What's the method for solving for a variable inside of a sin()?
453441
sin(300/A) = B/C\n300/A = arcsin(B/C)\nA = 300/arcsin(B/C)\n\nwhere y=arcsin(x) is function inverse to x=sin(y).\n\nWhat's a problem?
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.
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
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.
989233
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.
696016
sin^2 + cos^2 = 1\nsin^2 = 1 - cos^2\n\nsub in:\n\ncos^2 - 1 + cos^2 = 0\n2*cos^2 = 1\ncos = 1/sqrt(2) or -1/sqrt(2)\n\ncos(x) = +-1/sqrt(2)\ncos(x) = +-[1/sqrt(2)]*[sqrt(2)*sqrt(2)]\ncos(x) = +-sqrt(2)/2\n\nx = 45, 135, 225, 315 deg
704235
I would expand (sin 2x+1)^2, then replace (sin 2x)^2 by\n(1-cos 4x)/2, then integrate term by term.  If you can't do that, I expect another answerer will do the whole thing, or else you can email h_chalker@yahoo.com.au
931057
f(x) = sin½x + a \nSo x = sin(½f^(-1)( x)) + a\nx - a = sin(½f^(-1)( x))\nso ½f^(-1)( x) = arcsin(x - a)\nSo f^(-1)( x) = 2 arcsin(x - a) \nDomain: a -1 ≤ x ≤ a + 1\nRange: - π ≤ f^(-1)( x) ≤ π
983185
Take the first few derivatives of sin -- it is cyclical and you will see a pattern.  After you see the pattern, you will be able to say exactly what the 53rd derivative is, without having to actually take the derivative 53 times!\n\n~ ♥ ~
729006
3sin(x) - 2[cos(x)]^2 = 0\n\nNow [cos(x)]^2 = 1 - [sin(x)]^2 from the rule [sin(x)]^2 + [cos(x)]^2 =1\n-->\n\n3sin(x) - 2{1 - [sin(x)]^2} = 0\n2[sin(x)]^2 + 3sin(x) - 2 = 0\n\nsubstitute sin(x) = t\n\n2t^2 + 3t - 2 = 0\n(2t -1) (t + 2) = 0\n\nSolving ...t = -2, t = 0.5\n\nSubstituting back for sin(x)\n\nSin(x) = -2, 0.5\n\nSin(x) = -2 is not in the range since -1<sin(x)<1\n\nTherefore sin(x) = 0.5\n\nx = 30 degrees, 150 degrees for 0<x<360 degrees
699547
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
748469
The first thing I would do is rewrite both a and s in terms of h.\n\nThen you use the following rules to find the derivative.\n\n1.  Use the quotient rule to find the derivative of  sin(a)/s^2.  If you don't know the quotient rule -- look it up and memorize it.\n\n2.  The derivative of the sin is the cosin -- use the chain rule to get cos(a)*da/dh\n\n3.  The derivative of the bottom is  2*s * ds/dh (again using the chain rule).\n\nOnce you get the derivative, solve for zero.\n\nGood luck.
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.
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
511465
sin x = (1/2i)(e^ix-e^-ix)\ncos x = (1/2)(e^ix+e^-ix)\n\nsolutions of a simple harmonic oscilator are\ny = A sin x + B cos x or\ny = C e^ix + D e^-ix
888297
You can of course use a brute force approach to get the derivative but a little finesse often can simplify things.  You have:\n\ny=(sin(x)/1-cos(x))^2 = (sin(x)^2)/(1-cos(x))^2\n\nSince sin(x)^2 = 1 - cos(x)^2 it follows that:\n\ny = (1 - cos(x)^2)/(1 - cos(x))^2\n\nBy doing a little algebra up fornt, the final result is quite a bit simpler than some of the others given.\nFactoring the numerator:\n\ny = (1 - cos(x))*(1 + cos(x))/(1- cos(x))^2 = (1 + cos(x))/(1 - cos(x)) \n\nNow your function is of the form\n\ny = (1 + u)/(1 - u) where u = cos(x)\n\ndy/dx = dy/du * du/dx = (1/(1 - u) + (1 + u)/(1 - u)^2)(-sin(x))\n\ndy/dx = -2*sin(x)/(1 - u)^2 = -2*sin(x)/(1 - cos(x))^2\n\nBy doing a little algebra up front, the final result is quite a bit sinpler than those above.
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj