How to prove sin(2x)=2sin(x)cos(x)?
610758
sin(2x) = 2sinxcosx is a trigonometric identity called the double angle identity. It was derived from another identity:\n\n   sin(x + y) = sinxcosy + cosxsiny\n \nhere x and y are two different angles but what if x = y\nreplace y with x to make all the equations in terms of x, thus the equation becomes:\n\n   sin(x + x) = sinxcosx + cosxsinx\n   \nnote that sinxcosx is also equal to cosxsinx, so simplifying the equation it becomes:\n\n  sin(2x) = 2sinxcosx or sin(2x) = 2cosxsinx
562730
cos2t=cos^2t-sin^2t or 2cos^2t-1 or 1-2sin^2t\n\nso you can arrive at something like 2cost-sect (remember 1/cost=sect). that's the simplest answer that you can get.
160524
This is not a calculus problem, this is merely a trigonometric identity.  No derivatives, integrals, or numerical approximation were used.\n\nDefinition:\n\ntanh(2x) = sinh(2x)/cosh(2x)\n\nYou know the double angle formulas for sinh and cosh:\n\nsinh(x + x) = sinh(x)cosh(x)+cosh(x)sinh(x) = 2cosh(x)sin(h)\n\ncosh(x + x) = cosh(x)cosh(x)+sinh(x)sinh(x) = cosh(x)^2 + sinh(x)^2\n\nDivide:\n\n(2cosh(x)sinh(x))/(cosh(x)^2 + sinh(x)^2)\n\nDivide numerator and denominator by cosh(x)^2:\n\n((2cosh(x)sinh(x))/cosh(x)^2) / ((cosh(x)^2 + sinh(x)^2)/cosh(x)^2)\n\nSimplify:\n\n(2 sinh(x)/cosh(x))/(1 + sinh(x)/cosh(x)) \n\nSimplify more to solution:\n\n2tanh(x)/(1 + tanh(x)
952344
(1+tanx)/(1+cotx)\n=(1+sinx/cosx)/(1+cosx/sinx)\n=[(cosx+sinx)/cosx]/(sinx+cosx)/sinx]\n=(cosx+sinx)(sinx)/(cosx+sinx)cosx)\n=sinx/cosx\n=tanx\nhence proved
938253
2cos²(45º-0.5x) – 1 = sinx\n\nUsing: cos 2a = 2cos² a -1\nSo: 2cos²(45º-0.5x) – 1 = cos 2(45º-0.5x) = cos (90º - x)  = sinx
900488
cos(x) +sin(x)tan(x)=sec(x)\nmultiply by cos(x) on both sides\ncos^2(x)+sin^2(x)=1\n1=1\nI believe they are probably asking you to prove the identity
906876
2cos^2x-1+1/2sinxcosx\n=2cos^2x/2sinxcosx\n=cosx/sinx\n=cotx
726975
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...
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
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
1135417
u can factor it out to be (2sinx+1)(cosx)=0.. then find x for which either sinx=-1/2 or cosx=0\n\n(sinx+1)cosx=0 same method applies
276083
Yes, it can be rewritten as 2sin2xcos2x.  Or, you can use the addition formula for sineto prove this.  sin(x + y) = sinxcosy + cosxsiny.  In your case, sin(2x + 2x) = sin2xcos2x + cos2xsin2x = 2sin2xcos2x.
699547
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
898033
cos2x = 1 - 2 (sin x) ^2 = sin x\n\nthen 2 ( sin x)^2 + sin x  - 1 = 0\n or (2sin x - 1 )(sin x + 1) = 0\nsin x = 1/2 , - 1\nx = 30 deg, or - 90 deg
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
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