How do you derive the Sum-to-product identities using the Product-to-sum identites?
814204
To quote the wikipedia:\n\n"Replace x by (x + y) / 2 and y by (x – y) / 2 in the product-to-sum formulae."
650513
shoot me if i am wrong but product means multiply...  so wouldn't the answer be to multiply your n terms together...\n\nwithout further information i can't give you a formula\n\nare you trying to multiply the first n terms from 1 to n???\n\ndo you have a series with a formula that describes each of the terms???\n\nin short be specific...   or any answer you will get to this question will likely be too general or just plain wrong
463511
Use the double angle identities.
958614
#1.)-----------------------------------\n  \n25b^ -20by + 4y^\n\nMultiply 1st and 3rd term together to get 100b^y^.\n\nFactor 100b^y^ into two factors whose sum = the middle term of -20by.\n\n1by   X 100by\n2by   X 50by\n4by   X 25by\n5by   X 20by\n10by X 10 by   ....................BINGO!\n\n-10by + (-10by) = - 20by   &   (-10by)(-10by) = 100by\n\nThis is a really kewl factor by grouping trick, I LOVE it!\n\nBring down the first term of: (25b^)\n\nNext, write the SMALLER of the two factors 1st.  Followed by the LARGER of the two factors.  In this case, they're both the same, so it doesn't matter which one goes first: (-10by)\n\nFinally, bring down the third term of: (+4y^)\n\n25b^ -10by -10by + 4y^\n\nNow, lets factor by grouping! \nFactor out the GCFs.\nGCF#1 = 5b\nGCF#2 = -2y\n\n(25b^ -10by) + (-10by +4y^)\n5b(5b - 2y) -2y(5b - 2y)\n\nGCF#3 = (5b - 2y)\n\n(5b - 2y)(5b - 2y)\n\nSimplifies further as: (5b - 2y)^   ....Whatdayaknow?! \n.......................................................................................................It's a Perfect Square Trinomial.\n\n#2.)------------------------------------------------------------------\n\n100 - 9b^\n\nAh, the difference of two perfect squares.\n\n---------------------------------------------\n\nFORMULA:\na^ - b^ = (a-b)(a+b)\n\n---------------------------------\n\nSOLUTION:\n100 - 9b^\n(10)^ - (3b)^\n\n-----------------------\n\nANSWER:\n(10-3b)(10+3b)
775420
you can write \nP=d product((x-a_i)^k_i)) poduct((x-b_j)^2 + c_j^2)\n Then I give you 3 hints:\n1) d must be positive\n2) all k_i are even numbers\n3) use the formula (a^2+b^2)(c^2+d^2) =\n= (ac-bd)^2 + (ad+bc)^2
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)
224978
associative, commutative, identity, inverse
170974
Prove that for  u,v, and w, p1=s(u)+t(v)+r(w), i.e that you can create a linear combination of vectors such that v=a1v1+a2v2+...+anvn.
913234
it's "pi" not "pie" dear ... :-)\n\nAnd, yes I do know how to prove those identities.\n\nAnd, no I won't do them all, as it will deprive you of the wonderful opportunity to learn and practice for your test. \n\nBut, I will do one for you!  I'll do #8:\n\n(sinx + cosx) ² =\nsin²x + 2sinxcosx + cos²x = \nsin²x + cos²x + 2sinxcosx = \n1 + 2sinxcosx  {{ using trig ID #1 }} = \n1 + sin2x {{ using trig ID #2 }}\n. . . . .\nTrig ID #1 :  sin²x + cos²x = 1\nTrig ID #2 :  sin2x = 2sinxcosx\n\nNRT
723293
you have to use the couchy schuartz method\nim nit sure on the spelling \nbasicly you take one vector (V1) and use it then you take the second vecter (U1) minous vi times(V1 interproduct U1)/(magnatude V1^2) and use that total as V2 then you do a simular thing for the third and fourth vector and up if you have them by subtrating V1<Ua,V1>/(mag V1)^2 plus V2<Ua,V2>/(mag V2)^2 .... till you have done that for each vector \nin other words every vector will have a term for all of the new previous vectors subtracted from it then finaly you take all the new vectors and devide them by their magnatude to get unit vectors and they will then all be orthOGANOL UNIT VECTORS
1010848
First of all, u do not need Lagrange's identity to prove it. u just need to expand everything.\n\nSo, u want to use Lagrange's identity, fine. set a^m=a^2, a^n=x^2, substitute these into the Lagrange's identity and u get exactly [a^{(m+n)/2}+b*y]^(2) + [y*{a^(m/2)} - {a^(n/2)}*b]^(2) without even needing to do anything else.\n\nThis is not called proving!
899944
first, I would put it in an easier to express format g(t)=[t^(1/2)](1-t)  then take the derivative g'(t)=[(1/2)t^(-1/2)](1-t)-t^(1/2)   now just solve for g'(t)=0, so 0=[(1/2)t^(-1/2)](1-t)-t^(1/2)      now we distribute to get 0=[(1/2)t^(-1/2)]-(3/2)t^(1/2)     we add (3/2)t^(1/2) to both sides to get (3/2)t^(1/2)=(1/2)t^(-1/2)   simplfy this to [3sqrt(t)/2]=[1/2sqrt(t)]   we multiply both sides by 2sqrt(t) to get 6t/2=1 or 3t=1, which means that t=1/3\n\nAnd by the way, this isn't the chain rule, this is the Product Rule
221721
Product rule:\n\n(fg)' = (f ')g + f(g ')\n\nSo,\n\n(-2x^2 * ln(x))' = (-4x * ln(x)) + (-2x^2 * 1/x)\n\nMultiply and clean up:\n\n-4xln(x) - 2x
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
243814
although the equation is not complete and therefore the full answer can not be given. i believe this equation will need to be differentiated using the product of a product rule.
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.