express 10sin(w.t+pi/4) in the form A sin w.t + B cos w.t?
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
641643
First off, the phase difference between cos(wt) and sin(wt) is pi/2... (i.e. cos(wt) = sin(wt+pi/2) ) Hence, cos(wt+pi/3) = sin(wt+pi/2+pi/3) = sin(wt+5pi/6)... THEREFORE, the answer is 5pi/6.... The multiplying terms 9 and 4 mean NOTHING when considering phase... They are amplitude terms... For example, 3*cos(wt) and 10000*cos(wt) are IN PHASE... but one of them obviously has a much larger amplitude... You can always graph these things out using Matlab or a graphing calculator to get a better feel for what's going on.
855395
1 + jwt = 1 + 4.8j\nLet r = √(1² + 4.8²) = √(24.04) = 2√(6.01)\n\nSo 1 + 4.8j = m(1/m + j *4.8/m)\n\nSo cos θ = 1/r and sin θ = 4.8/r OR tan θ = 4.8\n\nSo θ ≈ 1.3654 rad\n\nSo 1 + 4.8j = 2√(6.01)(cos(arctan(4.8)) + j sin(arctan(4.8)))\n\n≈ 4.903 (cos(1.3654) + j sin(1.3654))
564945
4*sin(x)cos^2(x) \n= 2*{2*sin(x)cos(x)}cos(x) \n= 2*sin(2x)cos(x)\n= sin(3x) + sin(x)\nTherefore, sin(x)cos^2(x) = [sin(3x) + sin(x)]/4\n\n4*sin^2(x)cos(x)\n= 2*{2*sin(x)cos(x)}sin(x)\n= 2*sin(2x)sin(x)\n= cos(x) - cos(3x)\nTherefore, sin^2(x)cos(x) = [cos(x) - cos(3x)]/4\n\nHence the expression inside the integral sign is\n[sin(3x) + sin(x) - cos(x) + cos(3x)]/4\n\nThus the integral becomes\n(1/4)*∫[sin(3x) + sin(x) - cos(x) + cos(3x)]dx\n= (1/4)*[∫sin(3x)dx + ∫sin(x)dx - ∫cos(x)dx + ∫cos(3x)dx]\n= (1/4)*[{-cos(3x)}/3 - cos(x) - sin(x) + {sin(3x)}/3] + C\nwhere C = arbitrary constant of integration\n= (1/4)*[{sin(3x) - 3*sin(x)} - {cos(3x) + cos(x)}]/3 + C\n= (1/12)*[- 4sin^3(x) - 4cos^3(x)] + C\n[N.B. cos(3x)=4cos^3(x)-3cos(x) and sin(3x)=3sin(x)-4sin^3(x)]\n= - [sin^3(x) + cos^3(x)]/3 + C
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
1176396
sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
592160
i think the answer wil be (tanX)/(tanY)\n\nSIN(X+Y)=sin x cos y+cos x sin y\n\nsin(x-y)=sin x cos y-cos x sin y\n\n use cos x cos y to divide the numerator and denominator... there is the answer...
743513
it's A\n\n4st+6t -10s-15\n2t(2s+3) -5(2s+3)    \n(I factored out the negative for both numbers in the -10s-15)\n\n(2t-5)(2s+3) or (2s+3)(2t-5)
276671
(e^(cose^e+sin^e)/2)^pi\nhahaha ..just kidding
935340
= cosx[cosxcosy+sinxsiny]+sinx[sinxcosy-cosxsiny]\n\n= (Cosx)^2cosy+cosxsinxsiny+(sinx)^2cosy-cosxsinxsiny\n\n= (Cosx)^2cosy + (Sinx)^2cosy\n\n= Cosy [ (cos)x^2 + (sinx)^2]\n\n= Cos(y)
913883
(sin7xcos3x)\nmultiply and divide by 2\n2(sin7xcos3x)/2\nuse formula\n2sinAcosB\n=sin(A+B)+sin(A-B)\nit becomes\n(sin10x+sin4x)/2\nnow integrate u get\n[(-cos10x)/10+(-cos4x)/4]/2\n\nfor (2)part\nuse formula\n2cosAcosB\n= cos(A+B)+cos(A-B)
699547
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
913234
it's "pi" not "pie" dear ... :-)\n\nAnd, yes I do know how to prove those identities.\n\nAnd, no I won't do them all, as it will deprive you of the wonderful opportunity to learn and practice for your test. \n\nBut, I will do one for you!  I'll do #8:\n\n(sinx + cosx) ² =\nsin²x + 2sinxcosx + cos²x = \nsin²x + cos²x + 2sinxcosx = \n1 + 2sinxcosx  {{ using trig ID #1 }} = \n1 + sin2x {{ using trig ID #2 }}\n. . . . .\nTrig ID #1 :  sin²x + cos²x = 1\nTrig ID #2 :  sin2x = 2sinxcosx\n\nNRT
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
388285
equate the two equations: \nsin 2x = pi/4 \n2x = inverse sine pi/4 + k2pi or (pi - inverse sine pi/4) + k2pi\n2x = 0.90 + k2pi or 2.24 + k2pi\nx = 0.45 + kpi or 1.12 + kpi\nx = 0.45, 1.12, 3.59, 4.26 ( k = 0,1)
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???
566422
cos150=\n\ncos(180-30)=\n\ncos(p-30)=\n\n-cos30=\n\n-sin60