prove that:c^2=(a-b)^2 (cosC/2)^2 +(a+b)^2 (sinC/2)^2?
826654
we start on RHS as it is more complex\n\n(a-b)^2 (cos C/2)^2 + (a+b)^2 (sin C/2)^2 ... 1\n\ncos t = 2 cos^2 t/2 -1 so cos^2 t/2 = (cos t+1)/2\ncos t = 1- 2 sin^2 t/2 so sin ^2 t/2 = (1-cos t)/2\nputting in1 RHS\n= (a-b)^2 (cos C +1 )/2 + (a+b)^2 (1- cos C)/2\n= cos C((a-b)^2-(a+b)^2)/2 + ((a+b)^2 +(a-b)^2)/2 (rearranging terms)\n= -2ab cos C + a^2 + b^2\n= c^2 \n\nproved
830946
c)\n\n= x^3 + x^2 + x - x^2 - x -1\n= x^3 -1
697247
C = (W^2*L^2) / (24T^2)\nC(24T^2) = W^2*L^2\n[C(24T^2)]/L^2 = W^2\nAnswer:\nW = sq.rt {[C(24T^2)]/L^2}
419071
You can cheat and use a calculator to get the values, but I don't think this is what your teacher is looking for. \n\nFirst, use the relationship sin^2(x)+cos^2(x)=1 to find |sin(a)| and |cos(b)|. Pay close attention to the ranges of a and b to get the sign right-- sin(a) and cos(b) should both be negative. Doing this you should find:\n\nsin(a) = -12/13 \ncos(b) = (-2 sqrt(2))/3\n\nHere are two useful identities:\n\ncos(a-b) = cos(a)cos(b) + sin(a)sin(b)\ncos(2b) = 1 - 2sin^2(b)\n\nSince you now know cos(a),cos(b),sin(a) and sin(b), you can find that:\n\ncos(a-b)=(12-10sqrt(2))/39\ncos(2b)=7/9
728989
we have in a triangle\n\nsin B/b = sin C/c = k (say)\n\nsin B= bk  ...1\nsin C = ck  ...2 \nadditionally\n\nb^2 = c^2+a^2-2ac cos B\nso cos B = (c^2+a^2-b^2)/2ac ...3\n\ncos  C = (a^2+b^2-c^2)/2ab ...4\nfrom1 and 3 by deviding 3 by 1\ncot B = (c^2+a^2-b^2)/2abck\n\nor 2abck = (c^2+a^2-b^2) tan B  ..5\nsimilarly from 2and 4\n2abck =(a^2+b^2-c^2) tan C\n\nfrom above 2 we get the result as both are same
880711
sin(2A)=sin(A+A)=sinAcosA + sinAcosA=2sinAcosA
698714
Assuming you mean S=(c-b)/(b-a) rather than c-1-a...\n\nMultiply both sides by (b-a) to get S(b-a)=c-b\nMultiply the S through to get Sb-Sa=c-b\nAdd Sa+b to both sides to get Sb+b=c+Sa\nFactor out b to get b(S+1)=c+Sa\nDivide both sides by (S+1) to get b=(c+Sa)/(S+1)\n\nKeep in mind this only holds true for b not equal to a because then S would have no value.\n\nIs that what you're looking for?
1051369
Given a+b+c=0\na+b=-c\nCubing both sides, we get\n(a+b)^3=(-c)^3\na^3+3a^2 b+3ab^2+b^3=-c^3\na^3+b^3+c^3=-3a^2 b-3ab^2=-3ab(a+b)=-3ab(-c)=3abc\nTherefore a^3+b^3+c^3=3abc
255958
Substitute 2 for x and solve for c.\n2^2-3(2)+c=0\nRemember do powers, then multiplication, then subtraction.
962866
Anami's answer was correct and very simple.  You should give him credit for it.\n\nHere it is:\n\nID: cos(A+B)-cos(A-B)= -2sinAsinB\nso RHS: A=3x and B=x\n=> -2sin3xsinx = cos4x -cos 2x\nproved.\n\nWhat he did is this (only working with the left side):\n\ncos4x - cos2x\n------------------\n2sin3x\n\ncos(3x + x) - cos(3x - x)\n------------------------------\n2sin3x\n\ncos3x cosx - sin3x sinx - (cos3x cosx - sin3x sinx)\n-----------------------------------------------------------------\n2sin3x\n\n-2sin3x sinx\n---------------\n2sin3x\n\n-sinx\n\nis that more clear?
