IF sinA=12/13, cosA=5/13, and tanA=12/5 WHAT IS sin2A???????
461039
sin2A = 2sinA.cosA\nSo,\nsin2A = 2*12/13*5*13\n      = 120/169\n\ngood luck!!
324333
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.
710712
i took  trig and dont remember much but it should go like this\n(a)\n3x^2+12x=-12\n9x+12x=-12\n21x=-12\nx=-12/21\nx= -0.57\n\n(b)\n7x^2=28\n49x=28\nx=28/49\nx=0.57\n\n\nc im clueless\n\ni was wrong i just fixed it
817617
tan(A-B)+tan(B-C)+tan(C-A)\n= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)\n= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]\n= 0\nie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0\nOn expanding\n(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C) \n+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)\n+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)\n\n= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C \n+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C\n+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC  + tanAtan²Btan²C - tan²Atan²BtanC\n= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C\n=0 \nDivide all by tanAtanBtanC\nSo tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0\n\nie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC\n\nWhence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA\n\nSince a triangle then tan(A + B) = tan(π - C) = - tanC\nSo (tanA + tanB)/(1 - tanAtanB) = -tanC\nie tan A + tan B = tanAtanBtanC - tanC\nso tanA + tanB + tanC = tanAtanBtanC
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj
886349
SOHCAHTOA \nSOH =sine=Opposite /Hypotenuse\nCAH cos=adjacent/hypotenuse\nTOA tan=opposite/adjacent\nIn finding the adjacent,we use CAH\ncos19=adjacent/13.8\nadjacent=cos19x13.8\nadjacent=13.048
470919
I think you need to find A, if it is so then:\ntan A + sec A=2     square both sides\ntan^2 A + 2 tan A sec A + sec^2 A=4  since, sec^2 A= 1+tan^2 A  then:\n2 tan^2 A + 2 tan A sec A=3      since tan A=sin A/ cos A & sec A=1/cos A   then:\n2 sin^2 A/ cos^2 A + 2 sin A/ cos^2 A=3\n sin^2 A/ cos^2 A + sin A/ cos^2 A=3/2\n sin^2 A + sin A= 3/2* cos^2 A = 3/2* (1-sin^2 A)    since sin^2 A + cos^2 A =1\n5/2* sin^2 A + sin A -3/2= 0\n5 sin^2 A + 2 sin A – 3 = 0\n(5 sin A -3) (sin A +1)= 0 then,\neither sin A= 3/5 → A=36.87o \nor sin A=-1 → A=-90o (to be neglected)\ncheck: tan 36.87o=0.75 & cos 36.87o = 0.8 → sec 36.87o =1.25 then,\ntan 36.87o  + sec 36.87o =2  Q.E.D
746734
cos2x-cos4x = 2sin½(4x + 2x)sin½(4x - 2x)\n= 2sin3xsinx (using cosAcosB = 2sin½((B+A)sin½((B-A))
873429
cos5x = sinx\ncos5x = cos(pi/2 - x)\n5x = 2npi +- (pi/2 - x)\n6x = (4n + 1)pi/2   ,   4x= (4n-1)pi/2\nx= (4n+1)pi/12    ,  x = (4n-1)pi/8
241372
Csc (t) = 1/sin(t)\ntan (t) = sin(t)/cos(t)\n\nTherefore, Csc^2(t) = 1/sin^2(t)\n\ncsc^2(t)tan(t)sin(t) = [1/sin^2(t)][sin(t)/cos(t)]sin(t)\n                     = [sin^2(t)]/[sin^2(t)*cos(t)]\n                     = 1/cos(t)= sec(t)\n\nI do not know why they always make it complicated, why not just right sec(t)!
