This is proven using identities and formulas, such as the sum and difference formulas, double angle formulas, half angle formulas, pythagorean identities, ratio identities and reciprocal identities.  Please help!
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The first answer will be made by a girl. I promise.\n\nBut a negro is also OK.
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check out a trignometric tutorial in the internet...simply type "trignometric tutorial" on www.google.com\n\nthat way you can learn this properly, after all I can give you the answer - but you need to be able to do this sum in an exam...right
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1. All angles are equal\n2. All angles are less than 180° \n3. Size of each angle = (180 - 360/n)° where n = number of sides
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tanA + tanB + tanC = tanA tanB tanC is true for any non-right angle triangle\n\nrefer the link below . adequately solved
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You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
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Pythagoras:\nAC squared = AB squared + BC squared\nTherefore\nAC = square root of quantity (AB^2 + BC^2)\nYou can do the math.
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I have to go teach my class in about 4 minutes, but when I return, if no one has done it, I can show you how.  It requires factoring out a specific value that makes the coefficients "a" and "b" turn into two values that satisfy x^2 + y^2 = 1.  In other words, I know it starts out by factoring out sqrt(a^2 + b^2) from \n\na sin(Bx)+b cos(Bx)\n\n\nto get \n\n\nsqrt(a^2 + b^2) [ a/sqrt(a^2 + b^2) * sin (Bx) + b/sqrt(a^2 + b^2) * cos(Bx)]\n\n\nEDIT: Okay here we go.  And the "orthogonal rotating vectors" proof below is usually over most people's heads, or at least, most people who would ask this question on Yahoo answers.\n\n\nFirst, you factor out SQRT(a^2+b^2) as I mentioned above.  For the sake of simplicity, I will let A = SQRT(a^2+b^2).  So factor A out of  \n\na sin(Bx)+b cos(Bx)\n\nto get\n\nA [a/A sin(Bx) + b/A cos(Bx)]\n\n\n\n\nSince (a/A)^2 + (b/A)^2 = 1 (you might need to actually verify this for your purposes), there exists some angle C such that\n\ncos (C) = a/A and sin (C) = b/A \n\n(also, you should point out that both a/A and b/A have absolute values less than one, so that C is real and not complex with a nonzero imaginary part).\n\nSo make the substitutions to get\n\nA [cos(C) sin(Bx) +  sin(C) cos(Bx)].\n\n\n\n\n\nAccording to the formula \n\nsin(y+z) = sin y cos z + cos y sin z, \n\nthe expression in the brackets must equal sin(Bx + C).  So now the whole expression becomes\n\nA sin (Bx + C).\n\n\n\n\n\nSo in conclusion, \n\na sin(Bx)+b cos(Bx) = Asin(Bx+C),\n\nwhere A = SQRT(a^2+b^2), and C = arccos(a/A).  \n\n\n\n\nThere is a better representation for C, however.  \n\nSince cos C = a/A and sin C = b/A, you have\n\ntan C = (sin C)/(cos C) = (b/A)/(a/A) = b/a.\n\nThus, tan C = b/a, and C = arctan (b/a).  This definition of C does not rely on the creation of a third variable (A).\n\n\n\n\n\nHope this helps!
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The answer to your question is NO!  That is assuming a rectangle that fails to pass the test as a square, i.e. two sides of one length and two sides of another length.\n\nSince you have a difference of opinions in preceding answers, may I suggest an experiment.\n\nDraw a rectangle that is 1/2 inch wide and 11 inches long.  Now draw a diagonal in that rectangle.  It should be obvious that you are NOT bisecting the angles.  \n\nIf you are trying to prove this one way or the other, look to isosolese triangles with a 90 degree vertex.  If an isosolese triangle has a 90 degree angle the base can only be opposite that angle.  By definition an isosolese triangle has equal sides and equal base angles.  That means when you apply the Pythagorean theorem the square of the sides must equal each other.  So after you have drawn the diagonal in your rectangle, that diagonal must be opposite a 90 degree corner.  That means the length of the rectangle is one side of the triangle with the width being the other.  Since the length and width of the rectangle are not equal, the sides of the triangle are not equal.  Since the sides of the triangle are not equal, the triangle is not isosolese.  Since the triangle is not isosolese, the base angles are not equal.  And since all the angles of a triangle must equal 180 degrees and one angle is known to be 90 degrees, and the other two angles are known to be unequal neither of the other two angles can equal 45 degrees.\n\nThis is a bit involved, I hope you can follow the logic.
