How do you solve this equation: cos (theta) = 2 - 2 cos (theta)?
145134
I will use Q for theta here:\ncosQ = 2 - 2cosQ\nNow we add 2 cosQ to both sides to get rid of it on the right.\nSo we get:\n3cosQ = 2\nDividing both sides by 3 gives:\ncosQ = 2/3\nAnd solving for Q, we take the inverse cosine of both sides:\nQ = cos(-1)(2/3)\nHope this is helpful :-)
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
730213
Let u, v be the rotated x, y, rotated counter-clockwise by an angle t.\n\nThen x = u*cos(t) + v*sin(t), and y = -u*sin(t) + v*cos(t).\n\nPlug these into the equation (abbreviating c=cos(t), s=sin(t))\n\n(uc+vs)^2 + 4(uc+vs)(vc-us) + 4(vc-us)^2 + (5*sqrt(5))(vc-us) + 5\n= u^2*c^2 + 2*uvcs + v^2*s^2 + 4*uvc^2 + 4*v^2*cs - 4*u^2*cs - 4*uv*s^2 + 4*v^2*c^2  - 8*uvcs + 4*u^2*s^2 + 5*sqrt(5)(vc-us) + 5\n= u^2(c^2-4cs+4s^2) + uv(2cs + 4c^2 - 4s^2 - 8cs) + v^2(s^2 + 4cs + 4c^2) + 5*sqrt(5)(vc-us) + 5\n\nso to eliminate the xy term, we must have 2cs + 4c^2 - 4s^2 - 8cs = 0, but 2cs = sin(2t) and c^2 - s^2 = cos(2t), so we have\n4cos(2t) = 3sin(2t), or tan(2t) = 4/3, so t = arctan(4/3)/2 = 0.4636 radians.\n\nThe resulting equation is 5v^2 - 5u + 10v + 5 = 0.
761231
[cos(x)]^2 - 1.2cos(x) - 0.28 = 0\n\nMultiply through by 100 to get rid of decimals.\n100[cos(x)]^2 - 120cos(x) - 28 = 0\n\nDivide by the common factor = 4.\n25[cos(x)]^2 - 30cos(x) - 7 = 0\n\nThis expression factors, so we don't need the quadratic formula.\n[ 5cos(x) - 7 ] [ 5cos(x) + 1 ] = 0\n\nEquating each factor to zero gives :\ncos(x) = 7/5 = 1.4, but cos(x) is always <= 1,\nso this is not a solution.\n\nor\n\ncos(x) = -1/5 = - 0.2\nThis is a solution because -1 <= cos(x) <= 1.\n\nTaking the arccos by calculator gives :\nx = 101.54º\n\nThis angle is in the second quadrant.\n\nFor cos, there must also be an angle in the third quadrant,\nbecause the cos of all angles there, are also negative.\n\nThis angle is greater than 180º by the same\namount that the other angle is less than 180º.\n\nSo we find that 180º - 101.54º = 78.46º\nThus, 101.54º is less than 180º by 78.46º.\n\nTherefore, the other angle must be 78.46º more than 180º.\nSo the answer to this is 180º + 78.46º = 258.46º.\n\nTherefore, x = 101.54º or 258.46º.\n\nI'll let you calculate those values in radians.\nI still find it easier to work with degrees!!!
109151
It looks like your equation got cut off, but I think you were writing\n\nR = (V^2/2g)(sin (2 theta) +sin (2 theta))\n\nwhich can be immediately simplified, by adding the sin (2 theta) terms, as \n\nR= (V^2/2g )(2)(sin(2 theta) = (V^2/g )sin (2 theta)\n\nwhich is the formula that the other correspondent already gave.  To solve it, simply multiply both sides by g, divide by V^2, and you get \n\n(R g)/v^2 = sin (2 theta)\n\n[Arcsin (Rg/V^2) ]/2 = theta
404800
sec = (1/cos); cot = (1/tan) = (cos/sin)\n\ncos(-pi/3) = 1/2 \n\n1/1/2 = 2 = sec(- pi/3)\n\ncos(-5pi/4) = -sqrt(2)/2 ;      sin(-5pi/4) = sqrt(2)/2\n\n(-sqrt(2)/2)/(sqrt(2)/2) = -1 = cot(- 5pi/4)\n\n2 - (-1) = 3 There's your answer.
