {sin(3x)}/3] + C\nwhere C = arbitrary constant of integration\n= (1/4)*[{sin(3x) - 3*sin(x)} - {cos(3x) + cos(x)}]/3 + C\n= (1/12)*[- 4sin^3(x) - 4cos^3(x)] + C\n[N.B. cos(3x)=4cos^3(x)-3cos(x) and sin(3x)=3sin(x)-4sin^3(x)]\n= - [sin^3(x) + cos^3(x)]/3 + C
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]
610758
sin(2x) = 2sinxcosx is a trigonometric identity called the double angle identity. It was derived from another identity:\n\n   sin(x + y) = sinxcosy + cosxsiny\n \nhere x and y are two different angles but what if x = y\nreplace y with x to make all the equations in terms of x, thus the equation becomes:\n\n   sin(x + x) = sinxcosx + cosxsinx\n   \nnote that sinxcosx is also equal to cosxsinx, so simplifying the equation it becomes:\n\n  sin(2x) = 2sinxcosx or sin(2x) = 2cosxsinx
825804
First, let's simplify sin(2x) / sin(x)\nRemember that sin(2x) = 2sin(x)cos(x), so\nsin(2x) / sin(x) \n= 2sin(x)cos(x) / sin(x)\n= 2cos(x)\nThe integral of 2cos(x) dx\n= 2sin(x) + C