Prove the identity. (For Trigonometry)?
969104
okay,\n\ncos(x-y)      cos(x)*cos(y) + sin(x)*sin(y)\n----------  =   ----------------------------------- \ncos(x+y)     cos(x)*cos(y) - sin(x)*sin(y)\n\n(multiply and divide the fraction by cos(x)*cos(y))\n\n     1 + tan(x)*tan(y)\n=   ---------------------\n     1 -  tan(x)*tan(y)
925769
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2
714450
Use the identity cos(ø/2) = √[(1+cos(ø))/2]; then you get\n\nsinø = √[(1+cosø)/2]  Square both sides\n\nsin^2(ø) = (1+cosø)/2\n\n2sin^2(ø) = 1+cosø;  Expand sin^2(ø) = 1 - cos^2(ø) \n\n2 - 2*cos^2(ø) = 1 + cosø\n\n2*cos^(ø) + cosø -1 = 0\n\nSubstitute x = cosø\n\n2x^2 + x - 1 = 0  factor this (2x - 1)*(x + 1); so x = -1 or x = .5\n\nso cosø = -1 or cosø = .5\n\nø = (2n+1)π   and  ø = π/3+2nπ\nwhere n = 0,1,2...
610758
sin(2x) = 2sinxcosx is a trigonometric identity called the double angle identity. It was derived from another identity:\n\n   sin(x + y) = sinxcosy + cosxsiny\n \nhere x and y are two different angles but what if x = y\nreplace y with x to make all the equations in terms of x, thus the equation becomes:\n\n   sin(x + x) = sinxcosx + cosxsinx\n   \nnote that sinxcosx is also equal to cosxsinx, so simplifying the equation it becomes:\n\n  sin(2x) = 2sinxcosx or sin(2x) = 2cosxsinx
851033
(cosx-sinx)^2=cos^2x+sin^2x-2sinxcosx so\nit can be 1-2sinxcosx+cosx+sinx
749606
L.H.S. (tanx - cotx)/(tanx + cotx) + 2cos^2x = (sinx/cosx - cosx/sinx)/(sinx/cosx + cosx/sinx) + 2cos^2x = (sin^2x - cos^2x)/(sin^2x + cos^2x) + 2cos^2x = (sin^2x - cos^2x)/1 + 2cos^2x = sin^2x - cos^2x+ 2cos^2x = sin^2x + cos^2x = 1 = R.H.S.
958733
sin²y + cos²y = 1\nSo cos²y = 1 - sin²y\n= (1 - siny)(1 + siny)\nTherefore\ncos²y/(1 + siny) = 1 - siny .................... QED
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.
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 !!!
857809
Hi, on your first one,\nsin^3 θ cos θ - sin θ cos^3 θ = -1/4 \n\nFactor out sin θ cos θ (sin^2 θ - cos^2 θ) = -1/4\n You will notice that if the part in parentheses were reversed it would be an identity for cos 2θ so multiply both sides by -1 to have this happen:\n\nsinθ cosθ (cos^2θ - sin^2θ) = 1/4\n\nYou will also notice if there were a 2 in front you'd have an identity for sin 2θ so multiply both sides by 2 to get\n\n2 sinθcosθ (cos^2θ - sin^2θ) = 1/2\n\nreplace with the identities: sin 2θ cos2θ = 1/2\n\nNow, if this had a 2 in front it would be an identiti for sin 2(2θ) so mult. both sides by 2 to get\n2 sin 2θ cos 2θ = 1\n\nsin 2(2θ) = 1\n\nsin 4θ = 1\n\nYou know that sin pi/2 = 1 so 4θ = pi/2 and θ = pi/8.  You can figure the additional answers as θ increases.\n \n\n If I get the second I'll edit this unless someone else already got it\n\nHere's what I think:\nsec 4θ + 2 sin 4θ = 0\n\n1/cos4θ + 2 sin 4θ = 0\n\nget common denominators\n\n1/cos 4θ + 2 sin4θ cos4θ/cos 4θ\n\nyou will see that the top of the second fraction is an identity for sin 2(4θ) so replace it\n\n1/cos4θ + sin 2(4θ)/cos4θ = 0\n\nThis is 1 + sin 8θ all over cos 4θ = 0\n\nIf a fraction = 0, its numerator = 0 so 1 + sin 8θ = 0\n\nsin 8θ = -1\n\nsin of 3pi/2 = -1 so 8θ = 3pi/2 and θ = 3pi/16\n\nagain there are infinitely many more answers unless you were given a specific domain for θ\n\nHope that helps
592160
i think the answer wil be (tanX)/(tanY)\n\nSIN(X+Y)=sin x cos y+cos x sin y\n\nsin(x-y)=sin x cos y-cos x sin y\n\n use cos x cos y to divide the numerator and denominator... there is the answer...
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)
397638
Use these identities:\nsin(2x) = 2*sin(x)*cos(x)\nsin(4x) = 4*cos^3(x)*sin(x) - 4*cos(x)*sin^3(x)\ncos(2x) = -1 + 2*cos^2(x)\ncos(4x) = 1 - 8*cos^2(x) + 8*cos^4(x)\ntan(3x) = (3*tan(x)-tan^3(x))/(1-3*tan^2(x))\n\nAfter you expand the left hand side of the equation with the above identities, you can factor out a 2*sin(x)*cos(x) from the numerator and a 2*cos^2(x) from the denominator, which will turn into a tan(x). Now divide the numerator and denominator of the fraction by cos^2(x). You should get the right hand side of the tan(3x) identity above.\n\nTry this and if you can't figure it out, I'll write it out. (I don't really want to since equations look like garbage on this site).\n\nCheck out this site for some usefull identities.
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.
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
72362
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.
287405
Use the identity cos 2x = 2 cos^2 x - 1\n\nThen you have 3(2cos^2 x -1) - 5cosx - 1 = 0\n\n6 cos^2 x - 3 - 5cosx - 1 = 0\n\n6 cos^2 x - 5 cosx - 4 = 0\n\n(3 cosx - 4)(2 cosx + 1) = 0\n\ncosx = 4/3 (impossible) or cosx = -1/2\n\nIn degrees, x = 120, 240, or if you need, 120+360n, 240+360n\nIn radians, x = 2pi/3 or 4pi/3 etc.\n\nHope that helps you see what to do.