X\nlim  ------------ = ?????\nx->0 (sqrt)1-cosx
215813
JohnyBoy almost has it right.  When you multiply both numerator and denominator by sqrt(1+cosX), the numerator becomes    X sqrt(1+cosX)        , not X (1+cosX),\nThe rest is correct.  So the limit is sqrt(2).
515341
Are you sure this is the correct identity?\n\n 1+ tanx.tan2x = 1 + (sinx/cosx)(sin2x/cos2x)\n                        = 1 + (sinx/cosx)(2sinxcosx/cos2x)\n                        = 1 + 2sinx.sinx/cos2x\n\n cos2x = 2cosx.cosx - 1  = 2cos^2x - 1        (Note:  ^2 = square)\n\n then..\n      1+tanx.tan2x = 1 + [2sin^2x/(2cos^2x - 1)]\n                            = (2cos^2x + 2sin^x -1) / (2cos^2x - 1)\n                            = [2(cos^2x + sin^2x) -1] / (2cos^2x - 1)\n                            = 1/cos2x    (because 2cos^2x - 1 = cos2x)\n\n     Then 1/1+tanxtan2x = cos2x
476695
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
384342
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
28448
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
384365
cosx=tanx\ncosx=sinx/cosx\ncos^2(x)=sinx\n1-sin^2(x)=sinx    {pythagoras}\nsin^2(x)+sinx-1=0\nuse quadratic formula\nsinx=(-1+or-sqrt(5))/2\n=0.6180339888\nor -1.6180339888\nsine has to lie between +/-1\nsinx= 0.6180339888\nx=arcsin(0.6180339888)\n  =38.17270763\n+ve value for sine lies in 1st and 2nd\nquadrants \ntherefore,x=38.17270763 degrees or\n141.8272924 degrees\n\nthese two values were checked in the \noriginal equation and were found \nto be correct\n\ni hope that this helps
523708
Did you make a typo? 1 - 1 = 0. Maybe you mean lim x --> 0 of \n\n     ( 1 - 1/x ) sin x cos( x/2 )\n\n     = sin x cos (x/2) - ( sin x / x ) cos( x/2 ) \n\nLimits as x --> 0 of sin x, cos( x/2 ) and sin x / x are 0, 1 and 1, so answer is\n\n     0 × 1 - 1 × 1 = -1
421004
(sin(x))/(csc(x)-1) + (sin(x))/(csc(x)+1)\n=[(sin(x)*(csc(x)+1))+(sin(x)*(csc(x)-1))]/[(csc(x)+1)(csc(x)-1)]\n=[sinx((csc(x)+1)+(csc(x)-1))]/[((csc(x))^2)-1]\n=[2sin(x)csc(x)]/[(cot(x))^2]\n=2/[(cot(x))^2]\n=2(tan(x)^2)
480031
cot(-x)cos(-x) + sin(-x) = -csc x\n\nLHS = cot(-x)cos(-x) + sin(-x)\n= cos(-x)/sin(-x) * cos(-x) + sin (-x)\n= cos²(-x)/sin(-x) + sin²(-x)/sin(-x)\n= (cos²(-x) + sin²(-x))/sin(-x)\n= 1/sin(-x)\n= 1/-sin(x)\n= -csc(x)\n= RHS
489270
cos2x = 1 - 2 (sin x) ^2 = sin x\n\nthen 2 ( sin x)^2 + sin x  - 1 = 0\n or (2sin x - 1 )(sin x + 1) = 0\nsin x = 1/2 , - 1\nx = 30 deg, or - 90 deg
637727
Just plug in 1 for x.\n√(1-1) = √0 = 0
374338
Working above is good until this step:\n\n2 + 0.8x >0.9(x + 6)\n2 + 0.8x >0.9x + 5.4)\n0.1x > -3.4 ← Symbol error in this line\n\nShould read 0.1x < -3.4\nie x < -34
72188
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.
495907
2cos^2x-1+1/2sinxcosx\n=2cos^2x/2sinxcosx\n=cosx/sinx\n=cotx
92095
3cosx-cosx-2=0\n2cosx=2\ncosx=1\nx=0 or x=2π (on domain from o to  2π)
553212
Hello again dear Math student \n\nas you know ;\n▪sgn(x) = \n+1 , x >0\n 0  , x = 0\n- 1 , x <0\n\nso \n▪ D sgn(x)  = IR   \n▪ R sgn(x) = { -1 , 0 , +1 }\n\nGood Luck
443184
function Cos^ (-1)x is an inverse of function Cos x\n(like square(x) and squareroot(x) act oppostite\nsquare(sqrootx) = x or sqroot(square(x) = x\nTherefore Cos[Cos^(-1)x = x\nNow to your question\nLet [Cos^(-squareroot3/2) = x\nTaking Cos on both side\nCos[Cos^(-1)(-squareroot3/2)] = Cos x\nor -squareroot3/2 = Cos x\nThis equation has many solutions as you can see (for every cycle of 360 deg you will have two solutions - one each in second quadrant and  third quadrant where Cosine (or cos) has negative values\nPrimarily we take the solution which between zero and 360 degrees\nTwo solutions are\nCos (180 -30) = Cos (180+30) = (- squareroot 3/2)\ntherefore 150 deg and 210 deg\nconverted to 150xpi/180 = 5/6 pi radians\nand 210xpi/180 = 7/6 pi radians\nSimilarly we can find a set of two solutions between 360 &720 deg, between 720 &1080 deg etc and so on\neg [360 + (180+30)] deg and [360 +180-30] deg [720 + (180+30)] deg [720+ (180-30)] deg and so on\nCheers\nSubhash