Need help with Adv. Algebra Trig?
821547
f(x)=4x^2 + 9x - 2\n\nhere, this means the function (f) of x=4x^2 + 9x - 2\n\nso if there was f(2)=4x^2 + 9x - 2  that would mean solve the function if x=2
128596
im great with algebra, ive been doing it for the past 4 years now non stop, email me and ill be more than happy to show you how to do it step by step and answer you questions\nzinger_38@yahoo.com
278456
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
592160
i think the answer wil be (tanX)/(tanY)\n\nSIN(X+Y)=sin x cos y+cos x sin y\n\nsin(x-y)=sin x cos y-cos x sin y\n\n use cos x cos y to divide the numerator and denominator... there is the answer...
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
714084
plug in cos^-1(30/50) in a calculator\nto get cos^-1 just hit 2nd cos
710712
i took  trig and dont remember much but it should go like this\n(a)\n3x^2+12x=-12\n9x+12x=-12\n21x=-12\nx=-12/21\nx= -0.57\n\n(b)\n7x^2=28\n49x=28\nx=28/49\nx=0.57\n\n\nc im clueless\n\ni was wrong i just fixed it
854468
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
743138
join AC\nnow youhavetwo triangles ABC and ADC\nin triangle ABC <CAB+<B+<ACB=180*.............(1)<s of a triangle\nin triangle ADC <CAD+<D+<ACD=180*............(2)--------do--------\nadding (1) and (2)\n<ABC+<ADC+<B+<D+ACB+<ACD=360*\n=><DAB+<B+<BCD+<D=360*\nhence proved
764150
I presume you want to simplify it, though you didn't say that :P\nWith any trig expression like this, you have two options:\n1) Look for some cute trick. This is usually what you are shown when you get given the answer. However, these are often tricky to find.\n2) Turn everything into sin and cos, and simplify. Then use the fact that (sin x)^2 + (cos x)^2 = 1. This *always* works!\n\nSo, turn tan into sin/cos, and sec into 1/cos:\n((sin B)^2 - (sin B)^2/(cos B)^2) / (1 - 1/(cos B)^2)\nNow, multiply top and bottom by (cos B)^2:\n((sin B)^2 * (cos B)^2 - (sin B)^2) / ((cos B)^2 - 1).\nNow, factorise the top:\n(sin B)^2 ((cos B)^2 - 1) / ((cos B)^2 - 1)\nAnd cancel:\n(sin B)^2.
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
299283
first draw a right triangle and label the angles and sides. now angle B cut in half is 15 degrees and angle A is 90 degrees.  lets assign AM a value of x.  so the tangent function is equal to the opposite leg over ajacent leg=> tan(15degrees)=x/(2*sqrt(3)).  take the tan of 15 and then multiply that by 2*sqrt of 3 and that will be your answer for the length of AM which is approx=.9282  to find MC you must first find the length of AC with the same tangent function using 30degrees as your angle so lets assign AC a value of y. so tan(30degrees)=y/(2*sqrt(3)) which is = 2 so MC=AC-AM= 1.0718
756840
cosx=tanx\ncosx=sinx/cosx\ncos^2(x)=sinx\n1-sin^2(x)=sinx    {pythagoras}\nsin^2(x)+sinx-1=0\nuse quadratic formula\nsinx=(-1+or-sqrt(5))/2\n=0.6180339888\nor -1.6180339888\nsine has to lie between +/-1\nsinx= 0.6180339888\nx=arcsin(0.6180339888)\n  =38.17270763\n+ve value for sine lies in 1st and 2nd\nquadrants \ntherefore,x=38.17270763 degrees or\n141.8272924 degrees\n\nthese two values were checked in the \noriginal equation and were found \nto be correct\n\ni hope that this helps
982041
Well dear\n\n-If the first root is " α " and the second roots is " α + 3"  so have  \n▪x ² - (α + β) + α β = 0 \n▪( x - α )(x - β ) = 0 \n-b/a = α+ β  \n&\nc/a = α β\nso we have ; \nc/a = m/4  \n-b/a = - ( -16) / 4 = 4 \nα + α + 3 = 4 , 2α + 3 = 4 \nα = 1/2  ,  β = 3 1/2 \n2 /a = α β  =>  m/4 =  1/2 * 7/2  , m = 7\n\nGood Luck dear
329665
Well I'm not much help, but to solve trig. identies you basically just use the given identies and equal them to each other. You substitue whutever equation is possible to simply the identity down to it's simplest form. \n\nLook at the bottom of this website for more: http://library.thinkquest.org/20991/alg2/trigi.html\n\nExcept that website might actually be more focused on trig. Equations not specifically Identities... lol So try using this website instead =) :\nhttp://www.intmath.com/Add_Trig/1_TrId.php\n\nHave fun.