does anyone know how to prove these identities?
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
595375
divide the first row by 2\n1 1/2\n1  -1\n\nSubtract the first row from the second row:\n1  1/2\n0  -3/2\n\nMutiply the second row by -2/3\n1  1/2\n0   1\n\nSubtract 1/2 of the second row from the first row:\n1  0\n0  1\n\nAnd you have the identity matrix
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
928980
u^4-2u^2v^2 +v^4 -2u^2w^2-2uvw^2-v^2w^2)
167162
I do agree with Anniemc46.\nIf the denominator is 1+(tanhx)^2 then it is easy to do.\n\nI have done and uploaded to my web site.\nPlease check the url I provided in the sourse.\n\nThanking U,\nThuvalpakshi.
104581
It takes practice!  And its like a puzzle.  Just use the main trig identities as reference and a good math tutor.  Unfortunately, this place isn't the best place to ask that kind of question because we are limited by text explainations and trig identities need a lot more than that!
856009
false,\nuse the identity and -identity,\nboth are invertible, but the sum is the zero matrix, that is NOT invertible.....
940602
You're correct. Differentiating them and showing they are equal to zero will indicate that they are constant. The equation\n\nA(w_1)^2 + B(w_2)^2 + C(w_3)^2\n\ndifferentiated with respect to time yields\n\n2*A*w_1*(d/dt w_1) + 2*B*w_2*(d/dt w_2) + 2*C*w_3*(d/dt w_3)\n\nIf you substitute from the system of equations that shows the evolution then\n\n2*w_1*w_2*w_3*(B-C) + 2*w_1*w_2*w_3*(C-A) + 2*w_1*w_2*w_3*(A-B)\n\nyou can see that the terms cancel leaving you with zero.\n\nFollow the same technique for the second equation. It will look similar except you will wind up with an additional constantt, e.g., A*(B-C), etc.
154380
I posted these sources earlier to someone else who was looking for math help.\n\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
667025
You should verify this.\n_0_2_4_6_8_10_12\n0|0 0 0 0 0  0  0\n2|0 4 8 12 2 6 10\n4|0 8 2 10 4 12 6\n6|0 12 10 8 6 4 2\n8|0  2  4 6 8 10 12\n10|0 6 12 4 10 2 8\n12|0 10 6 2 12 8 4
950350
LHS=sin^2x-cos^2x=(sinx+cosx)(sinx-cosx)\nRHS=1-2sinxcosx\n=(sin^2x+cos^2x)-2sinxcosx as sin^2x+cos^2x=1\n=(sinx-cosx)^2\nLHS and RHS can be equal only ifsinx+cosx=sinx-cosx\nso it is not an identity\nan identity must hole for all values
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
733318
I think your identity should be :\ncsc^2(x) / [1 + tan^2(x)] = cot^2(x)\nas you forgot to square the tan(x).\n\nWorking on the 1 + tan^2(x) gives:\n1 + sin^2(x) / cos^2(x)\n= [sin^2(x) + cos^2(x)] / cos^2(x)\n= 1 / cos^2(x)\n\nNow we have : csc^2(x) / [1 / cos^2(x)]\n= [1 / sin^2(x)] / [1 / cos^2(x)]\n= cos^2(x) / sin^2(x)\n= cot^2(x), which was to be shown.
1028166
i know how to derive tis equation by means of calculus but i'm not sure that it involves algebra. i can give u the calculus proof though.
329665
Well I'm not much help, but to solve trig. identies you basically just use the given identies and equal them to each other. You substitue whutever equation is possible to simply the identity down to it's simplest form. \n\nLook at the bottom of this website for more: http://library.thinkquest.org/20991/alg2/trigi.html\n\nExcept that website might actually be more focused on trig. Equations not specifically Identities... lol So try using this website instead =) :\nhttp://www.intmath.com/Add_Trig/1_TrId.php\n\nHave fun.
40893
Remember this identity?\ncos A - cos B = - 2 sin ½(A + B) sin ½(A - B)\n\nIn this case, A is 5X and B is 5X\n3sin(3X)sin(2X) = -1.5(-2sin(½(5X+X))sin(½(5X-X)))\n\nUsing the identity I gave at first you get:\n-1.5(cos(5X)-cos(X)) = 1.5cos(X)-1.5cos(5X)\n\nYou could also write this using only additions as:\n1.5cos(X) + 1.5cos(5X+pi)