cosine 2 theta + 2sine square theta eauals whalt?
107615
t=theta\n^=to the power of...\n\ncos (2t) + 2 sin ^2 (t) = cos (2t) + 2 - 2 cos ^2 (t)
215627
well from what i learn we have a formulae for that right??!! then the formulae to find theta is.... to find cosine the formulae is adjecent over hypothenus in shortform its CAH or C=A/H. and to find sine is opposite over hypothenus or SOH or S=O/H.. and to fine tangent is opposite over adjecent or TOA or T=O/A.. so to remember it you just remmber it in the short form.. as if you have a chinesse friend name SOH CAH TOA.. soh for sine n cah for cosine n toa for tangent.. i hope thats what you want for the answer.. if not.. im sorry.. sooo sorryy..
791534
(2asinx*cosx)^2 + a^2(cos^2x - sin^2x)^2\n=4a^2 sin^2xCos^2x + a^2(cos^4x+sin^4x -2sin^2xCos^2x )\n=2a^2 sin^2xCos^2x + a^2cos^4x+a^2sin^4x\n=a^2(cos^2x + sin^2x)^2\n=a^2
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
1149867
We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1
872911
Let Z = 3 - 3i\n|Z| = √(3² + 3²)\n= 3√2\nThus Z = 3√2(1/√2 - 1/√2i)\nNow swwk an angle such that cosθ = 1/√2 and sinθ = -1/√2\nVlearly this is a 4th quadrabt angle\nTherefore θ = 2π - π/4\n= 7π/4\nThus Z = 3√2(cos7π/4 + i sin7π/4)\n[= 3√2e^(i7π/4)]
849646
-\na=(2, 1, 1)\nb=(1, root3-1, -root3 -1)\n\ncos(theta)= a.b/|a||b|\n= (2 +sqrt(3)-1 -sqrt(3)-1 )/|a||b| = 0\nso theta = pi/2 (90 degrees)
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
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
830578
Adding these fraction expressions (common denominator  etc) gives\n(sin3θsinθ + cos3θcosθ)/(cosθsinθ)\n\nThe top is the expansion of\ncos(3θ - θ) which is cos2θ\n\nThe bottom is half of \nsin2θ = 2sinθcosθ,\nso multiply top and bottom by 2 and you have the result, since\ncos2θ/sin2θ = cot2θ
150092
t=to\n_____\n√1-v2/c2
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
901670
(sin x +cos x)^2 =1    [0=<x=<360]\nsin^2x + 2sinx * cosx +cos^2x =1 or 2sinxcosx=0      Because sin^2x+cos2^x =1 \n2sinxcosx is zero when x =0, 90, 180, or 270 degrees\n\n1-sinx = cosx     [0=<x=<360]\nsinx + cosx = 1\nThis occurs when x= 0, 360
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
959978
use pythagorean thereom a^2 + b^2 = c^2. (c being teh longest side or the hypotenuse) the two legs are a and b. so you have the sq rt of 169. (sqrt of 12^2 + 5^2) which is 13
707200
F = uN\nF = ma\nuN = ma\n0.42*m*g*cos(theta) = m*g*sin(theta)\n0.42*cos(theta) = sin(theta)\n0.42 = sin(theta)/cos(theta)\n0.42 = tan(theta)\ntheta = 22.78 deg
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