please...help me in provin this......cos (a+b) cos (a-b)=(cos a+sin b)(cos a-sin b)?
464966
Its actually easy:\n\ncos(a+b) = cos a cos b - sin a sin b\ncos(a-b) = cos a cos b + sin a sin b\n\nso your lhs = cos^2 a cos^2 b - sin^2 a sin^2 b\n=cos^2 a ( 1 - sin^2 b) - ( 1 - cos^2 a) sin^2 b\n=cos^2 a - cos^2 a sin^2 b - sin^2 b + cos^2 a sin^2 b\n=cos^2 a - sin^2 b\n= (cos a - sin b)(cos a + sin b)\n=rhs\n\nHope I am clear here.
907452
This is derived quickest from de Moivre's Thm, which is \n\ncos(nx) + isin(nx) = (cosx + isinx)^n\n\n(where i^2 = -1)\n\nExpand the RHS, separate the Re and Im terms to get both cos(nx) and sin(nx) as polynomials in cos(x) and sin(x). Your case is n=3.\n\nOr you can plough through cos(x+x+x) applying f-las for cos(a+b) and sin(a+b).
336698
2. (2sin^2 x - 1)^2 / sin^4 x - cos^4 x = 1 - 2cos^2 x\n\nAssuming you meant\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\nthen\n\n(2sin^2 x - 1)^2 / (sin^4 x - cos^4 x)\n = (2sin^2 x - 1)^2 / ((sin^2 x + cos^2 x)(sin^2 x - cos^2 x))\n = (2sin^2 x - 1)^2 / (sin^2 x - cos^2 x)\n = (2(1 - cos^2 x) - 1)^2 / ((1 - cos^2 x) - cos^2)\n = (1 - 2cos^2x)^2 / (1 - 2cos^2 x)\n = 1 - 2cos^2 x\n\n\n\nUpdate (added after wiildmunkii's answer)\n= = = = = = = = = = = = = = = = = = = = =\n\n1. sin^2 x + sin^4 x / cos^2 x + cos^4 x = tan 3x\n\nIf you meant\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x)) = tan3x\nthen yes.\n\nCombining\n\nsin(a+b) = sin(a) cos(b) + cos(a) sin(b)\ncos(a+b) = cos(a) cos(b) - sin(a) sin(b)\ntan(a + b) = sin(a + b) / cos(a + b)\n = (sin(a) cos(b) + cos(a) sin(b)) / (cos(a) cos(b) - sin(a) sin(b))\n = (tan(a) + tan(b)) / (1 - tan(a) tan(b))\n\nyou get\n\nsin(2x) = sin(x) cos(x) + cos(x) sin(x)\n = 2 sin(x) cos(x)\ncos(2x) = cos(x) cos(x) - sin(x) sin(x)\n = cos^2(x) - sin^2(x)\n = cos^2(x) - (1 - cos^2(x))\n = 2 cos^2(x) - 1\n [similarly = 1 - 2 sin^2(x)]\ntan(2x) = (tan(x) + tan(x)) / (1 - tan(x) tan(x))\n = 2 tan(x) / (1 - tan^2(x))\n\n\ne^it = (cos(t) + i sin(t))\n\nSo (e^it)^n = (cos(t) + i sin(t))^n\n\nBut (e^it)^n = (e^int) = cos(nt) + i sin(nt)\n\nSo cos(nt) + i sin(nt) = (cos(t) + i sin(t))^n\n\nFor any given n, expand the right hand side using the binomial theorem and equate real and imaginary parts to get:\n\ncos(2t) + i sin(2t) = cos^2(t) - sin^2(t) + 2 i sin(t) cos(t)\ncos(2t) = cos^2(t) - sin^2(t)\n = 2 cos^2(t) - 1\n = 1 - 2 sin^2(t)\nsin(2t) = 2 sin(t) cos(t)\n\ncos(3t) + i sin(3t) =\n cos^3(t) + i (3 cos^2(t) sin(t)) - (3 cos(t) sin^2(t)) - i sin^3(t)\ncos(3t) = cos^3(t) - 3 cos(t) sin^2(t)\n = 4 cos^3(t) - 3 cos(t)\nsin(3t) = 3 cos^2(t) sin(t) - sin^3(t)\n = 3 sin(t) - 4 sin^3(t)\n\ncos(4t) + i sin(4t) =\n cos^4(t) + i (4 cos^3(t) sin(t)) - (6 cos^2(t) sin^2(t)) - i (4 cos(t) sin^3(t)) + sin^4(t)\ncos(4t) = cos^4(t) - 6 cos^2(t) sin^2(t) + sin^4(t)\n = cos^4(t) - 6 cos^2(t) (1 - cos^2(t) + (1 - cos^2(t))^2\n = 8 cos^4(t) - 8 cos^2(t) + 1\nsin(4t) = 4 cos^3(t) sin(t) - 4 cos(t) sin^3(t)\n = 4 cos(t) sin(t) (cos^2(t) - sin^2(t))\n = 4 cos(t) sin(t) (2cos^2(t) - 1)\n = 4 cos(t) sin(t) (1 - 2 sin^2(t))\n\nFinally...