946513
Subject: Re: Proving trig identiies...small error..thisis the right one\nFrom: \nDate: Monday June 14, 2004 at 08:35:54\n\nRHS= 2csc^2x=2(cot^2x+1) \n\nnote :-(csc^2x=cot^2x+1) \n\n=2cot^2x+2 \n\n=2cos^2x/sin^2x +2 \n\nnote :-((cot^2x=cos^2x/sin^2x) \n\n=2cos^2x/(1-cos^2x) +2 \n\nnote :-((sin^2x=1-cos^2x) \n\n=2cos^2x +2(1-cos^2x) /(1-cos^2x) \n\n=2cos^2x +2 -2cos^2x) /(1-cos^2x) \n\n=2/(1-cos^2x) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-((1-cos^2x=(1-cosx)(1+cosx)) \n\n\nTHEREFORE lHS =2/(1-cosx)(1+cosx) (1-cos^2x=(1-cosx)(1+cosx)) \nnow for the RHS we get \n\nRHS= 1/1+COSX + 1/1-COSX \n\n=(1-cosx) + ( 1 + cosx)/(1-cosx)(1-cosx) \n\nnote :-(( through expansion) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-(cosx on the numerator cancel out) \n\ntherefore RHS= LHS \n\nTHEREFORE 2csc^2x= 1/1+cosx + 1/1-cosx \n\nas required\n\n\n\nCurrent Thread\n   Proving trig identiies   a girl - Monday June 14, 2004 at 08:06:52 \n   Re: Proving trig identiies   - Monday June 14, 2004 at 08:31:28 \n--> Re: Proving trig identiies...small error..thisis the right one   - Monday June 14, 2004 at 08:35:54 <--
870598
(sin x+ cos x)^2 = sin^2x + 2sin x cos x + cos^2 x using (a+b)^2\n= sin^2 x + cos ^2 x + 2 sinx cos x\n= 1 + sin 2x\nbecuase sin^2x+ cos^2x =1 \nand sin 2x = sin(x+x) = sin x cos x + cosx sin x = 2sin xcos x\nusing the formula sin(A+B) = sin A cos B + cos A sin B
1073595
(cosx)^2 - 2sinx = 2\ncosx^2 = 1 - sinx^2\n(1 - sinx^2) - 2sinx = 2\n1 - sinx^2 - 2sinx = 2\n-sinx^2 - 2sinx + 1 = 2\n-sinx^2 - 2sinx - 1 = 0\n-(sinx^2 + 2sinx + 1) = 0\n-(sinx + 1)(sinx + 1)\n\n-sinx - 1 = 0\n-sinx = 1\nsinx = -1\nx = sin^-1(-1)\nx = 270°\n\n--------------------------------------\n\n1.5^2 = 1.7^2 + 1.8^2 - 2(1.7)(1.8)cosA\n2.25 = 2.89 + 3.24 - 6.12cosA\n2.25 = 6.13 - 6.12cosA\n-6.12cosA = -3.88\ncosA = (388/612)\ncosA = (97/153)\nA = 50.6551127\n\n1.7^2 = 1.5^2 + 1.8^2 - 2(1.5)(1.8)cosB\n2.89 = 2.25 + 3.24 - 5.4cosB\n2.89 = 5.49 - 5.4cosB\n-2.6 = -5.4cosB\ncosB = (13/27)\nB = 61.2177953\n\n1.8^2 = 1.5^2 + 1.7^2 - 2(1.5)(1.7)cos(C)\n3.24 =  2.25 + 2.89 - 5.1cosC\n3.24 = 5.14 - 5.1cosC\n-1.9 = -5.1cosC\ncosC = (19/51)\nC = 68.1270919\n\nAs you can see, the angle opposite of side A, is the smallest angle.\n\nANS : about 51°\n\n----------------------------------------------\n\ncos(A) = 4/7\nx = 4\nr = 7\n\nr = sqrt(x^2 + y^2)\n7 = sqrt(4^2 + y^2)\n49 = 16 + y^2\ny^2 = 33\ny = sqrt(33)\n\nsin(A) = y/r\nsinA = (-sqrt(33))/7\n\ntanA = y/x\ntanA = (-sqrt(33))/4\n\ncscA = r/y\ncscA = -7/(sqrt(33))\ncscA = (-7sqrt(33))/33\n\nsecA = r/x\nsecA = 7/4\n\ncotA = x/y\ncotA = -4/(sqrt(33))\ncotA = (-4sqrt(33))/33
706305
R^2=A^2+B^2 -2(A*B)cos(180-acute angle)
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
402319
ya it is a^2 +b ^2 +2abcos(c)