sin^4A-cos^4A=1-2cos^2A?
967873
I assume you want a proof of this fact?\n\nWe know that sin(x)^2+cos(x)^2 = 1 for any x,\nand also any function of the form x^2-y^2 can be factored as (x+y)(x-y) these two facts allow us to prove the above.\n\n\nHere if you let x=sin(A)^2 and y=cos(A)^2, then\n\nsin(A)^4-cos(A)^4 = (cos(A)^2+sin(A)^2)(sin(A)^2-cos(A)^2)\n                             = (sin(A)^2-cos(A)^2)\n                             = (1-cos(A)^2-cos(A)^2)\n                             = 1-2cos(A)^2\nlet me know if you need more assistance
413333
let suppose that \nr(@) = 4/ sqrt( 4cos^2(@) +sin^2(@) )\nthen x(@(t) ) = r(@) . cos (@) emplies\n\nx^2 (t) = r^2 (@) . cos^2(@)\n\nx^2 = 16 / (4 cos^2 + sin^2)  . cos^2\n\nso x^2(t) = 16/ (4+ tan^2(t) )\n\nBut tan( @(t) ) = y(t) / x(t)\n\nThus: x^2 = 16 / (4+ y^2 / x^2 )\nThus: \nEQ : x^2/4 + y^2/16 =  1\n\nCheck @=0  --> r(@=0) = 2\nAnd (x=2 , y=0 ) is solution of EQ
958733
sin²y + cos²y = 1\nSo cos²y = 1 - sin²y\n= (1 - siny)(1 + siny)\nTherefore\ncos²y/(1 + siny) = 1 - siny .................... QED
873429
cos5x = sinx\ncos5x = cos(pi/2 - x)\n5x = 2npi +- (pi/2 - x)\n6x = (4n + 1)pi/2   ,   4x= (4n-1)pi/2\nx= (4n+1)pi/12    ,  x = (4n-1)pi/8
895946
There is a trig identity you need: cos(2x) = 2cos^2(x)-1.\nIf you want to derive it, you can use the cosine of sum formula:\ncos(x+x) = cos(x)cos(x)-sin(x)sin(x) and go from there...\n\nYou can apply this to your equation cosx+cos(2x)=0\n\nThis gives cos(x)+2cos^2(x)-1 = 0\n\nThen yes, you can substitute p for cos(x) giving\n\n2p^2 + p - 1 = 0\n\n(2p-1)(p+1) = 0\n\np = 1/2, p = -1\n\ncos(x) = 1/2, -1\n\nx = Arccos(1/2) or Arccos(-1)\n\nx = 60 degrees or 180 degrees
484958
sin 2x=2 sinx . cos x\ncos 2x=(cos x)^2-(sin x)^2\ncos 2x=2 (cos x)^2 -1\ncos 2x=1- 2 (sin x)^2\ntan 2x=(sin 2x)/(cos 2x)\n\nor you can \nsin 2x=1\nlet 2x=y\nsin y=1\ny=90 degree + 360 degree\n2x=90 degree + 360 degree\nx=45 degree + 180 degree\nso x={45,225,....}degree\n\ntan x+cot x=2/sin 2x:\ntan x+cot x\n=(sin x/cos x )+(cos x/sin x)\n=((sin x)^2+(cos x)^2)/(sin x . cos x)\n=1/(sin x . cos x)\n=2/2(sin x . cos x)\n=2/sin 2x\nso it is an identity
329959
1-CosA = 2 sin^2 A/2     -----1\n\nFirst whe have to find the value of the cos \n By Cos A = Root(1-Sin^2A)\nCos A = 5/13\nAnd \nThe apply the formaula of the 1\n1- 5/13 = 2Sin^2 A/2\nSinA/2 = Root(8/26) it gives the answer\n And the value of the cos A/2\nIs get by equation 1+ cos A = 2 Cos ^2 A/2
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!!!