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kay,here is the proof\n\n1^3=3\n1^3+2^3=9 \n1^3+2^3+3^3=36  etc\n \n1,9,36,100,225,441 etc\n\nusing difference theory;\n\nfor n =1,2,3........\nnth term=1/4n^4+1/2n^3+1/4n^2\n=n^2(1/4n^2+1/2n+1/4)\n=n^2(1/2n+1/2)^2\n={ n(1/2n+1/2)}^2\n=k, say\n\nnow,for n=1,\nk=1\nfor n=2\nk=3^2 =(1+2)^2\nfor n=3\nk=6^2 =(1+2+3)^2\nfor n=4,\nk=(10)^2=(1+2+3+4)^2\nfor n=n,\nk=(1+2+3+......n)^2\n\nbut,it has been shown that k is the\nnth term for the sequence\n\n1,9,36,100,225,441 etc\n\ntherefore,\n \n(1^3+2^3+3^3+.......n^3)\n=(1+2+3+......n)^2\n\nif you don't follow the reasoning,\nplease contact me\n\nnote:the sum of 1 to n is n/2(n+1)\nk is simply the square of that\n\nk is the sum of all the cubes \nfrom 1 to n
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View source of the that page.\nYou will see the formulas.
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Yes. Many. Here's a way to get more:\n\nPick and two numbers A and B with A more than B.\n\nLet x=A^2 -B^2\nLet y=2*A*B\nLet z=A^2 +B^2.\n\nThen it will turn out that x^2+y^2=z^2.\n\nAll primitive Pythagorean triples are of this form, by the way. Primitive means that there are no common divisors between the three numbers.\n\nFor example, if A=5 and B=2, then you get x=21, y=20 and z=29.
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Wow, what an interesting question. I'm surprised no one has solved it yet, but that may be a consequence of the fact that no one has understood it yet. Clearly, schools need to focus more on reading comprehension ^_^.\n\nTo answer your question - not only can we prove that there exist integers a and b satisfying your assumptions, but we can even show a method of finding them. The proof is long, so sit back, relax, grab your favorite drink, and prepare to read.\n\nFirst, consider that any solutions of these equations must obey the quadratic formula: x=(-b±√(b²-4ac))/2a. Substituting the coefficients of the  actual equations in, we find that the solutions of the first equation are (-p±√(p²-4q))/2 and the solutions of the second equation are (-p±√(p²+4q))/2. Let us consider only the greater of the solutions in each pair - that is, replace each ± with a +. The difference between these solutions is then (-p+√(p²-4q))/2 - (-p+√(p²+4q))/2. Call this number K. We can then perform some simple algebra as follows:\n\n(-p+√(p²-4q))/2 - (-p+√(p²+4q))/2=K\nmultiply both sides by 2 and simplify\n√(p²-4q)-√(p²+4q)=2K\nsquare both sides\n(p²-4q) - 2√((p²-4q)(p²+4q)) + (p²+4q)=4K²\nsimplify\n2p²-2√(p^4-16q²)=4K²\ndivide both sides by 2\np²-√(p^4-16q²)=2K²\nsubtract 2K² and add √(p^4-16q²) to both sides\np²-2K²=√(p^4-16q²)\nsquare both sides\np^4-4K²p²+4K^4=p^4-16q²\nmultiply both sides by -1 and simplify\n16q²=4K²p²-4K^4\ndivide both sides by 4\n4q²=K²p²-K^4\nadd K^4 to both sides\nK^4+4q²=K²p²\ndivide both sides by K² and rearrange\np²=K²+4q²/K²\n\nThus, if we let a=K and b=2q/K, then p²=a²+b² as required by the problem. All that remains is to show that a and b are nonzero integers.\n\nLemma 1: K is an integer\nProof: All solutions of both quadratic equations are integers. K is the difference of two of these solutions - that is, K is the difference of two integers. Therefore, K is an integer.\n\nLemma 2: K is nonzero\nProof: K is the difference between a solution of the first equation and a solution of the second equation. Suppose K=0. Then these equations have a common solution - that is, for some x, x²+px+q=0=x²+px-q. This would imply that q=-q. This can only be true if q=0. But q≠0. Therefore, the equations do not have a common solution and K≠0.\n\nLemma 3: 2q/K is an integer\nProof: Earlier we showed that √(p²-4q)-√(p²+4q)=2K. We can perform some simple algebra as follows:\n\n√(p²-4q)-√(p²+4q)=2K\nmultiply both sides by (√(p²-4q)+√(p²+4q)) and simplify\n2K(√(p²-4q)+√(p²+4q)) = (p²-4q)-(p²+4q)\nsimplify further and rearrange:\n-8q=2K(√(p²-4q)+√(p²+4q))\ndivide both sides by -4:\n2q=-K(√(p²-4q)+√(p²+4q))/2\ndivide both sides by K:\n2q/K=-(√(p²-4q)+√(p²+4q))/2\n\nNote that 2K=√(p²-4q)-√(p²+4q) Adding 2√(p²+4q) to both sides of that identity gives 2K+2√(p²+4q) = √(p²-4q)+√(p²+4q). We can then substitute to get:\n\n2q/K=-(2K+2√(p²+4q))/2\nSimplifying at last yields:\n2q/K=-K-√(p²+4q)\n\nWe already know K is an integer, so if √(p²+4q) is also an integer, then 2q/K is the difference