Trigonometric transformation?
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)
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)
394432
sin(x) - cos(x) - tan(x) = -1\nsin(x) - cos(x) - ((sin(x))/(cos(x))) = -1\n(sin(x)cos(x)) - cos(x)^2 - sin(x) = -cos(x)\n(sin(x)cos(x)) - sin(x) = cos(x)^2 - cos(x)\n((sin(x))(cos(x) - 1) = (cos(x))(cox(x) - 1)\n\ndivide both sides by (cos(x) - 1)\n\nsin(x) = cos(x)\nor\nsin(x) = 1cos(x)\n\ndivide both sides by cos(x)\n\n(sin(x))/(cos(x)) = 1\n\ntan(x) = 1\n\ninverse tan both sides\n\nx = 45° or 225°\n\nAlso i found out that "x" also equals 0\n\nIf you go to www.quickmath.com, it will give you answers of (pi/4) which equals 45°, (-3pi/4) which equals 225°, and 0.
983394
Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
835259
I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)
801525
This is a tricky one. You have to use a couple of steps to get this one. First you use a u-substituion to "simplify" it and then use a trig substitution.\n\nIf you let u=cos(x) then du=-sin(x) then the integral becomes\n&#92;int -1/(1+u^2) du.\n\nNow let u=tan(s) and du=sec^2(s) and the integral is\n&#92;int -sec^2(s)/(1+tan^2(s)) ds = &#92;int -sec^2(s)/sec^2(s) ds = &#92;int -1 ds.\n\nThe integral is -s + C where C is any constant. We now have to undo the steps to get it back in terms of x. First, we undo the trig substituion. Since u=tan(s) then the integral is\n\n-arctan(u)+C\n\nSince u=cos(x) we get\n\n-arctan(cos(x))+C
107615
t=theta\n^=to the power of...\n\ncos (2t) + 2 sin ^2 (t) = cos (2t) + 2 - 2 cos ^2 (t)
1122517
sin^2+cos^2=1\nsin=1/cosec\nsin=opp side/hypotheses side\ncos=adj side/hypotheses side\nhyp^2=opp^2+adj^2\n\nSin*cosec*cos=cos
129693
1) square both sides\n\n2) then make trig substitutions to get same 'variable'. Get sin^2[x] throughout equation. \n\ncos^2[x] + sin^2[x] = 1, so\n\n1 - cos^2[x] = sin^2[x].\n\n3) sin[x] cos[x] = ...\n\n4) Set equal to zero\n\n5) Solve.
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.
542056
Start by factoring:\n\n(cos^2(x))(sin^2(x))-(cos^2(x)) = (cos^2(x))(sin^2(x) - 1)\n= -(cos^2(x))(1 - sin^2(x))\n= -(cos^2(x))(cos^2(x))\n= -cos^4(x)
814944
we can  Use the trig identity\nsin(a+b)=sin(a)cos(b) + sin(b)cos(a) \nand we have\n\n10[sin(w.t)cos(π/4) + sin(π/4)cos(w.t)] \nwhich can be simplified as\n 10[√(2)/2sin(w.t) + √(2)/2cos(w.t)]\n= 5(√2)[sin(w.t) + cos(w.t)] \n\n a=5√2\n b=5√2
419071
You can cheat and use a calculator to get the values, but I don't think this is what your teacher is looking for. \n\nFirst, use the relationship sin^2(x)+cos^2(x)=1 to find |sin(a)| and |cos(b)|. Pay close attention to the ranges of a and b to get the sign right-- sin(a) and cos(b) should both be negative. Doing this you should find:\n\nsin(a) = -12/13 \ncos(b) = (-2 sqrt(2))/3\n\nHere are two useful identities:\n\ncos(a-b) = cos(a)cos(b) + sin(a)sin(b)\ncos(2b) = 1 - 2sin^2(b)\n\nSince you now know cos(a),cos(b),sin(a) and sin(b), you can find that:\n\ncos(a-b)=(12-10sqrt(2))/39\ncos(2b)=7/9
1050391
EDIT: I was looking at the answer above mine, and the cos(0) and sin(0) values are reversed, so it is not x=0.\n\nOK!! No problem. First of all, let me just say I teach math, but there is no way I would want to memorize all of the trig identities, so open up your precalc book, and if it is a good book, the major identities will be located on the front or back cover. If not, then go to the applicable page.\n\nStart with what is given:\n(sq root:3)sinx + cosx = 1\n\nAll I did was go through the identity lists and see what seemed to match closest to what you have there, and I noticed that:\n\ntan(x/2) = (1 - cosx) / sinx\n\nThen I visualized doing a little bit of algebra with your problem, and saw that if you subtracted cosx from both sides you would have:\n\nsqrt(3) sinx = 1 - cosx\n\nWhich is kind of similar to tan(x/2), except we need a sinx in the denominator. And if you divide both sides by sinx, that is what you get! So now we have:\n\nsqrt(3) = (1 - cosx) / sinx\n\nBut we already know that this is the same as tan(x/2)\n\nSo, we have:\n\nsqrt(3) = tan(x/2)\n\nFrom here, since you want to find the actual value for x which will be an angle measure, you go to inverse tangent:\n\ntan(-1) sqrt(3) = (x/2)\n\nI also have a handy table of all the major trig values open (it is in the back cover of my precalc book) and notice that tan(60) = sqrt(3)\n\nWe are almost done:\n\n60 degrees = x / 2\n\nMultiply by 2,\n\nx = 120 degrees.\n\nOf course, you should check that answer and make sure it is correct!!
788868
cosθ = 2cos²θ/2 - 1\nSo cos²θ/2 = ½(1 + cosθ)\nSo cos²(π/12) = ½(1 + cos(π/6))\n= ½ + √3/4\n= ¼(2 + √3)    \nSo cos(π/12) = + √{¼(2 + √3)}\n= ½√(2 + √3)\n\nLet √(2 + √3) = a + b√3\nSquare both sides\n2 + √3 = a² + 3b² +2ab√3\n\nSo a² + 3b² = 2 and 2ab = 1\na = 1/√2 and b = 1/√2 \nSo √(2 + √3) = ½(√2 + √6)\n\nSo cos π/12 = ¼(√2 + √6)