Trigonometry (complex numbers)
124612
Well, the square root is 27i, that makes it easy, cause 3X3X3 = 27, so 3 to the 6th is 729, and i X i is -1, so the sixth root of -729 is 3i.
434259
its -3sin(3x)\n\nnote:\nderivate of sin is cos\nderivate of cos is -sin\n\nit would be (-1/3) sin(3x) if you were integrating..
257462
The simplest solution to this problem can be done by using the "Sine Rule". If you do not know this rule you can find it in any standard textbook of trigonometry. The rule is stated below. \n\nIf the side BC, CA and AB of a triangle be represented by a, b and c respectively and if angles <BAC, <CBA and <ACB be represented by A, B and C respectively, then the area (say, A0) of triangle ABC is given by\n A0 = 0.5*a*b*sin(C) = 0.5*b*c*sin(A) = 0.5*c*a*sin(B).\n\nNow, in this case BF = (2/5)a and CF =(3/5)a [since BC = a; BF and CF are in the ratio 2:3]. Similarly, AE = (1/8)b; CE = (7/8)b; AD = (1/4)c; BD = (3/4)c.\n\nApply sine rule to triangles ADE, BDF and CEF and denote their respective areas as A1, A2 and A3. \n\nFor first triangle we can write\nArea of triangle ADE i.e. A1 = 0.5*(b/8)*(c/4)*sin(A)____(1)\nRecall, area of triangle ABC i.e. A0 = 0.5*b*c*sin(A)____(2)\nDividing equation (1) by equation (2) we get A1/A0 = 1/32.\n\nRepeating similar procedure with triangles BDF and CEF we obtain A2/A0 = 3/10 and A3/A0 = 21/40 respectively.\n\nNow, area of triangle DEF is equal to \n(A0 - A1 - A2 - A3) \n= {1-(1/32)-(3/10)-(21/40)}*A0\n= (23/160)A0\n\nAs A0 = 160 units so area of triangle DEF is 23 units.
470508
Let Z = 3 - 3i\n|Z| = √(3² + 3²)\n= 3√2\nThus Z = 3√2(1/√2 - 1/√2i)\nNow swwk an angle such that cosθ = 1/√2 and sinθ = -1/√2\nVlearly this is a 4th quadrabt angle\nTherefore θ = 2π - π/4\n= 7π/4\nThus Z = 3√2(cos7π/4 + i sin7π/4)\n[= 3√2e^(i7π/4)]
454344
In this case involving trigonometric operation on angles, you will need to carry the error over to your operation results. What I mean by that is:\n\nYou have an angle known only to 1 degree of precision, i.e.\n\nθ = 32° ± 1°\n\nTherefore, when you compute Cos (32°) = 0.848 \n\nYou will also need to perform the trig operation including the error:\n\nCos(31°) = 0.857 and Cos(33°) = 0.839\n\nSo your measurement is 0.848 ± 0.009 or 0.85 ± 0.01 depending on how many significant digits you want to keep.
439947
just put in the exact values for the unit circle sin(30) is 1/2, cos(30) is square root 3/2, cot(30) is cos(30)/sin(30) and csc(30) is 2.
356780
The only possible answer for your fixed number would be zero.  So you just have to do some inverse trig functions to get the value for x.  \n\nSolved:\nLet x=1.0038848219 [radians]\nTan(1.0038848219)=1.570796327=pi/2\nCos(pi/2)=0\nSin(0)=0
425444
no! S=sin(14)\njust look S=[sin(14)-sin(14/2)]+[sin(14/2)-sin(14/3)]+\n+[sin(14/3-sin(14/4)]+.... & so on!
401587
1. cosθ=A*B / (|A|*|B|)=(1*5+0*2+3*6) / 5√ 26\n            =81.171\n\n2. assume the angle of vectors of a and b is "m"  \n   so cos m= a*b/(|a|*|b|)=0   because cos90=0\n   m=90...so it is a right triangle...\n   |a| and |b| are representing the right-angled sides of the triangle.\n   |a|=√17; |b|=√26\n   S=|a|*|b| / 2 =10.5\n\n3. P(1,3,5) to cylindrical coordinates.......\n                         r=√1+9 =√10....\n                         θ= tan^(-1)3=71.6\n                         z= 5 \n       In terms  of x , y , and z\n                         x= r.cosθ =√10*71.6 =226\n                         y= r.sinθ = 0.95*√10= 3\n                         z=5\n                        so its (226,3,5)\n            ditto, T 's doesn't have cylindrical coordinates \n                     S 's cylindrical coordinates is (0.6, -0.8, -10)\n  ......................(that's all i know..)
