How do you prove this equation?
827948
To understand you should read over this section on Trigonometry  identities on the site www.wikipedia.org and form here you will be able to reagrane the equation since somethings equal others
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.
758594
2. The equation is (y-y1)/(y1-y2)=(x-x1)/(x1-x2)\n=>(y+2)(-2-0)=(x+1)(-1-0)\n=>(y+2)(-2)=(x+1)(-1)\n=>-y-2=-2x-2\n=>-y+2x=0=> 2x-y=0 which is the required equn.\n3.The equation is (y-4)(4+4)=(x-0)/(0-4)\n=>y-4/8 =x/-4\n=>8x=-4y+16  or,8x+4y-16=0  o4 2x+y-4=0 is the reqd.equation\n4.the eqn. is (y+3)/(-3-3)= (x+2)/(-2-0)\n=>(y+3)/-6=(x+2)/-2\n=>-6x-12=-2y-6\n=>-8x+2y-6=0 or,-4x+y-4=0 which is the equired equation
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
300657
let a = b  \nMultiply both sides by a \na² = ab  \n Add (a² - 2ab) to both sides \na² + a² - 2ab = ab + a² - 2ab  \n Factor the left, and collect like terms on the right \n2(a² - ab) = a² - ab  \n Divide both sides by (a² - ab)\n2 = 1\n(The place where the proof goes wrong is the very last step.  If a² - ab = 0, then you would be dividing by zero, which is undefined.)
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 <--
1004368
x^4+y^4+8>= 8xy? \n\nI redid the proof using the Am-Gm inequality.\nUsing Am-Gm\nlet x^4=(x²)² & y^4=(y²)²\n((x²)²+ (y²)²)/2 >= sqrt{(x²*y²)²} =>\n\n((x²)²+ (y²)²)>= 2*x²*y²\n\n((x²)²+ (y²)²)+8>= 2*x²*y²+8\nUsing Am-Gm again on 2*x²*y²+8 =>\n(2*x²*y²+8)/2>=sqrt(2*x²*y²*8) =>\n(2*x²*y²+8)>= 2*sqrt(16*x²*y²) =>\n(2*x²*y²+8)>= 2*4*x*y=8xy\nSince x^4+y^4+8>= (2*x²*y²+8) &\n(2*x²*y²+8)>=8*x*y\nthen x^4+y^4+8>= 8xy. \nend of proof using the Am-Gm twice
790751
you have to do it by induction:\nbasically:\nn=2:\nka_1 + ka_2 = k( a_1 +a_2)\n\nnow, assume it is true for n\n\nlet us prove it for n+1:\n\nsum_{i=1} ^{n+1} ka_i = sum_{i=1} ^{n} ka_i + ka_{n+1}\n= k ( sum_{i=1} ^{n} a_i )+ ka_{n+1} now you have only two summands, so:\n= k( sum_{i=1} ^{n} a_i  + a_{n+1} )\n= k sum_{i=1} ^{n+1}  a_i \n~
268888
I think I can do this one...I'll get right back to you when I figure it out... *grabs a pen and paper...and also tries to figure out why people are answering, saying simply that they can't or won't answer your question*\n\nEDIT...k, here's what I got...not sure if I understood what your question was asking, so I may be mistaken...\n\nm=h(x)-2x\nk=(2-3h(x))/6x
580733
First equation:\n(x+a+b)^2 = x^2 + (x+a)^2 - (x+b)^2\nor x^2 + 2x(a+b) + (a+b)^2 = x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2\nor x^2 + 2ax + 2bx + a^2 + 2ab + b^2 - [x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2] = 0\nor 4bx + 2ab + 2b^2 = 0\nor 2b(2x+a+b) = 0\nor  2x + (a + b) = 0\nor 2x = -(a+b)\nTherefore x = -(a+b)/2\n\nSecond equation:\n(x+a)^2=(x+b)^2\nor (x+a)^2 - (x+b)^2 = 0\nor (x+a+x+b)(x+a-x-b) = 0\nor (2x+a+b)(a-b) = 0\nIf a = b then the equation becomes identity. So a is not equal to b.\nHence 2x + a + b = 0 \nor x = -(a+b)/2\n\nThus from both equations we have x = -(a+b)/2
760807
Please be careful with your symbols. I believe you mean to prove\n\n    tan²x\n------------ = sin²x\n1 + tan²x\n\nwritten as tan²x / ( 1 + tan²x ) = sin²x\n\nReplace tan x with sin x / cos x to get\n\n( sin²x / cos²x ) / ( 1 +  ( sin²x / cos²x ) )\n\nMultiply numerator and denominator by cos²x to get\n\nsin²x / ( cos²x + sin²x ) = sin²x / 1 = sin²x
803771
sq.rt of x*y = Sq.rt X * sq.rt. Y\n\nIt can be proven that:\n √(x*y) = √x * √y\n\nConsider the first part → √(x*y).\nTwo number values, one is x and the other is y, they are multiplied together.  You then take the square root of this result.\n\nConsider the second part → √x * √y.\nYou take the square root of x and y and then you multiply the results together.\nIt can be shown that the results of the first part is the same as the result of the second part.
605879
(x+3)(x+12)(x+2)(x+8)= [(x+3)(x+8)][(x+12)(x+2)]\n                     = (x^2+11x+24)*(x^2+14x+24)\nlet t=x^2+24+(25X)&#92;2 . \nSo we have: x^2+11x+24=t-(3x)&#92;2 and x^2+14x+24=t+(3x)&#92;2                                 \nso the equation becomes: [t-(3x)&#92;2][t+(3x)&#92;2] = 4x^2\n                    <-->t^2-(9x^2)&#92;4=4x^2  \n                    <-->t^2=(25x^2)&#92;4\n                    <-->t=(5x)&#92;2 or t=-(5x)&#92;2\n                    <-->x^2+10x+24=0 or x^2+15x+24=0\ncontinue by yourself
67585
Formulaically it reduces down to x*z = x*z, so no matter which numbers you plug in for any of the values it will be a valid equation.  \n\n(x*(y+z))-(x*y)=x*z      becomes\nx*y + x*z - x*y = x*z     rewrite that as \nx*y + x*z + -(x*y) = x*z    the positive and negative x*y terms cancel,leaving\n\nx*z = x*z\n\nSince that is a truism, an infinite number of solutions exist.
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
282131
Note that although x1=x2 satisfy the equations but some distinct values for x1 and x2 can satisfy it too: \n\nsuppose x1 and x2 are distinct and so x1-x2 is not zero\nthen eq1 is:\n\n4(x1-x2)(x1+x2)-4(x1-x2)=0\n\n---> 4(x1+x2)-4=0\n---> x1+x2=1\n\nand eq2 is:\n\n-3(x1-x2)(x1+x2)(x1^2+x2^2)+2(x1-x2)(x1+x2)=0\n\nomit (x1-x2)\nand substitute x1 by 1-x2 to get:\n(1-x2)^2+x2^2=2/3\n\n---> x2=(3+-sqrt(3))/6\n\nso we have two values for x1 and x2 which satisfy the equation and x1 and x2 are distinct.