How do I prove this? Not solving for 0, but making a proof? (tanx + 1) ^2 = 1 + (2sinx)(cosx) / (cos^2x)?
826721
(1+tanx)^2\n=1+tan^2x+2tanx\n=sec^2x+2tanx\n=1/cos^2x+2sinx/cosx\n=1+2sinxcosx/cos^2x\nhenceproved
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
249210
if you mean (1+sinx)/cosx-cosx/(1-sinx)=0, as I'm sure you do, you add the fractions, cross-multiplying to get the numerators:\n\n[(1+sinx)(1-sinx)-cos²x]/[cosx(1-sinx)]=\n[1-sin²x-cos²x]/[cosx(1-sinx)]=\n[cos²x-cos²x]/[cosx(1-sinx)]=\n0/[cosx(1-sinx)]=\n0
880634
tan²x =\n\nsin²x\n------- =\ncos²x\n\n½(1 - cos2x)\n------------------ =\n½(1 + cos2x)\n\n1 - cos2x\n------------- =\n1 + cos2x\n\n2 - 1 - cos2x\n----------------- =\n1 + cos2x\n\n2 - (1 + cos2x)\n-------------------- =\n1 + cos2x\n\n2\n-------------  - 1\n1 + cos2x\n\nOR\n\ntan²x =\n\nsec²x - 1 =\n\n1\n--------  - 1 = \n cos²x\n\n1\n------------------  - 1 =\n½(1 + cos2x)\n\n2\n-------------  - 1\n1 + cos2x
475675
tan x = sin x/cos x\n\nsin 2x = 2 sin x cos x\n\nso tan x sin 2x = (sin x/cos x) * (2 sin x.cos x)\n                = 2 sin^2 x\nhalf of it is = sin^2 x\n\nDifferentiate half of tan x sin 2x\n= diff of sin^2 x\n= 2 (sin x).cos x  (refer to below)\n= sin 2x    (Q.E.D.)\n\n\nDifferentiating sin^2 x.\n\nLet's call sin x as A\nand B = A^2\nthen dB/dA = diff of B * diff of A\n           = 2A * diff of A\n           = 2 sin x * cos x\n           = sin 2x
841196
This is interesting....\n\nTry this.  First, change all tan(x) and cot(x) into sin(x)/cos(x) and cos(x)/sin(x) respectively.  Then combine the fractions using common denominator which is cos(x)sin(x).\n\nEventually, the fraction will reduce to sin^2(x) -cos^2(x)\n\nPlug this back into the original equation and you get:\nsin^2(x) - cos^2(x) + 2cos^2(x) = 1\nsimply and you get:\nsin^2(x) + cos^2(x) = 1\n1 = 1\n\nX can be any real number.....
881848
First apply the identities tan x = sin x/cos x and cot x = cos x/sin x:\n\n(tan x - cot x)/(tan x + cot x) + 2 cos^2 x = (sin x/cos x - cos x/sinx)/(sin x/cos x + cos x/sin x) + 2 cos^2 x.\n\nNow get a common denominator and combine some fractions:\n\n(sin x/cos x - cos x/sinx)/(sin x/cos x + cos x/sin x) + 2 cos^2 x =\n((sin^2 x - cos^2 x)/ sin x cos x)/((sin^2 x + cos^2 x)/ sin x cos x) + 2 cos^2 x\n\nWe can cancel the sin x cos x's on bottom, and get\n\n((sin^2 x - cos^2 x)/ sin x cos x)/((sin^2 x + cos^2 x)/ sin x cos x) + 2 cos^2 x = (sin^2 x - cos^2 x)/(sin^2 x + cos^2 x) - 2 cos^2 x.\n\nNow, applying the identity sin^2 x + cos^2 x = 1 gives\n\n(sin^2 x - cos^2 x)/(sin^2 x + cos^2 x) - 2 cos^2 x = (sin^2 x - cos^2 x) + 2 cos^2 x.\n\nSimplifying gives\n\n(sin^2 x - cos^2 x) + 2 cos^2 x = sin^2 x + cos^2 x.\n\nAnd finally\n\nsin^2 x + cos^2 x = 1.
652227
Taking RHS\n\ncot^2x(secx-1)\n=(secx-1)/tan^2x\n=(secx-1)/(sec^2x-1)\n=1/(secx+1)\n\nWhich is the LHS
591905
1+2(tanx)^2+(tanx)^4\n= (1+(tanx)^2)^2\n= ((secx)^2)^2\n= (secx)^4\n= (cosx)^-4
148647
x+sinx\nLim ------        \n    x+2sinx\nx->0\n \n    1+cosx\nlim-------   [according to Hospital's theory]\n    1+2cosx\nx->0\n\n   1+cos0\n=----------\n   1+2cos0\n\n  1+1\n=-----\n  1+2\n\n  2\n=----\n  3\n ans 2/3.\nThanks.\n  3
946513
Subject: Re: Proving trig identiies...small error..thisis the right one\nFrom: \nDate: Monday June 14, 2004 at 08:35:54\n\nRHS= 2csc^2x=2(cot^2x+1) \n\nnote :-(csc^2x=cot^2x+1) \n\n=2cot^2x+2 \n\n=2cos^2x/sin^2x +2 \n\nnote :-((cot^2x=cos^2x/sin^2x) \n\n=2cos^2x/(1-cos^2x) +2 \n\nnote :-((sin^2x=1-cos^2x) \n\n=2cos^2x +2(1-cos^2x) /(1-cos^2x) \n\n=2cos^2x +2 -2cos^2x) /(1-cos^2x) \n\n=2/(1-cos^2x) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-((1-cos^2x=(1-cosx)(1+cosx)) \n\n\nTHEREFORE lHS =2/(1-cosx)(1+cosx) (1-cos^2x=(1-cosx)(1+cosx)) \nnow for the RHS we get \n\nRHS= 1/1+COSX + 1/1-COSX \n\n=(1-cosx) + ( 1 + cosx)/(1-cosx)(1-cosx) \n\nnote :-(( through expansion) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-(cosx on the numerator cancel out) \n\ntherefore RHS= LHS \n\nTHEREFORE 2csc^2x= 1/1+cosx + 1/1-cosx \n\nas required\n\n\n\nCurrent Thread\n   Proving trig identiies   a girl - Monday June 14, 2004 at 08:06:52 \n   Re: Proving trig identiies   - Monday June 14, 2004 at 08:31:28 \n--> Re: Proving trig identiies...small error..thisis the right one   - Monday June 14, 2004 at 08:35:54 <--
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
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.
846581
Expand the exponent: (cos (x/2) - sin(x/2))^2 = (cos x/2)^2 - 2*cos(x/2)*sin(x/2) + (sin x/2)^2 = 1 - 2*sin(x/2)*cos(x/2) = 1-sin(x)\n\nusing the fact (cos a)^2 + (sin a)^2 = 1 and 2*sin(a)*cos(a) = sin(2a)\n\nSteve
181415
let that be x insted of *, it's easier to write\nsinx/(1+cosx) +1 + cosx/sinx = \n[(sinx)^2 + sinx + sinxcosx + cosx + (cosx)^2]/[(1+cosx)*sinx]=\n(sinx+cosx+sinxcosx+1)/(1+ cosx)sinx=\n[sinx+ 1 + cosx(1+sinx)]/[(1+cosx)sinx)]=\n[(1+sinx)(1+cosx)]/[(1+cosx)\nsinx]=\n(1+sinx)/sinx=\n1/sinx +1 = cscx+1\nI don't know maybe I have done something wrong but it doesn't look like 2cscx.
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 !!!
797886
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)