how do I write r = 4 / srqt 4cos^2x + sin^2x in cartesian form (x/a)^2 + (y/b)^2 = 1?
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
64064
put, for example: t=cos x, then sin^2(x)= 1-cos^2(x)= 1-t^2\n\nThus you get from the original equation, a quadratic equation:\n\nt-2(1-t^2)+1=0 or t-2+2t^2+1=0, or\n\n2t^2+t-1 = 0.\n\nAfter factorization of the left side \n\n(2t-1)(t+1)=0\n\nthen 2t-1=0 gives one solution for t;\n\nt+1= 0 gives another solution.\n\nSince this is quadratic equation, you have received all the possible solutions.\n\nT_1 = 1/2 ;    T_2 = -1\n\nBut we have assumed earlier, to get to that equation, that\n\nT = cos (x)\n\nSo now we have to consider both options for t;\n\n1)cos x = 1/2; 2)cos x = -1\n\nFrom it goes something like that, check those calculations for yourself of course:\n\nSolution of the first is x = PI/2+ PIK, K= 0, 1..\n\nSolution of the second is x= PI+ PIK, K=0,1...\n\nThat is how you turn a trig problem into an algebraic one.\n\nHope that helps.
1121270
This one can't be simplified further uness you just write\nsqrt{5+3sqrt{3}}.\n\nOn the other hand, sqrt{7+4sqrt{3}} can be rewritten as 2+sqrt{3}. Making such simplifications is not easy in general.
356153
using trig identities\nsin^2x+cos^2x=1, so 1-sin^2x=cos^2x and sin^2x=1-cos^2x\nsubstitue\ncos^2x(cosx)-2sin^2cosx\ncos^3x-2sin^2xcosx\ncosx(cos^2x-2*(1-cos^2x))\ncosx(cos^2x-2+2cos^2x)\ncosx(3cos^2x-2)
714157
One important rule is that cos^2+sin^2=1\n\nFirst of all, do this:\nsin^2c(1+(cos^2c/sin^2c))\n+cos^2c(1+(sin^2c/cos^2c))=\nsin^2c+cos^2c+cos^2c+sin^2c=\ncos^2c+sin^2c+cos^2c+sin^2c=\n1+1=2\n2=2\n\nI hope this helps!
949562
Use csx =1/sin and  1-cos^2=sin^2  and cot=cos/sin\n Ready...Go\ncscx - (cosx)(cotx) =  1/sin - cos^2/sin = (1-cos^2)/sin=sin^2/sin =sin\n       ole, ole , infree
741518
1+(cotx)^2=(cscx)^2\n\nLeft hand side = 1 + cot²x\n= 1 + cos²x/sin²x\n= (sin²x + cos²x)/sin²x\n= 1/sin²x\n=csc²x\n= right hand side ......... QED
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
830100
'x= rcost\ny=rsint\n\nand r= sqrt(x^2+y^2)\nt=arctan(y/x)\nso\nif x=2, and y=2, then r= sqrqt(4+4)=sqrt(8)\nand t=arctan(2/2) = arctan(1) = pi/4
924254
f(x) = 3cos(2x)+4sin(2x)\n√(3² + 4²) = 5\nDivide and mutlipy by 5\nSo f(x) = 5*(3/5 cos(2x) + 4/5 sin(2x))\nNow because (3/5)² + (4/5)² = 1 there exists an angle such that cos ε = 3/5 and sin ε = 4/5 (OR vice versa if you prefer)\n\nSo f(x) = 5*(cosεcos(2x) + sinεsin(2x)))\n= 5 cos(2x - ε) where ε ≈ 53°07'48" ≈ 0.9273 rad\n\nie f(x) = 5cos(2x - 0.9273)\n\nYou could also have got to f(x) = 5 sin (2x + 0.6435) if you used sinε = 3/5 and cosε = 4/5 It is easy to show that these are the same as (2x + 0.6435) - (2x - 0.9273) = 1.5708 = π/2\n\nand if x - y = x + (-y) = π/2\nthen x = π/2 - (-y)\nSo sin x = cos (-y) = cos y as the cosine function is an odd function (ie f(-x) = f(x))\n\nIn general:\nif(x) = AsinY + BcosY\n= √(A² + B²)(A/√(A² + B²) * sinY + B/√(A² + B²) * cosY\n= √(A² + B²)(sinY cosε + cosY sinε)\n= √(A² + B²)sin(Y + ε) where ε = arctan (B/A)\n(also = √(A² + B²)cos(Y - ε) where ε = arctan (A/B)\n\nand other variations can be more useful depending on the signs of A and B and a desire to have 0 < ε < π/2 (ie ε acute)\n\nThere are 4 possibilities f(x) = Rsin(Y + ε)\nf(x) = Rsin(Y - ε)\nf(x) = Rcos(Y + ε)\nf(x) = Rcos(Y - ε), the latter two if A and B are opposite in sign.
808695
If r and s are the roots of ax² + bx + c = 0, then we can rewrite it as\nx = r or x = s.\n\nTransposing, we get\nx - r = 0 or x - s = 0\n\nThen\n(x - r)(x - s) = 0\n\nMultiply\nx² - (r + s)x + rs = 0\n\nWe divide a from the quadratic\nx² + b/a x + c/a = 0\n\nTherefore,\nr + s = -b/a and rs = c/a\n\nNow, we need to find\nr^-1 + s^-1\n\nSince an exponent of -1 means the reciprocal, then\n= 1/r + 1/s\n\nGetting the LCD,\n= (r + s)/rs\n\nNow we substitute:\n= (-b/a)/(c/a)\n\nWe cancel the a's:\n= -b/c\n\nTherefore,\nr^-1 + s^-1 = -b/c\n\n^_^
263848
cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
831103
I would change cos 2 theta to 2 cos squared theta -1.\n\nYou then have a quadratic in cos theta.
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
910165
cos2A = 2cos²A - 1 → cos²A = ½(cos2A + 1)\n\ncos2A = 1 - 2sin²A → sin²A = ½(1 - cos2A)
766977
4cos2x +3cosx +3 = 0  [x=theta]\nLet y=2x\nThen 4cos y +3 cos(y/2) +3 = 0\nNow use the cos 1/2 angle formula to get:\n4cos y +3 [sqrt (1+cosy)/2] +3 =0\n3 [sqrt (1+cos y)/2] = -3 -4cos y\n9(1+cos y)/2= 9 +24 cos y +16cos^2y\n9+9cosy =18 +48cos y +32cos^2y\n32cos^2y +39cos y +9 =0\ncos y = [-39 + or - sqrt(39^2- 4*32*9)]/2\ncos y = -39/2 + 3/2sqrt (41), and \ncos y = -39/2 -3/2sqrt(41)\ny=arccos[-39/2 + 3/2sqrt (41)] , and \ny=arccos[-39/2 - 3/2sqrt (41)]\nx= y/2 =1/2*arccos[-39/2 + 3/2sqrt (41)], and\nx=1/2*arccos[-39/2 - 3/2sqrt (41)] \nUse your calculator to find final answer
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)