591362
I think...\n\n# means theta\n\nsin(90-#)*cos(90-#)=tan#/sec^2#\ncos#sin#                =(sin#/cos#)/(1/cos^2#)\ncos#sin#                =(sin#cos^2#)/cos#\ncos#sin#                =cos#sin#\n\nUsing: sin(90-#)=cos#\n           cos(90-#)=sin#\n           tan#= (sin#/cos#)\n           sec^2#=(1/cos^2#)\n\nThat work?
475814
2 - csc²z =\n2 - (1/sin²z) = \n(2sin²z - 1) / sin²z = \n(sin²z + sin²z - 1 ) / sin²z =\n(sin²z - cos²z) / sin²z =  {using trig id sin²+cos²=1}\n1 - (cos²z / sin²z) =\n1 - cot²z\n\n~ ~ ~ ~\n\ncot³t / csct =\n(cos³t/sin³t) / (1/sint) =\n(cos³t/sin³t) * sint =\ncos³t/sin²t =\ncost (cos²t/sin²t) =\ncost (cot²t) = \ncost (csc² -1)  {using trig id 1+cot²=csc²}\n\nHope that helped!\n\n~ ♥ ~
255367
1. Make sure the angle of interest is in radians.  Radians are the "natural" (non)-units of angles, because they relate the angle measurement to the length of arc the angle cuts off on a unit circle.  So 2*Pi radians are equivalent to 360 degrees.  Basically, if you have that sucker in degrees, multiply it by 2 * Pi / 360.  Unit conversion is done. \n \n2. Make sure the angle is between -Pi and Pi.  Because sine is a periodic function, with period 2*Pi, just subtract or add 2*Pi enough times to get the angle between those two points.\n\nnow for the meat of the problem:\n\nsin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ... + \n+ (-1)^(n-1)*x^(2n-1)/(2n-1)! + ...\n\nExample: sin(15)\n\nStep 1: 15 degrees = 15*(2*Pi/360) radians = Pi/12 radians \n \nStep 2: I need Pi/12 in decimal (let's see, Pi=3.14159 (memorize that much, it's pretty easy... "Pi when calculating sine, 3 point 1 4 1 5 9!")) \nPi/12 approx = .261799 (we only had 6 sig figs, so we can't add any more than that) \n \nNow to the series: \n         first term:  x=.261799 \n         1st estimate: .261799 \n         second term: -x^3/3! = -.0029905  \n         2nd estimate: .258809 \n         third term: x^5/5! = .0000102     \n         3rd estimate: .258819 \n         fourth term: -x^7/7! = -.167*10^-7  \n         stop here - past precision\n\nThe answer is positive \nsin(15) = 0.258819\nsin(18) = 0.309017
404829
I think the formula for sin(c-d) is sin c cos d - cos c sin d\n\nso sin (x - pi/3) = sin x cos pi/3 - cosx sin pi/3\n\nwhich is 1/2 sin x - sqrt3 / 2 cos x\n\nMultiplied by the -2 in front, this is -1 sin x + sqrt 3 cos x\n\nSo A = -1 and B = sqrt 3\n\nLook up the original formula, I may have messed up the signs
288521
