coumpute with out table: cos^/5 -cos2^/5 SHOW THE PROSIDURE u can do it with stemp jump!! do u able to solve ?
605433
cos( pi / 5 ) - cos( 2 pi / 5 ) = ?\n\nObserve that cos( pi / 5 ) is the real part of the complex number e^( i pi / 5 ), which is a fifth root of e^( i pi ) = -1.  Likewise cos( 2 pi / 5 ) is the real part of the complex number e^( i 2 pi / 5 ), which is a fifth root of e( i 2 pi ) = 1.\n\nSo let a = e^( i 2 pi / 5 ) = c + i d\nand b = e^( i pi / 5 ) = f + i g\n\nNote that a, a^2, a^3, a^4, and a^5 = 1 are all fifth roots of 1, which in a field (such as the complex numbers) can have at most five fifth roots.  So\n\n(x - a)(x - a^2)(x - a^3)(x - a^4)(x - a^5) = x^5 - 1\nDividing out the term x - a^5 = x - 1 yields\n\n(x - a)(x - a^2)(x - a^3)(x - a^4) = x^4 + x^3 + x^2 + x + 1\n\nBut (x - a)(x - a^2)(x - a^3)(x - a^4)\n = x^4 - (a + a^2 + a^3 + a^4) + ...\n\nso a + a^2 + a^3 + a^4 = -1\n\nNow cos(p - q) = cos(p) cos(q) + sin(p) sin(q)\nand sin(p - q) = sin(p) cos(q) - cos(p) sin(q)\nso\ncos(pi - z) = cos(pi) cos(z) + sin(pi) sin(z)\n = (-1) cos(z) + (0) sin(z)\n = - cos(z)\nsin(pi - z) = sin(pi) cos(z) - cos(pi) sin(z)\n = (0) cos(z) - (-1) sin(z)\n = sin(z)\n\nAnd cos(p + q) = cos(p) cos(q) - sin(p) sin(q)\nand sin(p + q) = sin(p) cos(q) + cos(p) sin(q)\nso\ncos(pi + z) = cos(pi) cos(z) - sin(pi) sin(z)\n = (-1) cos(z) - (0) sin(z)\n = - cos(z)\nsin(pi + z) = sin(pi) cos(z) + cos(pi) sin(z)\n = (0) cos(z) + (-1) sin(z)\n = - sin(z)\n\nAbove we defined a = e^( i 2 pi / 5 ) = c + i d\nso we know c = cos( 2 pi / 5) and d = sin( 2 pi / 5 )\nLikewise b = e^( i pi / 5 ) = f + i g\nso f = cos(pi / 5) and g = sin(pi / 5)\n\nSo a^2 = (c + i d)(c + i d)\n = (c^2 - d^2) + i 2 c d\n\nBut a^2 = (e^( i 2 pi / 5 ))^2\n = e^(i 4 pi / 5)\n = cos( 4 pi / 5 ) + i sin( 4 pi / 5 )\n = cos( pi - (pi / 5) ) + i sin( pi - (pi / 5) )\n = - cos(pi / 5) + i sin(pi / 5)\n = - f + i g\n\na^3 = (e^(i 2 pi / 5))^3\n = e^(i 6 pi / 5)\n = cos(pi + (pi/5)) + i sin(pi + (pi/5))\n = - cos(pi / 5) - i sin(pi / 5)\n = - f - i g\n\na^4 = (a^2)^2 = (- f + i g)^2\n = (f^2 - g^2) - i 2 f g\n\nBut a = b^2 = (f + i g)^2 = (f^2 - g^2) + 2 i f g\n\nSo -1 = a + a^2 + a^3 + a^4\n = ((f^2 - g^2) + i 2 f g) + (- f + i g) + (- f - i g) + ((f^2 - g^2) - i 2 f g)\n = 2(f^2 - g^2 - f)\n\nyielding\n\nf^2 - g^2 - f = -1/2\n\nNow (cos(z))^2 + (sin(z))^2 = 1 for any z and so f^2 + g^2 = 1, thus\n\nf^2 - (1 - f^2) - f = -1/2\n2 f^2 - 1 - f = -1/2\n\nNow we are looking for cos(pi / 5) - cos(2 pi / 5) = f - a\n = f - (f^2 - g^2)\n = f - (f^2 - (1 - f^2))\n = f - (f^2 - 1 + f^2)\n = f - (2 f^2 - 1)\n = - (2 f^2 - f - 1)\n = - (- 1/2)\n = 1/2\n\nSuch a simple result suggests that there was a simpler way to do this. :-)
150092
t=to\n_____\n√1-v2/c2
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
942547
Can you please check you haven't made a mistake somewhere? I have just calculated (with 99% certainty) that there are no solutions to those equations with those restrictions.\n\nEdit - OK, I assumed you made exactly one mistake. By removing each possible word and trying to solve it, I discovered that removing 'SIGH' was the only option that gave an answer.\nt 1\nh 8\na 3\nn 7\nk 6\ns 5\ng 9\ni 4\nv 2\nwhere SIGH = 26, not 25. Then THANKSGIVING = 65.\n\nThat was actually a cooler puzzle that the original, assuming you made one mistake :)
925629
polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
869806
213\n321\n132\n\nall rows, columns, diagonals = 6\nnice try rt11guru but that wasn't your original answer\nmirror image:\n312\n123\n231\nupside down:\n231\n123\n312\nmirror image of that:\n132\n321\n213\n\nfuntym888 what do you mean it cant be done? dont you read the other answers before you answer?  it certainly can be done, in 4 different ways no less, as I just showed!  you simply take a 1-9 magic square and replace 4,5,6 by 1,2,3 and 7,8,9 by 1,2,3 respectively.  And it is "really" a magic square, it just isnt a normal magic square.  You should read your own sources as well!