\n\n(sin(2x) + sin(4x)) / (cos(2x) + cos(4x))\n = (2 sin(x) cos(x) + 4 cos(x) sin(x) (1 - 2 sin^2(x))) / (2 cos^2(x) - 1 + 8 cos^4(x) - 8 cos^2(x) + 1)\n = (6 sin(x) cos(x) - 8 sin^3(x) cos(x)) / (8 cos^4(x) - 6 cos^2(x))\n # divide by 2 cos(x)\n = (3 sin(x) - 4 sin^3(x)) / (4 cos^3(x) - 3 cos^(x))\n = sin(3x) / cos(3x)\n = tan(3x)
1093073
A = 45 degrees or (pi / 4) radians\n\n1/sin^2A-1/cos^2A-1/tan^2A-1/cot^2A-1/sec^2A-1/csc^2A = -3\nIf i understand the way you wrote the problem above, it's 1/ each term, subtract each other right?\n1/ ((sinA)^2) - 1/ ((cosA)^2) - 1/ ((tanA)^2) - 1/ ((cotA)^2) - 1/ ((secA)^2) - 1/ ((cscA)^2) = -3\nI'm going to drop the A part, i assume you know that it will still be officially added in the final version\n\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cot^2)] + [-(tan^2)] + [-(cos^2)] +  [-(sin^2)] = -3\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cos^2) / (sin^2)] + [-(sin^2) / (cos^2)] + [-(cos^2) - (sin^2)] = -3\ncombine denominators\n[( 1 - (cos^2) ) / (sin^2)] + [( (-1) - (sin^2) ) / (cos^2)] + (-1) [(cos^2) + (sin^2)] = -3\n[(sin^2) / (sin^2)] + (-1)[( 1 + (sin^2) ) / (cos^2)] + (-1) [ 1 ] = -3\nTrig identity [(cos^2) + (sin^2)] = 1\n[ 1 ] + (-1)[( 1 + (sin^2) ) / (cos^2)] + [ -1 ] = -3\n(-1)[( 1 + (sin^2) ) / (cos^2)] = -3\n[( 1 + (sin^2) ) / (cos^2)] = +3\n1 + (sin^2) = 3 * (cos^2)\nTrig identity 1 = [(sin^2) + (cos^2)] =\n[(sin^2) + (cos^2)] + (sin^2) = 3 * (cos^2)\n2 * (sin^2) = 2 * (cos^2)\nsin^2 = cos^2\nsinA = cosA\nA= 45 degrees of (pi/4) radians
1028428
well..\n\nI'll give you one:\n\ntanx + cotx = 1/(sinx cosx)\n\ntanx = sinx/cosx\ncotx = cosx/sinx\n\nsubstituting:\n\nsinx/cosx + cosx/sinx\n\nmultiply each term to get a common denominator:\n\nsinx/cosx *(sinx/sinx) + cosx/sinx *(cosx/cosx)\n\nsin^2 x / (cosx*sinx) + cos^2x/(cosx*sinx)\n\nsince sin^2x + cos^2 x = 1, and you can add the two terms together because they have a common denominator, the entire expression equals:\n\n1/(sinxcosx)
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 <--
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)
843808
If sin A = x and sinB = -x  then \nA = .... , -4π - B, -3π + B, -2π - B, -π + B,.-B, π + B, 2π - B, 3π + B, 4π - B, .... \n= nπ + (-1)^nB (n is any integer)\n\nSo arcsin(x) - arcsin(-x) = nπ + (-1)^nB - B\n= nπ + ((-1)^(n) -1)B\n\nNo this looks too clumsy ....new strategy!!\n\nsinA + sinB = 0\nSo 2 sin(½(A + B))cos(½(A - B)) = 0\n\nThus sin(½(A + B)) = 0 or cos(½(A - B)) = 0\nie ½(A + B) = 0, ±π, ±2π, ±3π, .... , ±nπ, \nie A + B = ±2nπ\nie arcsinx + arcsin (-x) = ±2nπ\n\nOr ½(A - B) =  ±π/2, ±3π/2, ±5π/2, ... , ±(2n - 1)π/2\nSo A - B = ±(2n - 1)π\nSo arcsinx - arcsin(-x) = ±(2n - 1)π
324333
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.