of two integers and therefore an integer. Now we know that the solutions of the second equation are (-p+√(p²+4q))/2 and (-p-√(p²+4q))/2. The difference between them is √(p²+4q). However, since both of the solutions to the second equation are integers, their difference must be an integer. Therefore √(p²+4q) is an integer, and by extension, 2q/K is an integer. Q.E.D\n\nLemma 4: 2q/K is nonzero\nWe know 2q/k=-K-√(p²+4q). Suppose 2q/K=0. Then -K-√(p²+4q)=0 From this we would get:\n\n-K-√(p²+4q)=0\nadd K to both sides and rearrange:\nK=-√(p²+4q)\nmultiply both sides by 2\n2K=-2√(p²+4q)\nremembering 2K=√(p²-4q)-√(p²+4q), we substitute:\n√(p²-4q)-√(p²+4q)=-2√(p²+4q)\nadd √(p²+4q) to both sides and rearrange:\n-√(p²+4q)=√(p²-4q)\n\nHowever we specified in the beginning t
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1 - A is enough because with the lengths of 2 sides you can use pythagoras' theorem to work out the length of the other
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You know there are 8 axioms in mathematics which we cannot prove\n\n1) if 'a' and 'b' are integers then 'a+b' is an integer and 'a.b' is an integer (Closure law)\n2) a + b = b + a and a..b = b.a (cumulative law)\n3) a + (b + c) = (a + b) + c (Associative law)\n4) a (b + c) = a.b + a.c (Distributive law)\n5) a + 0 = a = 0 + a (Additive Identity)\n6) a.1 = a = 1.a (Multiplicitive Identity)\n7) for all integer 'a' there is a unique integer (-a) such that a + (-a) = 0\n8) I can't remember...\n\nSo we can prove your requirement using above axioms\n\n0 + 0 = 0 (Obivious) \na ( 0 + 0 ) = a.0 (Multiplying by same factor)\na.0 + a.0 = a.0 (Distributive over addition)\na is an integer and 0 is an integer so a.0 is an integer, hence there exits a unique -(a.0) such that -(a.0) + a.0 = 0\n\nso adding -(a.0) both sides\n\n-(a.0) + (a.0 + a.0) = -(a.0) + (a.0) \n(-(a.0) + a.0 )+ a.0 = 0 (associative law)\n0 + a.0 = 0\na.0 = 0 (additive identity)
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You **did** say that the Pythagorean Theorem was a given, right? I think that your INTENT is to prove the converse of the Pythagorean Theorem, right?\n\nLet x, y, and z be positive integers such that\nx^2 + y^2 = z^2.\nClearly z > x and z > y (otherwise either x^2 or y^2 must be less than zero; prove this for yourself if you must).\n\nLet x and y be the sides of a right triangle that form the right angle.\nThen, by the Pythagorean Theorem the length of the hypotenuse h is\nh = √(x^2 + y^2)\nh = √(z^2) . . . . . . . . (given equation above)\nh = z . . . . . . . . . . . (given z is positive) \n        \nTherefore, by side-side-side congruence, any triangle whose sides are of lengths, x, y, and z, where x^2 + y^2 = z^2, must be a right triangle with hypotenuse z.\n\n\nASIDE TO HOOD...: You are using the Pythagorean Theorem to prove the Pythagorean Theorem. Although the questioner's intent is not perfectly clear, I don't think the intent was to prove "if x, then x". If so there are easier ways to do so.
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Induri Kartic is almost right but she has missed a certain steps.\n\nlet x = tan A , y= tan B z = tan C (A B C are acute angles) \n\nnow tan(A+B+C) = tan (A+D) using B+C = D\n= (tan A + tan D)/(1- tan A tan D)\n= tan A + (tan B+tan C)/(1- tan B tan C)/denominator say a\n\nnumerator = tan A + tan B + tan C - tan A tan B tan C / some thing say b\nwe have not ecaluated ab \ntan (A+B+C) = (x+y+z-xyz)/ab = 0\nso evaluation of a b was not required\n\nso A+B+C is  pi (as A+B+C < 3pi/2)\n\nnow tan 3(A+B+C) = 0\nso tan 3A + tan 3B + tan 3C = tan 3A \n\ntan 3A = (3tan A -tan^3 A)/(1-3 tan^2 A) = (3x-x^3)/(1-3x^2)\nputting values we get the resultant immediately
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Please do not answer this question about math. It is part of an ongoing competition called the USAMTS, which gives you a month to write proofs for five problems. All work has to be your own, and this is a clear violation of that policy. You are just helping this user cheat on the competition if you answer this question.\n\nSite: http://usamts.org\nCurrent Problems: http://usamts.org/tests/usamtsproblems_18_2.pdf\nRules: http://usamts.org/about/u_abrules.php\n"Participants may not use 'live' online help, e.g. they may not ask for help with the problem on online forums or tutorial services."