874492
Here's what you have to do.\nReplace Q with 1.6x10^(-19)\nReplace C with 3.31x10^20\nReplace K with .49\nFor #1, replace x with 1.9\nF = (1.6x10^(-19)) (3.31x10^20) (.49) / (2(1+cosh(.49*1.9))\nEvaluate using a calculator.\n\nFor #2, replace F with 3 and solve for x.\n3 = (1.6x10^(-19)) (3.31x10^20) (.49) / (2(1+cosh(.49*x))\n3(2(1+cosh(.49*x)) = F = (1.6x10^(-19)) (3.31x10^20) (.49) \n6 + 6cosh(.49*x) = F = (1.6x10^(-19)) (3.31x10^20) (.49)\n6cosh(.49*x) = F = (1.6x10^(-19)) (3.31x10^20) (.49) - 6\ncosh(.49*x) = ((1.6x10^(-19)) (3.31x10^20) (.49) - 6) / 6\n.49*x = inverse cosh (((1.6x10^(-19)) (3.31x10^20) (.49) - 6) / 6)\nx = inverse cosh (((1.6x10^(-19)) (3.31x10^20) (.49) - 6) / 6) / .49
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
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)
864532
Set the tension of the left (T1) times the cos 53 degress equal to the tension of the right (T2) times the cos 37 degrees. Solve for either T1 or T2 in terms of teh other variable. Then set the sum of T1 times the sin 53 degees and T2 times the sin 37 degrees equal to 23 kg times 9.8 meters per second. Substitute the term from teh first equation into the second one, and solve for the variable that is left. Then just plug that variable into the first equation to get the other variable.
429835
ok, u have to use distance formula\n\nsqrt[(-4-(-7))^2+(c-7)^2]=5   \nsquare both sides\n[(-4-(-7))^2+(c-7)^2]=25      \nadd the stuff in the parentesies\n[(-3)^2+(c-7)^2]=25           \nexponents\n9+(c-7)^2=25                  \nsubtract from both sides\n(c-7)^2=16                    \nsquare root both sides\nc-7= +/-4                     \nadd to both sides\nc=11 or c=3
189924
just add\n\n$02E2\n+ $85\n-----\nyou tell me...\njust remember to carry the one after hitting 14 (or E)
1065017
I'm writing up a proof right now that this is not possible.\n\nCheck out the pdf at:\nhttp://www.geocities.com/euleratuo/
715417
Cos 25=0.906307787\nTan 25=0.466307658
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.
51334
cos(2wt)*cos(3wt) = 1/2 (cos(5wt) + cos(wt))\nL(f(t)) = 1/2(s/(s^2+25w^2) + s/(s^2 + w^2))\n\nby normal formula = L(cos(wt)) =  s/(s^2 +w^2)\n\nsimple.
757793
1. Using exact values, verify that Cos 60°= 1-2 sin² 30°\nRHS=1-sin^2 30*\n=1-(1/2)^2\n=1-1/4\n=3/4\nLHS=[(rt3)/2]^2\n=3/4\nLHS=RHS\n\n2. Using exact values, find the value of:\n\na. 12 sin 30° -6 tan 45° + sec45°/52 \n12(1/2)-6(1)+(rt2)/52\n\nb. sin 2A + tan 3A/2 – cos A + sec (A+15°), When A= 30°\n=sin60*+tan45*-cos30*+sec45*\n=rt3/2+1-rt3/2+rt2\n=1+rt2\n\n3. Write the other functions of x if cos x= 3/5..(show work)\nthe right triangle will bea 3,4,5 triangle using the pythagorean triplet\nsinx=4/5\ntanx=4/4\ncotx=3/4\nsecx=5/3\ncosecx=5/4 \n.\n4. In a right triangle sin A =3/4 and c =17. Find a and b. Be sure to draw a right triangle and letter it properly.\nc=17\nletthe oppositeside be 3x and the hypotenuse be 4x\nadjacent side=rt(16x^2-9x^2)\n=(rt7)x\n4x=17\nx=17/4\nrt7 x=17/4\nsquaring\n7x^2=289/16\nx^2=(289/16)*(1/7)\nx=rt[9289/16)(1/7)\n5. The vertex angle of an isosceles triangle is 57° 24’ and each of its equal sides is 375.5 \nft. long.Find the altitude of the triangle.\naltitude/side=sin of base angle\nbase angle=(180-57*24')/2\naltitude=375.5 times sin[(180*-57*24')/2]\n6. Solve the following right triangles:\na. a=117 ft.--B=16.35ft.\ncapital letters are used only for angles\nso B=16.35' is wrong\nb. B=8°29, --a=32.8 ft. \nare you taking <c as the right angle\nif so <a=(90*-8*29')\nuse a/sinA=b/sinB=c/sinC\n\n7. A man standing on top of a building sees an automobile on a street below him.The angle of depression is 36 degrees 22. If eye is 550 ft. above the level of the automobile ,how far is the automobile from him?\nlet the horizontal distance be x\nx/550=cot36*22'\nx=550cot 36*22'\n=>550tan(90-36*22')\n\n8. When the suns angle of elevation is 30 degrees. The shadow of a post is 6 ft.longer than when the angle is 45 degrees . Find the height of the post.(show work)\nlet the height of the post be x\nthe length of the shadow at 30*=xtan30*\nthe length of the shadow at 45*=x tan 45*\ngiven the difference x(tan45*-tan30*)=6\nsolve\n\n9. In a right triangle, given a= 32.8 ft. and A=28° 30’ 15’’,solve for B and C\ntaking <C=90* by convention\n<B=90*-28*30'15" \nnow use a/sinA=b/sinB=c/sinC