912572
sin^3 β + cos^3 β  = (sinβ + cosβ)(sin^2β + cos^2β - sinβcosβ)\n= (sinβ + cosβ)(1 - sinβcosβ)\n\nsin^3 β + cos^3 β / sin β + cos β\n = (sinβ + cosβ)(1 - sinβcosβ)/( sin β + cos β)\n= 1 - sinβcosβ
438454
Change cscx to 1/sinx and sin(-x) = -sinx\ncscx(cscx + sin-x)\n= (1/sinx)(1/sinx - sinx)\nGet a common denominator\n= (1/sinx)([1-(sinx)^2]/sinx)\n1-(sinx)^2 = (cosx)^2\nSo\n(1/sinx)([1-(sinx)^2]/sinx)\n= (1/sinx)((cosx)^2/sinx)\n= (cosx)^2/(sinx)^2\n= (cosx/sinx)^2\n= (cotx)^2
788468
let a be cos2x  | a | <= 1\n4a^2 - 4a + 1 = 0\n<=> (2a - 1)^2 = 0\n<=> 2a -1 = 0 \n<=> a=1/2\n<=> cos2x = cos (pi/3)\n<=> 2x = +- pi/3 + k2pi (k is a subset of Z )\n<=> x = +- pi/6 + kpi\naccording to the problem, we have these following roots :\nx = pi/6 or x = 7pi/6 or x = -pi/6  or x = 5pi/6
1073595
(cosx)^2 - 2sinx = 2\ncosx^2 = 1 - sinx^2\n(1 - sinx^2) - 2sinx = 2\n1 - sinx^2 - 2sinx = 2\n-sinx^2 - 2sinx + 1 = 2\n-sinx^2 - 2sinx - 1 = 0\n-(sinx^2 + 2sinx + 1) = 0\n-(sinx + 1)(sinx + 1)\n\n-sinx - 1 = 0\n-sinx = 1\nsinx = -1\nx = sin^-1(-1)\nx = 270°\n\n--------------------------------------\n\n1.5^2 = 1.7^2 + 1.8^2 - 2(1.7)(1.8)cosA\n2.25 = 2.89 + 3.24 - 6.12cosA\n2.25 = 6.13 - 6.12cosA\n-6.12cosA = -3.88\ncosA = (388/612)\ncosA = (97/153)\nA = 50.6551127\n\n1.7^2 = 1.5^2 + 1.8^2 - 2(1.5)(1.8)cosB\n2.89 = 2.25 + 3.24 - 5.4cosB\n2.89 = 5.49 - 5.4cosB\n-2.6 = -5.4cosB\ncosB = (13/27)\nB = 61.2177953\n\n1.8^2 = 1.5^2 + 1.7^2 - 2(1.5)(1.7)cos(C)\n3.24 =  2.25 + 2.89 - 5.1cosC\n3.24 = 5.14 - 5.1cosC\n-1.9 = -5.1cosC\ncosC = (19/51)\nC = 68.1270919\n\nAs you can see, the angle opposite of side A, is the smallest angle.\n\nANS : about 51°\n\n----------------------------------------------\n\ncos(A) = 4/7\nx = 4\nr = 7\n\nr = sqrt(x^2 + y^2)\n7 = sqrt(4^2 + y^2)\n49 = 16 + y^2\ny^2 = 33\ny = sqrt(33)\n\nsin(A) = y/r\nsinA = (-sqrt(33))/7\n\ntanA = y/x\ntanA = (-sqrt(33))/4\n\ncscA = r/y\ncscA = -7/(sqrt(33))\ncscA = (-7sqrt(33))/33\n\nsecA = r/x\nsecA = 7/4\n\ncotA = x/y\ncotA = -4/(sqrt(33))\ncotA = (-4sqrt(33))/33
226513
csc(x) is the same as 1/sin(x) and sec(x) is the same as 1/cos(x) so the left side of the equation can be simplified to \n\nsin(x)sin(x) + cos(x)cos(x) \nor sin^2(x) + cos^2(x)    {"sin squared x plus cos squared x"}\n\nUsing the same steps, the right side of the equation becomes sin(x)/sin(x) or 1.\n\nSo the identity has been reduced to  sin^2(x) + cos^2(x) = 1 which is another common trigonometry identity!
879956
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~ ♥ ~
842198
I think you mean\n\n(sin^4 t - cos^4 t )/ sin 2t where t is for O\n= (sin ^2 t+ cos^2t)(sin^2 - cos ^ 2 t) as (a+b)(a-b) = a^2 - b^2\n= 1. (- cos 2t)/sin 2t  using sin^2t + cos^2 t = 1 and cos^2 t- sin^2 t = cos 2t\n= - cot 2t
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
107615
t=theta\n^=to the power of...\n\ncos (2t) + 2 sin ^2 (t) = cos (2t) + 2 - 2 cos ^2 (t)