Can someone please help me with this math problem? (sin140°cos50°+cos140°sin50°)?
151637
sin 140cos 50 + cos 140sin50\n= sin (140 + 50)\n= sin 190\nIf the purpose is to express the answer as a single trigo function, then that is the answer.\nHowever, if the value is to be solved, then a scientific calculator has to be used. Also,\n\nsin 190 = sin (180-190) = sin (-10) = -sin 10
958733
sin²y + cos²y = 1\nSo cos²y = 1 - sin²y\n= (1 - siny)(1 + siny)\nTherefore\ncos²y/(1 + siny) = 1 - siny .................... QED
433978
Simply equate the functions as it asks you to;\nso you get 1-sinx = 2sin2x\nThen use the trig identity sin2x=2sinxcosx\nso you get 1-sinx=4sinxcosx\n\nCan you solve it from there?\nI'd next use cosx = sqrt (1-(sinx)^2)\nThen square the equation and solve it for x as you've probably done a million times before.\n\nYou'll probably get a quadratic equation in sinx. Use the quadratic formula to get some values out, and remember that you have to give answers between -360 and 360 degrees.
716883
Trying for wow #2---\n\nR = .15 m\nω = 2π/T = 13.963 rad/sec\nA little vector diagram shows that tanθ = rω²m/mg = rω²/g\nbut r = Rsinθ, so tanθ = sinθ/cosθ = Rsinθω²/g  from which\n\n1/cosθ = secθ = Rω²/g = .15*13.963²/9.8 = 2.984 → θ = 70.42°\n\nYou get to do b)\nGood luck..
1054686
Job one, convert everything to trig functions of a single angle, namely x.\n\nsin2x=2sinxcosx\ncos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\n\nThen substitute in and simplify:\n2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx\n2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0\n4cos^2(x)+2sinxcosx-sinx+4cosx-3=0\n\nFactor this into:\n(2cosx-1)(2cosx+sinx+3)=0\n\nset each parantheses equal to zero\n2cosx-1=0   \ncosx=1/2\nx=arccos(1/2)\nx={pi/3, 5pi/3} + 2pi*n  where n=0,1,2....\n\nAnd\n\n2cosx+sinx+3=0\n\nSolved numerically gives imaginary solutions of\n{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}\n\nSolved....
109605
Just use Cosinus Theorem: c^2=a^2+b^2-2abCos(gamma)\n1.\nc=sqrt(4+64-2*2*8*sqrt(3)/2))\nc=6.34\n\n2. Is not a right Triangle.. but anyway lets find the angle A:\n22^2=15^2+33^2-2*15*33*Cos(A)\nCos(A)=830/990\nA=ArcCos(83/99)\nA=33°1' 48.54"
912572
sin^3 β + cos^3 β  = (sinβ + cosβ)(sin^2β + cos^2β - sinβcosβ)\n= (sinβ + cosβ)(1 - sinβcosβ)\n\nsin^3 β + cos^3 β / sin β + cos β\n = (sinβ + cosβ)(1 - sinβcosβ)/( sin β + cos β)\n= 1 - sinβcosβ
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
894390
2sin3x*[-15sin3x]\n=-30*sin^23x
864887
14 sin² y = 13\nsin² y = (13/14)\n|sin y| = √(13/14)\nsin y = ±√(13/14)\ny = arcsin [±√(13/14)]\n\nThere are four solutions for this equation such that 0° ≤ y < 360°... one in each quadrant.  They are (to the nearest tenth of a degree) 68.2°, 111.8°, 248.2°, and 291.8°.
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
702851
the reference number for 135 degrees is 45 degrees and is found in quadrant 2\nsec 45 is also the same as \n1/(cos45)\nif cos 45 is \nsqrt(2)/2 (but since it is found in the second quadrant it would be\n-sqrt(2)/2 \n1/(-sqrt(2)/2) would be\n-2/(sqrt(2)\nwhich can also be written as\n-sqrt(2)\nthus your answer is\n\n-sqrt(2)
1008994
a*a(cos^2 B-cos^2 C)\n+b*b(cos^2 C-cos^2A)\n+c*c(cos^2 A-cos^2 B)\n\nare A,B, C angles in a triangular?\n\nif it is,  you may use the relation  A+B+C=180\nand cosine rule in a triangular.
880226
it is not what you specified\n\nIt cannot be solved without identities\n\nsin ^5 t = sin^4 t sin t\n= (1-cos^2 t)^2 sin t\n\nnow let cost = x\nso diffentiate to get - sint dt = dx\n\nso we need to integrate\n\n-(1-x^2)^2 dx\n\n= -dx + 2 x^2 dx - x^4 dx\n\nintegrating we get\n\n-x + 2x^3/3 - x^5/5 + C\n= - cos t + 2 cos^3 t/3 - cos^5 t /5 + C\n\nwhere C is constant of integration
543158
Yes.  By the law of sines, sin A/a = sin B/b, so B must be 45 degrees.  Thus C = 75 degrees, but maybe my geometry is a bit rusty.
751757
this is correct\n\nthis is cos pi/5 + i sin pi/5 + cos pi/5 - i sin pi/5\n= 2 cos pi/5\n\n=2 *phi/2 = phi\n\nfor cos pi/5 = phi/2 refer below
926741
sin^2(x) = ( 1 - cos(2x))/2\n\nso,\n\ncos(2*2y) = 1 - sin^2(2y)\n\nso we can rewrite sin(2y) = cos(4y) as:\n\n2sin^2(2y) + sin(2y) - 1 = 0\n\nlet u = sin(2y)\n\n2u^2 + u -1 =0\n\nu = 1/2\nor\nu = -1\n\nso,\n\nsin(2y) = 1/2\nor\nsin(2y) = -1\n\n2y = sin^-1(1/2)\nor\n2y = sin^-1(-1)\n\nso\ny = pi/12 = 15 degrees\nor\ny = -pi/4 = -45 degrees\n\nso the answer is y =15 degrees since we want only positive solutions.\n\ncheck:\nsin(2*15) = cos(4*15)\nsin(30) = cos(60)