846581
Expand the exponent: (cos (x/2) - sin(x/2))^2 = (cos x/2)^2 - 2*cos(x/2)*sin(x/2) + (sin x/2)^2 = 1 - 2*sin(x/2)*cos(x/2) = 1-sin(x)\n\nusing the fact (cos a)^2 + (sin a)^2 = 1 and 2*sin(a)*cos(a) = sin(2a)\n\nSteve
435670
lets assume b = a/2\ntherefore 2b = a\n\nnow sin 2b = 12/13\n\nsin 2b is also equal to 2sinbcosb\n\nso 2sinbcosb = 12/13\n\nsinb * cosb = 6/13\n\nnow u know that cosb = sqrt ( 1 - sin^2 b)\n\nso sinb * sqrt(1-sin^2 b) = 6/13\n\nor sin^2 b * (1 - sin^2 b) = 36/169\n\nsubstituting sin ^ b = x\n\nyou get x - x^2 - 36/169 = 0\n\nThis is a simple quadratic equation\n\nsolving for x you get\n\nx = 234/338 or x = 104/338\n\nso you get sin b is sqrt (234/338) or sin b is sqrt(104/338)\n\nsince a is in the first quadrant, the value of sin a/2 will be less than that of cos a/2 since a/2 is less than 45 degrees.\n\nSo, sin a/2 = sqrt of (104/338)= 0.554\n\ncos a/2 = sqrt of (234/338) = 0.832
756840
cosx=tanx\ncosx=sinx/cosx\ncos^2(x)=sinx\n1-sin^2(x)=sinx    {pythagoras}\nsin^2(x)+sinx-1=0\nuse quadratic formula\nsinx=(-1+or-sqrt(5))/2\n=0.6180339888\nor -1.6180339888\nsine has to lie between +/-1\nsinx= 0.6180339888\nx=arcsin(0.6180339888)\n  =38.17270763\n+ve value for sine lies in 1st and 2nd\nquadrants \ntherefore,x=38.17270763 degrees or\n141.8272924 degrees\n\nthese two values were checked in the \noriginal equation and were found \nto be correct\n\ni hope that this helps
916168
6 sin 2x != sin 12 x\n\nnow sin x = 6 sin 2x\nor sin x = 6(2 sin x cos x)\nor sin x = 12 sin x cos x\nor 12sin x cos x - sin x = 0\nor sinx(12cos x -1) = 0\nsin x = 0 or cos x = 1/12\n\nsin x =0 => x = npi\ncos x = 1/12 => x = 2npi +/- cos^-1(1/12)
675696
I agree with changing Cot(Theta) into Cos(Theta)/Sin(Theta)\n\nsqrt(3)*Cot(Theta) = 2*Cos(Theta)\n\nsqrt(3)*Cos(Theta)/Sin(Theta) = 2 * Cos(Theta)\n\nsqrt(3)/2 = Sin(Theta)\n\nTheta = pi/3 and 2pi/3, or 60 degrees and 120 degrees.
676730
Opening brackets \nLHS=SinA+ SinA. TanA+ CosA+ CosA.CotA\n=SinA+Sin^2A/CosA+CosA+Cos^2A/SinA\nGrouping\n=(SinA+Cos^2A/SinA)+(CosA+Sin^2A?CosA)\n=(Sin^2A+Cos^2)/sinA+(Cos^2A+Sin^2A)/CosA\n=1/sinA+1/CosA\n=SecA+CosecA\n\nEasy Pesy
96766
1) tan A = x/y\n2) sin A = x/r\n3) cos A = y/r\n4) sin A/cos A = (x/r)/(y/r)\n5) sin A/cos A = xr/yr\n6) sin A/cos A = x/y\n7) sin A/cos A = tan A\n8) tan A = sin A/cos A\nReasons\n1-3)DEFINITION OF CIRCULAR FUNCTIONS\n4)MPE (STATEMENTS 2 AND 3)\n5)SIMPLIFYING COMPLEX FRACTIONS\n6)CANCELLATION\n7)SUBSTITUTION (STATEMENT 1)\n8)SYMMETRIC PROPERTY\n\nQED