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
291468
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
910165
cos2A = 2cos²A - 1 → cos²A = ½(cos2A + 1)\n\ncos2A = 1 - 2sin²A → sin²A = ½(1 - cos2A)
736970
Here are some real tough ones.(Looked tough to me)\n\nProve that cot(90 - theta)sin(90 - theta) = sin theta\n\nFind Cos(3x)\n\nFind Sin(3x)
994111
First, factor the left hand side of the equation:\n\n2sin(x)*cos(x) + cos(x) = 0\n\ncos(x)*[2sin(x) + 1] = 0\n\nNow, this equation is satisfied whenever either cos(x) = 0 or (2sin(x)+1) = 0\n\ncos(x) = 0 at x = pi/2 and 3*pi/2  (You should memorize when the cosine and sine functions are equal to zero!)\n\nThese are the first two solutions to the original problem.\nNow deal with the factor containing the sine term.  We need to find the values of x when:\n\n2sin(x) + 1 = 0\n\nRearrange this equation to read:\n\n2sin(x) = -1\n\nthen\n\nsin(x) = -1/2\n\nx = 11pi/6 and 7pi/6 .  These are the other two solutions to the original problem.\n\nYou should also memorize when the sine and cosine functions are equal to 1/2.  For the sine function, this happens at x = pi/6 = 60 degrees and 5pi/6 = 150 degrees.  If the sine or cosine function has a certain value at x, then it will have -1 times that value at x + 180 degrees (x + pi radians), so in this case if sin(pi/6) = 1/2, then sin(pi/6 + pi) = sin(7pi/6) = -1/2.  Similarly, if sin(5pi/6) = 1/2, then sin(5pi/6 + pi) = sin(pi/6+pi) = sin(7pi/6) = -1/2.\n\nNote that despite what the previous answerer said, the sin(2pi/3) and the sin(4pi/3) do not equal -1/2.  [It's the cosine that's equal to -1/2 at these points.)
966696
Arc=r.theta=8*pi/8=pi
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
28429
Basics first:\ncosec(theta) = 1/sin(theta)\ncot(theta) = cos(theta)/sin(theta)\nsec(theta)= 1/cos(theta)\n\nNow, the given expression:\nNumerator = 1+cosec(theta) = (sin(theta)+1)/sin(theta)   \nDenominator = cot(theta)+cos(theta) = cos(theta)[1+sin(theta)]/sin(theta)\n\nDividing Numerator by Denominator, we get:\n1/cos(theta) = sec(theta)\n\nThere you have it!
1122517
sin^2+cos^2=1\nsin=1/cosec\nsin=opp side/hypotheses side\ncos=adj side/hypotheses side\nhyp^2=opp^2+adj^2\n\nSin*cosec*cos=cos
830578
Adding these fraction expressions (common denominator  etc) gives\n(sin3θsinθ + cos3θcosθ)/(cosθsinθ)\n\nThe top is the expansion of\ncos(3θ - θ) which is cos2θ\n\nThe bottom is half of \nsin2θ = 2sinθcosθ,\nso multiply top and bottom by 2 and you have the result, since\ncos2θ/sin2θ = cot2θ
326529
As sin(4t) + sin(2t) = 0,\n\n=> 2sin(2t)cos(2t) + sin(2t) = 0,\n=> sin(2t).[2cos(2t) + 1] = 0.\n\n=> sin 2t = 0; or, 2cos2t + 1 = 0.\n\nNow, sin2t is 0 at 2t = 0, pi, 2pi.\nor, t= 0, pi/2, pi.  (note: pi=3.14)\n\nAnd 2cos2t + 1 = 0 means\ncos2t = -1/2, that is 2t = 5pi/6, 7pi/6.\nor, t= 5pi/12, 7pi/12.\n\nhence answrs are: t= 0, 5pi/12, 7pi/12, pi/2 and pi.