VeryifingTrigonometric Identities?
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
77120
It doesn't appear to simply significantly from the original expression.  I was able to get it to the following:\n\n(tan^3-tan)*sec\n\nDo the multipling by csc-1 top-n-bottom and then use the fact that cot^2=csc^2-1.  Bring the cot^2=cos^2/sin^2 to the top to get the above.\nGood luck.
1132938
I can help you with the part of  "Identify the expression that completes the equasion so that it is an identity.\nsin^2x/(1-cosx)"\n\nYou got sin^2x/(1-cosx). Then to complete the equation you multiply[(1+cosx )/(1+cosx)] to the equation.  \n\nThen you get {[(sin^2x)(1+cosx)]/[(1-cosx)(1+cosx)]}...Then by factorization you get that {[(sin^2x)(1+cosx)]/(1-cos^2)]}....The (1-cos^2) is the same as to sin^2x.  \n\nThen you cancel the terms sin^2x/(1-cos^2)...It leaves the equation as following ==> (1+cosx)(1) = (1+cosx)\n\nHope someone helps you with the rest SOrry
769357
It can be helpful here to express this in terms of complex exponentiation. Note that sin x = (e^(ix) - e^(-ix))/(2i) and cos x = (e^(ix) + e^(-ix))/2. So:\n\n(sin x)^3 * (cos x)^4\n(e^(ix) - e^(-ix))^3/(2i)^3 * (e^(ix) + e^(-ix))^4/2^4\n(e^(3ix) - 3e^(ix) + 3e^(-ix) - e^(-3ix))/(-8i) * (e^(4ix) + 4e^(2ix) + 6 + 4e^(-2ix) + e^(-4ix))/16\n-1/(128i) * (e^(7ix) - e^(-7ix) + e^(5ix) - e^(-5ix) - 3e^(3ix) + 3e^(-3ix) - 3e^(ix) + 3e^(-ix))\n-1/64 (sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x))\n\nThus, all that tedious mucking about in the complex numbers has allowed us to reduce a hard-to-integrate product of powers of trigonometric functions into a sum of very to simple integrate functions. Thus:\n\n∫sin^3 x cos^4x dx\n-1/64∫sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x) dx\n-1/64 (- cos (7x)/7 - cos (5x)/5 + cos (3x) + 3 cos x) + C\ncos (7x)/448 + cos (5x)/320 - cos (3x)/64 - 3 cos x/64 + C
278141
http://mathworld.wolfram.com/
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
1010848
First of all, u do not need Lagrange's identity to prove it. u just need to expand everything.\n\nSo, u want to use Lagrange's identity, fine. set a^m=a^2, a^n=x^2, substitute these into the Lagrange's identity and u get exactly [a^{(m+n)/2}+b*y]^(2) + [y*{a^(m/2)} - {a^(n/2)}*b]^(2) without even needing to do anything else.\n\nThis is not called proving!
521923
Let 'I' denote the integral sign.\nWe will use the method of substitution.\nLet e^2x = z.\nThen taking differentials of both sides 2*e^2x*dx = dz or 2*z*dx = dz or dx = dz/(2*z).\n\nNow 1/((e^-2x)+( e^2x)) can be written as (e^2x)/[1 + (e^2x)^2]\n\nHence given integral = I[(e^2x)/{1 + (e^2x)^2}]dx\n= I[z/(1 + z^2)]{dz/(2*z)} [substituting z = e^2x]\n= 0.5*I[1/(1 + z^2)]dz\n= 0.5*tan-1(z) + C [tan-1(z) denotes tan inverse z]\nwhere C = arbitrary constant of integration\n= 0.5*tan-1(e^2x) + C
815598
a+b+ c+ab+cb+ac = abc + 1\nor a+b+c - abc = 1- ab-cb-ac\nor (a+b+c - abc)/(1-ab-bc-ac) = 1\n\nputting a = TAN A , b= tan B, c = tan C\n\nwe get LHS = tan(A+B+C) = 1\nor A+B+C = tan ^-1 = pi/4+npi\n\n chose any a and b then c can be chosen as\n\ntan^1(pi/4 - tan A - tan B)
276083
Yes, it can be rewritten as 2sin2xcos2x.  Or, you can use the addition formula for sineto prove this.  sin(x + y) = sinxcosy + cosxsiny.  In your case, sin(2x + 2x) = sin2xcos2x + cos2xsin2x = 2sin2xcos2x.
947622
Try the number theory group at mymathforum.com .
929231
(a^2-b^2)=(a-b)(a+b)\nHence,\nsec^4y-tan^4y = (sec^2y-tan^2y)(sec^2y+tan^2y)\n\nSubstitute that back into the initial equation...\n\n(sec^2y-tan^2y)(sec^2y+tan^2y)\n------------------------------------------\n           (sec^2y+tan^2y)\n\nand cancel common terms\n\n= (sec^2y-tan^2y) \n= 1 (a standard trig identity)\n\nhope that solves it.
224978
associative, commutative, identity, inverse
736970
Here are some real tough ones.(Looked tough to me)\n\nProve that cot(90 - theta)sin(90 - theta) = sin theta\n\nFind Cos(3x)\n\nFind Sin(3x)
316406
Depends on how fast you're counting...\n\nhttp://hub1.worlded.org/docs/maththing/ny1p5.htm\nhttp://mathforum.org/library/drmath/view/58739.html
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
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