814691
Did you get the question correct?\n\nStart from right-hand side:\n   cos^2x sinx - cos^4x\n= cos^2x [sinx - cos^2x]\n= cos^2x [sinx - (1 - sin^2x)]\n= cos^2x [sin^2x + sinx -1]\n\nYou can't transform (sin^2x + sinx - 1) into sin^3x.
728989
we have in a triangle\n\nsin B/b = sin C/c = k (say)\n\nsin B= bk  ...1\nsin C = ck  ...2 \nadditionally\n\nb^2 = c^2+a^2-2ac cos B\nso cos B = (c^2+a^2-b^2)/2ac ...3\n\ncos  C = (a^2+b^2-c^2)/2ab ...4\nfrom1 and 3 by deviding 3 by 1\ncot B = (c^2+a^2-b^2)/2abck\n\nor 2abck = (c^2+a^2-b^2) tan B  ..5\nsimilarly from 2and 4\n2abck =(a^2+b^2-c^2) tan C\n\nfrom above 2 we get the result as both are same
313252
This relies mostly on knowing these two identities:\n\nSin (A + B) = Sin A Cos B + Cos A Sin B\nSin (A – B) = Sin A Cos B – Cos A Sin B\n\nso the first half is Sin x Cos pi/4 + Cos x Sin pi/4\n\nbut since both Cos pi/4 and Sin pi/4 are (sqrt 2)/2 this reduces to \n\n(sqtr 2)/2 ( Sin x + Cos x)\n\nIn the same way the second half reduces to\n\n(sqrt 2)/2 (Sin x - Cos x)\n\nNow we plug these back into the original and square (because we want sin^2 of each):\n\n2/4( Sin^2 x + 2Sin x Cos x + Cos^2 x) + 2/4 ( Sin^2 x - 2Sin x Cos x + Cos^2 x) = 1\n\nYou should know Sin^2 x + Cos^2 x = 1 so we simplify:\n\n1/2(1 + 2Sin x Cos x) + 1/2(1 - 2Sin x Cos x) = 1\n1/2 + 1/2 + 2Sin x Cos x - 2Sin x Cos x = 1\n1 = 1
429515
The site below, around one-third of the way down, shows the geometry needed to answer your question.\n\nOnce you have the formula for cos(A+B), namely cosAcosB - sinAsinB, the proof is easy. Just substitute -B for B, and you get\n\ncos(A-B) = cosAcos(-B) - sinAsin(-B)\n\nNow, cos(-B) = cosB and sin(-B) = -sin(B). Thus\n\ncos(A-B) = cosAcosB + sinAsinB
904134
sin3x = sin (x+ 2x) \n          = sinx.cos2x + cosx.sin2x\n          = sinx (2cos^2x -1) + cosx (2sinx.cosx)\n          = 2sinx.cos^2x - sinx + 2sinx.cos^2x\n          = sinx (2cos^2x - 1 + 2cos^2x)\n          = sinx (4cos^2x - 1)\n \n But, cos^2x = 1 - sin^2x\nso then,\n\n   sin3x = sinx [ 4 - 4sin^2x - 1]\n             = sinx (3 - 4sin^2x)\n\n Therefore,\n\n  sin3x/sinx =  sinx(3 - 4sin^2x)/sinx\n                   =  3 - 4sin^2x  --- answer.
729562
1/sin(x) - cos(x) * cos(x)/sin(x)\n1/sin(x) - cos^2(x)/sin(x)\n(1-cos^2(x))/sin(x)\nsin^2(x)/sin(x)\nsin(x)
908544
sin 2a / (1 + cos 2a)\n\n(2 sin a cos a) / (1 + (2 cos ^2 a - 1) )\n\n2 sin a cos a / (2 cos ^2 a)\n\nsin a / cos a\n\ntan a
763314
cot (A + B) = 1/tan(A + B) =\n\n1 - tanA tanB\n----------------- =\ntan A + tan B\n\n1 - 1/(cotAcotB) .. cotAcotB\n-------------------- X ------------ =\n1/cotA + 1/cotB .. cotAcotB\n\ncotAcotB - 1\n---------------- ............ QED\ncotB + cotA