I already know the answer, just offering a challenge.  There's a geometric answer as well as an algebraic answer.
133424
sin(2x) = sin(3x)\nsin(2x) = 2sin(x)cos(x)\n\nsin(3x) = -sin(x)^3 + 3cos(x)^2*sin(x)\nsin(3x) = (sin(x)) (-1 + 4cos(x)^2)\n\nsin(2x0 = sin(3x)\n2sin(x)cos(x) = (sin(x)) (-1 + 4cos(x)^2)\ndivide both sides by sin(x)\n2cos(x) = -1 + 4cos(x)^2\nsubtract 2cos(x) from both sides\n4cos(x)^2 - 2cos(x) - 1 = 0\n\nNow just think of this problem like this\n\n4x^2 - 2x - 1 = 0\n\nUsing the quadratic formula\n\nx = (-b ± sqrt(b^2 - 4ac))/2a\nx = (2 ± sqrt(4 + 16))/8\nx = (2 ± sqrt(20))/8\nx = (2 ± sqrt(4 * 5))/8\nx = (2 ± 2sqrt(5))/8\nx = (1/4)(1 ± sqrt(5))\n\nnow just plug in cos(x) for x\n\ncos(x) = (1/4)(1 ± sqrt(5))\ninverse cosine both sides\n\nx = 36° and 108°
519915
No, it would have to be the square of something like ax+b, whose square is a^2x+2abx+b^2.  You would have to have a=1 or -1 and b=i or -i, but then 2ab would be imaginary, and not -2.
526041
Yes, graph this. (0,0),(1,0),(1,1),(0,2)\nThere are two right angles and two sides of length 1, but the figure is not a parallelogram. It is only a trapezoid.
540250
I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
352801
See my answer to the exact same question asked in the mathematics forum. You have it wrong.
359885
sidharth has solved the problem correctly and given you the two values for x that satisfy the conditions of the problem.\n\nHowever, he didn't specify the POINTS on the curve.\nThey are (0,0) and (-2,-2).
445024
Simple math & logic problem. You can do this, & gain a little wisdom in the process. Remember, it is simple. Don't complicate it.
521866
I am a math major-- but i am not going to help you. It seems that you are simply trying to get a free answer rather than doing the question yourself.
520435
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
551700
This website looks like it's got a lot of proofs.\n\nhttp://www.cut-the-knot.org/pythagoras/index.shtml
67168
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
642364
(√3)² + 1 ² = 2²\nsin α = √3/2 α = 60º;  ß = 90 - α;  ß = 30º\n\nAnd your second question is\naⁿ = c² - b².  I'm going to assume that you wanted to put in a, b and c are all integers and a > 1.  Because if not, your theorem is not true and cannot be proven.  With that stipulation, this theorem passes the "intuitive" test.  It makes sense.\nWe know for certain that, for all even n, it must be true by the Pythagorean theorem.  So only odd n are of consequence to this theorem.\n\nI'm going to take a moment here and say that if you're trying to use this theorem as a FLT lemma, you must understand that it won't fly.  The lemma must use the carry the same stipulations as the successor theorem.  Said the other way, the corollary is bound by the restrictions of the theorem.\n\na=2, n=3 gives us c=3 and b=1.  8=9-1.  I took some time trying to find a counter-example and could not.  I did find some neat patterns emerging.\n\nThe proof to your theorem is simple: aⁿ = a²[(½a+½)² - (½a-½)²] because the expression in brackets will always be equal to a (a is odd prime; composite a can be simplified to prime a).\n\nI also found that 2ⁿ= [2^(n-2)+1]² - [2^(n-2)-1]².  So, in addition to all the solutions from the above expression, there will be another set of solutions for 2ⁿ; n is odd integer > 5.
401848
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
343722
sorry i am no good at math, but this may help:\nhttp://www.fourmilab.ch/earthview/pacalc.html
211922
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.
463830
The only thing I remember about geometry is...a2 + b2 = c2.\n\nSomehow, I doubt that helps you.....sorry.
229478
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/