If cosine x is equal to (-4/5) and x is between 180 and 270 degress. solve for sin(x/2), cos(x/2),& tan(x/2).?
931616
COSx = -4/5\nx = ARCCOS(-4/5)\n\nThen just plug in the x value to the rest of the equations.
1159612
1) You meant to have a minus sign not "over". I've used x in place of "theta"\n    secx = 1/cosx and tanx = sinx/cosx \nSo 1 becomes (1/cosx)*sinx - tanx = 0 \nor tanx - tanx = 0, which is true.\n\n2) tanx = 1/cotx, so tanx*cotx = 1 \nSo 2 becomes 1- tanx*cosx = 1-sinx\nBut tanx = sinx/cosx so we have:\n1- (sinx/cosx)*cosx = 1-sinx, so\n1-sinx =1-sinx, which is true.\n\nI'll do one more:\n\n3) tan^2 x - 3sinx*tanx*secx = - 2tan^2 x\nAgain secx = 1/cosx and tanx = sinx/cosx\nSo 3. becomes:\ntan^2 x - 3(sinx/cosx)*tanx = - 2tan^2 x, so\ntan^2 x - 3tan^2 x = -2tan^2 x, so\ntan^2 x = tan^2 x, which is true.\n\nYou need to learn to do these yourself: Here is a link which gives you all the trig identities to solve the rest. They are not difficult.\n \nhttp://www.sosmath.com/trig/Trig5/trig5/trig5.html
938253
2cos²(45º-0.5x) – 1 = sinx\n\nUsing: cos 2a = 2cos² a -1\nSo: 2cos²(45º-0.5x) – 1 = cos 2(45º-0.5x) = cos (90º - x)  = sinx
784816
sin x - cos (2x) + 1 = 0\nsin x - (cos² x - sin² x) + 1 = 0\nsin x - (1 - 2 sin² x) + 1 = 0\n2 sin² x + sin x = 0\n(2 sin x + 1) sin x = 0\n\nSo either sin x = 0 or 2 sin x + 1 = 0. In the latter case:\n\n2 sin x + 1 = 0\n2 sin x = -1\nsin x = -1/2\n\nSo either sin x equals 0 (in which case x is coterminal to either 0 or π) or sin x = -1/2 (in which case x is coterminal to either 7π/6 or 11π/6). As before, we must test these solutions:\n\nsin 0 - cos 0 + 1 = 0 → 0 - 1 + 1 = 0, so x=0 is valid\nsin π - cos (2π) + 1 = 0 → 0 - 1 + 1 = 0, so x=π is valid\nsin (7π/6) - cos (14π/6) + 1 = 0 → -1/2 - 1/2 + 1 = 0, so x=7π/6 is valid\nsin (11π/6) - cos (22π/6) + 1 = 0 → -1/2 -1/2 + 1 = 0, so x=11π/6 is valid\n\nErgo, all of the solutions we just found are valid, and x is coterminal to any one of 0, π, 7π/6, of 11π/6\n\nEdit: You're welcome, Jessica.
1162151
Use the fact that tan^2(x) = sec^2(x ) - 1 and integrate that.  You will get tanx -x +c.
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
274244
Put Y=tan x and solve the equation for Y.\nThis gives two solutions Y=tan x = 1,7808 and Y =tan x = -0,2808.\nThis leads to x = 60.68 degrees and x = -15.68 degrees.\n\nAdd 360 degrees to the second solution and transform to degrees and minutes for the final answer.
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.
1086900
Yes it is true for x=200.62 degrees.\n\nCarefuly schetch tan(x) and sec(x) [sec(x)=1/cos(x)] and you will see the answer or let your fingers do the calculator walk :).\n\n\nWas that helpfull?
1159205
pair up the  opposite sines, for example sin(10) and sin(350) and add them:\nsin(10) + sin(350) = 0\nsin(20) + sin(340) = 0\n...\nuntil you only have sin(180) and sin(360) which both equal 0\n\nPS. I assumed you were using degrees (sexagesimal). Won't work in radians (RAD) or metric (decimal) angles.
809968
(tan x- cot x)/( tan x + cot x)+ 2cos^2 x=1\n\nSenx / cos x  - cosx/senx\n---------------------------------  + 2cos^2x\nSenx / cos x  + cosx/senx\n\n(sen^2 x - cos^2x) / senx cosx\n--------------------------------------  + 2cos^2 x\n(sen^2 x + cos^2x) / senx cosx\n\nThe fraction reduces on the above:\n\nsen^2 x - cos^2 x                       + 2 cos^2 x\n\nsen^2x + cos^2x = 1\n\nproved!
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
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.
766844
1/tanx - 1/x=\n(x-tanx)/(xtanx)=0/0\nso use l'hopital's rule\n(1-sec^2(x))/(tanx+xsec^2(x))=\n(1-1)/(0+0)= 0/0\nso use l'hopital again\n(-2secxsecxtanx)/(sec^2(x)+\nsec^2(x)+x(2secxsecxtanx))=\n0/(2+0)=0
291250
Factor the left side: 2cosx(tanx+1)=(tanx + 1)\n\nThis is true whenever tanx+1 = 0, or 2cosx = 1\n\ntanx = -1 in degrees, -135 or 315\ncosx=1/2 in degrees, 60 or 300\n\n(or multiples of 360 plus the above)
821835
Problem 1.\nu = sin x\nu^2 + 3u + 2 = 0\nu = -3/2 +/- 1/2\nu = -1, -2\nsin x = -1  [because the sine is never -2]\nx = 270 degrees\n\nProblem 2.\nu = cos x\nu^2 + 2 u + 1 = 0\n(u+1)(u+1) = 0\nu = -1\ncos x = -1\nx = 180 degrees
842500
Use the sum/product identity. sin(a) + sin(b) = 2 sin((a+b)/2)cos((a-b)/2).\nSo sin(x+pi/4)+sin(x-pi/4) = 2sin(x)cos(pi/4) = -1. So 2sin(x)*sqrt(2)/2 = -1, so sin(x) = -1/sqrt(2).\nThat gives x = 5pi/4 or 7pi/4.