cos( pi / 5 ) - cos( 2 pi / 5 ) = ?\n\nObserve that cos( pi / 5 ) is the real part of the complex number e^( i pi / 5 ), which is a fifth root of e^( i pi ) = -1.  Likewise cos( 2 pi / 5 ) is the real part of the complex number e^( i 2 pi / 5 ), which is a fifth root of e( i 2 pi ) = 1.\n\nSo let a = e^( i 2 pi / 5 ) = c + i d\nand b = e^( i pi / 5 ) = f + i g\n\nNote that a, a^2, a^3, a^4, and a^5 = 1 are all fifth roots of 1, which in a field (such as the complex numbers) can have at most five fifth roots.  So\n\n(x - a)(x - a^2)(x - a^3)(x - a^4)(x - a^5) = x^5 - 1\nDividing out the term x - a^5 = x - 1 yields\n\n(x - a)(x - a^2)(x - a^3)(x - a^4) = x^4 + x^3 + x^2 + x + 1\n\nBut (x - a)(x - a^2)(x - a^3)(x - a^4)\n = x^4 - (a + a^2 + a^3 + a^4) + ...\n\nso a + a^2 + a^3 + a^4 = -1\n\nNow cos(p - q) = cos(p) cos(q) + sin(p) sin(q)\nand sin(p - q) = sin(p) cos(q) - cos(p) sin(q)\nso\ncos(pi - z) = cos(pi) cos(z) + sin(pi) sin(z)\n = (-1) cos(z) + (0) sin(z)\n = - cos(z)\nsin(pi - z) = sin(pi) cos(z) - cos(pi) sin(z)\n = (0) cos(z) - (-1) sin(z)\n = sin(z)\n\nAnd cos(p + q) = cos(p) cos(q) - sin(p) sin(q)\nand sin(p + q) = sin(p) cos(q) + cos(p) sin(q)\nso\ncos(pi + z) = cos(pi) cos(z) - sin(pi) sin(z)\n = (-1) cos(z) - (0) sin(z)\n = - cos(z)\nsin(pi + z) = sin(pi) cos(z) + cos(pi) sin(z)\n = (0) cos(z) + (-1) sin(z)\n = - sin(z)\n\nAbove we defined a = e^( i 2 pi / 5 ) = c + i d\nso we know c = cos( 2 pi / 5) and d = sin( 2 pi / 5 )\nLikewise b = e^( i pi / 5 ) = f + i g\nso f = cos(pi / 5) and g = sin(pi / 5)\n\nSo a^2 = (c + i d)(c + i d)\n = (c^2 - d^2) + i 2 c d\n\nBut a^2 = (e^( i 2 pi / 5 ))^2\n = e^(i 4 pi / 5)\n = cos( 4 pi / 5 ) + i sin( 4 pi / 5 )\n = cos( pi - (pi / 5) ) + i sin( pi - (pi / 5) )\n = - cos(pi / 5) + i sin(pi / 5)\n = - f + i g\n\na^3 = (e^(i 2 pi / 5))^3\n = e^(i 6 pi / 5)\n = cos(pi + (pi/5)) + i sin(pi + (pi/5))\n = - cos(pi / 5) - i sin(pi / 5)\n = - f - i g\n\na^4 = (a^2)^2 = (- f + i g)^2\n = (f^2 - g^2) - i 2 f g\n\nBut a = b^2 = (f + i g)^2 = (f^2 - g^2) + 2 i f g\n\nSo -1 = a + a^2 + a^3 + a^4\n = ((f^2 - g^2) + i 2 f g) + (- f + i g) + (- f - i g) + ((f^2 - g^2) - i 2 f g)\n = 2(f^2 - g^2 - f)\n\nyielding\n\nf^2 - g^2 - f = -1/2\n\nNow (cos(z))^2 + (sin(z))^2 = 1 for any z and so f^2 + g^2 = 1, thus\n\nf^2 - (1 - f^2) - f = -1/2\n2 f^2 - 1 - f = -1/2\n\nNow we are looking for cos(pi / 5) - cos(2 pi / 5) = f - a\n = f - (f^2 - g^2)\n = f - (f^2 - (1 - f^2))\n = f - (f^2 - 1 + f^2)\n = f - (2 f^2 - 1)\n = - (2 f^2 - f - 1)\n = - (- 1/2)\n = 1/2\n\nSuch a simple result suggests that there was a simpler way to do this. :-)
47461
http://mathworld.wolfram.com/ComplexNumber.html\n\nfind definitions of absolute value (||), power and product.\nIt should be straightforward from there.
362067
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...
108902
cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.