proe that the differencial value of half of tanx sin2x  is sin2x?
475675
tan x = sin x/cos x\n\nsin 2x = 2 sin x cos x\n\nso tan x sin 2x = (sin x/cos x) * (2 sin x.cos x)\n                = 2 sin^2 x\nhalf of it is = sin^2 x\n\nDifferentiate half of tan x sin 2x\n= diff of sin^2 x\n= 2 (sin x).cos x  (refer to below)\n= sin 2x    (Q.E.D.)\n\n\nDifferentiating sin^2 x.\n\nLet's call sin x as A\nand B = A^2\nthen dB/dA = diff of B * diff of A\n           = 2A * diff of A\n           = 2 sin x * cos x\n           = sin 2x
482644
Let us denote the integral sign by I.\nWe will use substitution method.\nLet x = 2*tan(t)\nThen dx = 2*sec^2(t)dt\nThus, I[1/(x^2+4)^2]dx\n= I[1/{4*tan^2(t)+4}^2]*2*sec^2(t)dt\n= (1/8)*I[1/{tan^2(t)+1}^2]*sec^2(t)dt\n= (1/8)*I[1/{sec^2(t)}^2]*sec^2(t)dt  [since tan^2(t) + 1 =sec^2(t)]\n= (1/8)*I[1/{sec^2(t)}]dt\n= (1/8)*I[cos^2(t)]dt\n= (1/16)*I[2*cos^2(t)]dt\n= (1/16)*I[1+cos(2t)]dt [since 2*cos^2(t) = 1 + cos(2t)]\n= (1/16)*I[dt] + (1/16)*I[cos(2t)]dt\n= (1/16)*t + (1/16)*[{sin(2t)}/2] + C where C = arbitrary constant of integration\n= t/16 + {sin(2t)}/32 + C\n\nNow sin(2t) \n= {2*tan(t)}/{1+tan^2(t)}\n= 4x/(4+x^2) [since x = 2*tan(t)]\n\nAlso t = tan-1(x/2) [tan-1(x/2) means tan inverse (x/2)]\n\nHence the given integral equals to\n{tan-1(x/2)}/16 + x/{8(4+x^2)} + C
1162151
Use the fact that tan^2(x) = sec^2(x ) - 1 and integrate that.  You will get tanx -x +c.
624769
You can use sin and cosine.
1155939
It's something like:\n\n2x + 4 = 2(x +2)\n\nWhen you distribute the right hand side and then put the x's together and the constants together, you end up with 0=0.\n\nThat means any value of x would make this a true equation. There is no specific solution to this equation.\n\nLater, when you get into trigonometry, identities become very important and aren't as simply and obvious as this example. For example you learn that tan(x)=sin(x)/cos(x) and that is always true. (Of course, you don't know what that means yet.) These identities mean we can always substitute what's on one side of the equation for what's on the other side of the equation without changing anything.
972809
Consider the left-hand side(LHS) [cot(x) + 1] / [cot(x) – 1]. We want to prove that this is equal to the right-hand side (RHS), that is, [1 + tan(x)] / [1 – tan(x)]. Thus, we have the following:\n\nLHS = [cot(x) + 1] / [cot(x) + 1]\n        =  { [cot(x) + 1] / [cot(x) – 1] }* { tan(x) / tan(x) }.\n\nNOTE: We multiply tan(x) to cot(x) + 1 and the other tan(x) is multiplied to cot(x) – 1.\n\n        =  { tan(x) [cot(x) + 1] } / { tan(x) [cot(x) – 1] }\n\n        =  { tan(x) cot(x) + tan(x)] } / { tan(x) cot(x) – tan(x)] }, by distributive property of multiplication over addition (subtraction)\n\n        =  { 1 + tan(x)] } / { 1 – tan(x)] }, since tan(x) cot(x) = 1 by reciprocal identity.\n\n        =  RHS.\n\nIndeed, [cot(x) + 1] / [cot(x) – 1] = [1 + tan(x)] / [1 – tan(x)]
982428
LHS\n= (sin^2B-tan^2B)/(1-sec^2B)\n= (sin^2B-tan^2B)/(-tan^2B)\n= (sin^2B/-tan^2B) + (-tan^2B/-tan^2B)\n[splitting the numerator up]\n= (-cos^2B) + 1\n[cosx= sinx/tanx]\n= 1 - cos^2B\n= sin^2B\n= RHS
932836
Are you sure this is the correct identity?\n\n 1+ tanx.tan2x = 1 + (sinx/cosx)(sin2x/cos2x)\n                        = 1 + (sinx/cosx)(2sinxcosx/cos2x)\n                        = 1 + 2sinx.sinx/cos2x\n\n cos2x = 2cosx.cosx - 1  = 2cos^2x - 1        (Note:  ^2 = square)\n\n then..\n      1+tanx.tan2x = 1 + [2sin^2x/(2cos^2x - 1)]\n                            = (2cos^2x + 2sin^x -1) / (2cos^2x - 1)\n                            = [2(cos^2x + sin^2x) -1] / (2cos^2x - 1)\n                            = 1/cos2x    (because 2cos^2x - 1 = cos2x)\n\n     Then 1/1+tanxtan2x = cos2x
59699
36.4 %\n\npercent difference is x1-x2/[(x1+x2)/2]
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
245398
sinx = a with a Є [-1;1] or -1 </= a </= 1\n\na^2 - 2a - 3 = 0\nwe have 2 roots: a = -1 (accept) & a = 3 (reject) because a Є [-1;1]\n=>  sinx = -1\n<=>  x = -∏/2 + k2∏ with k Є Z
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
851033
(cosx-sinx)^2=cos^2x+sin^2x-2sinxcosx so\nit can be 1-2sinxcosx+cosx+sinx
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
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
890950
Did they tell you that\nIf x is very close to 0, then sin(x)=tan(x)=x;\n(sin(x/2))^2 / x^2 = (x/2)^2/x^2=1/4\n(tan(2x))^3 / x^3 = (2x)^3/x^3=8