how can solve this " sin(X)+sin(2x)+sin(3x)+....+sin(nx) " ?
815664
Imaginary part of\n(e^(i*x)+ e^(2*i*x) + ...)\n=Imag(e^(i*x)/(1- e^(i*x))\n=Imag(e^(i*x)/\n  (1-cos(x)-i*sin(x)))\n=Imag(e^(i*x)\n   (1-cos(x)+i*sin(x))/\n   ((1-cos(x))^2+(sin(x)^2))\n=((cos(x))\n(sin(x))+(sin(x))\n(1-cos(x)))/(2-2*cos(x)\n=sin(x)/(2(1-cos(x)))
932663
Let a = A cos C\nb= A sin C\n\nsquare and add a^2+b^2 = A^2\n\nthen given expression\n= A cos C sin Bx + A sin C cos Bx = A sin(Bx+C)
617038
sin(pi/2 + x) = sin (pi/2)cosx + cos(pi/2)sinx\n\n= 1cosx + 0 = cosx\n\ncos(-x)=cos(x)\n\nSo cos(x)cos(x) = cos^2(x)
901088
4sinxcosx =1 (assumed for 0 ≤ x ≤ 2π)\nSo 2 sin2x = 1 (since sin2x = 2sinxcosx)\nSo sin2x = ½ (0 ≤ 2x ≤ 4π)\n2x = π/6, 5π/6, 13π/6, 17π/6\nThus x = π/12, 5π/12, 13π/12, 17π/12\n\nsin4x= -2sin2x (assumed for 0 ≤ x ≤ 2π)\nSo 2sin2xcos2x + 2sin2x = 0\nie 2sin2x(cos2x + 1) = 0\nso sin 2x = 0 or cos 2x = -1/2 (0 ≤ 2x ≤ 4π)\n2x = 0, π, 2π, 3π, 4π or 2x = 2π/3, 4π/3, 8π/3, 10π/3\nwhence x = 0, π/3, π/2, 2π/3, π, 4π/3, 3π/2, 5π/3, 2π
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)
1156022
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out. \nIf you get stuck, send me an e-mail or add additional details.
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
397638
Use these identities:\nsin(2x) = 2*sin(x)*cos(x)\nsin(4x) = 4*cos^3(x)*sin(x) - 4*cos(x)*sin^3(x)\ncos(2x) = -1 + 2*cos^2(x)\ncos(4x) = 1 - 8*cos^2(x) + 8*cos^4(x)\ntan(3x) = (3*tan(x)-tan^3(x))/(1-3*tan^2(x))\n\nAfter you expand the left hand side of the equation with the above identities, you can factor out a 2*sin(x)*cos(x) from the numerator and a 2*cos^2(x) from the denominator, which will turn into a tan(x). Now divide the numerator and denominator of the fraction by cos^2(x). You should get the right hand side of the tan(3x) identity above.\n\nTry this and if you can't figure it out, I'll write it out. (I don't really want to since equations look like garbage on this site).\n\nCheck out this site for some usefull identities.
355137
first, this equation is wrong, it should look likle this:\nsin2x=2(sinx (multiplied by) cosx). If yu want, i'll try to solve the true equation for yu.
356153
using trig identities\nsin^2x+cos^2x=1, so 1-sin^2x=cos^2x and sin^2x=1-cos^2x\nsubstitue\ncos^2x(cosx)-2sin^2cosx\ncos^3x-2sin^2xcosx\ncosx(cos^2x-2*(1-cos^2x))\ncosx(cos^2x-2+2cos^2x)\ncosx(3cos^2x-2)
591905
1+2(tanx)^2+(tanx)^4\n= (1+(tanx)^2)^2\n= ((secx)^2)^2\n= (secx)^4\n= (cosx)^-4
181415
let that be x insted of *, it's easier to write\nsinx/(1+cosx) +1 + cosx/sinx = \n[(sinx)^2 + sinx + sinxcosx + cosx + (cosx)^2]/[(1+cosx)*sinx]=\n(sinx+cosx+sinxcosx+1)/(1+ cosx)sinx=\n[sinx+ 1 + cosx(1+sinx)]/[(1+cosx)sinx)]=\n[(1+sinx)(1+cosx)]/[(1+cosx)\nsinx]=\n(1+sinx)/sinx=\n1/sinx +1 = cscx+1\nI don't know maybe I have done something wrong but it doesn't look like 2cscx.
704759
tan(a+b)=sin(a+b)/cos(a+b)=\n(sinacosb+sinbcosa)/\n(cosacosb-sinasinb)\ndivide top and bottom by cosacosb\n(tana+tanb)/(1-tanatanb)
756591
Given\n\nsin 3x = 1/2\n\n\n\nGet arcsin of both sides\n\narcsin (sin 3x) = arcsin (1/2)\n\n\n\nThe left side simplifies as 3x.  The right side varies.  \nLet k be any integer.  Thus,\n\n3x = π/4 + 2kπ and\n3x = 3π/4 + 2kπ\n\n\n\nThus,\n\n3x = (π + 8kπ)/4 and\n3x = (3π + 8kπ)/4\n\n\n\nTherefore,\n\nx = π/12 (8k + 1) and\nx = π/12 (8k + 3)\nwhere k is ANY integer. Enumerating, we have,\n\nx = ... -5π/4, -7π/12, π/12, 3π/4, 17π/12, 25π/12, 11π/4, ...\nx = ... -13π/12, -5π/12, π/4, 11π/12, 19π/12, 9π/4, 35π/12, ...\n\n^_^
160524
This is not a calculus problem, this is merely a trigonometric identity.  No derivatives, integrals, or numerical approximation were used.\n\nDefinition:\n\ntanh(2x) = sinh(2x)/cosh(2x)\n\nYou know the double angle formulas for sinh and cosh:\n\nsinh(x + x) = sinh(x)cosh(x)+cosh(x)sinh(x) = 2cosh(x)sin(h)\n\ncosh(x + x) = cosh(x)cosh(x)+sinh(x)sinh(x) = cosh(x)^2 + sinh(x)^2\n\nDivide:\n\n(2cosh(x)sinh(x))/(cosh(x)^2 + sinh(x)^2)\n\nDivide numerator and denominator by cosh(x)^2:\n\n((2cosh(x)sinh(x))/cosh(x)^2) / ((cosh(x)^2 + sinh(x)^2)/cosh(x)^2)\n\nSimplify:\n\n(2 sinh(x)/cosh(x))/(1 + sinh(x)/cosh(x)) \n\nSimplify more to solution:\n\n2tanh(x)/(1 + tanh(x)
760641
x + 6 + 3x = 1 + 3x + 1\nCombine like terms.  x+3x = 4x and 1+1 = 2\n4x + 6 = 3x + 2\nSubtract 3x from each side\nx + 6 = 2\nSubtract 6 from each side\nx = -4
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