Question #146110
TanA × Tan2A = 1

Prove sin3A + cos3A = 1
1
Expert's answer
2020-11-29T19:26:36-0500

tanxtan2x=1sinxcosxsin2xcos2x=1 sinxcosx2sinxcosxcos2xsin2x=1  2sin2xcos2xsin2x=12sin2x=cos2xsin2xtherefore cos2x=3sin2xcosx=±3sinx  sin3x+cos3x==sinxcos2x+sin2xcosx+cosxcos2xsinxsin2x==sinxcos2xsin3x+2sinxcos2x++cos3xsinxcos2x2sin2xcosx==cos3xsin3x+2sinxcos2x2sin2xcosx==(cosxsinx)(cos2x+sinxcosx+sin2x)++2sinxcosx(cosxsinx)=(cosxsinx)(1+3sinxcosx)==cosxsinx+3sinxcos2x3sin2xcosx=1)3sinxsinx+3sin3x33sin3x==3sinx(3sin2x1)+sinx(3sin2x1)=(3sin2x1)(sinx3sinx)12)3sinxsinx+3sin3x+33sin3x==3sinx(3sin2x1)+sinx(3sin2x1)=(3sin2x1)(sinx+3sinx)1 answer:sin3x+cos3x1\tan x * \tan 2x = 1\\ \dfrac{\sin x}{\cos x} * \dfrac{\sin 2x}{\cos 2x} = 1\\ \ \\ \dfrac{\sin x}{\cos x} * \dfrac{2\sin x *cosx}{\cos^2 x-\sin^2x} = 1\\ \ \\ \ \dfrac{2\sin^2 x}{\cos^2 x-\sin^2x} = 1\\ 2\sin^2 x = \cos^2 x - \sin^2 x\\ therefore \ \cos^2 x = 3\sin^2 x\\ \cos x = \pm \sqrt3\sin x\\ \ \\ \ \\ \sin 3x +\cos 3x = \\ = \sin x*\cos 2x +\sin 2x * \cos x + \\ \cos x *\cos 2x -\sin x *sin2x = \\ = \sin x *\cos^2x -\sin^3x + 2\sin x*\cos^2x + \\ +\cos^3 x -\sin x*\cos^2x - 2 \sin^2x* \cos x = \\ = \cos^3x-\sin^3x+2\sin x\cos^2x-2\sin^2 x\cos x = \\ = (\cos x-\sin x)(\cos^2x+\sin x\cos x +\sin^2 x) +\\ +2\sin x\cos x(\cos x-\sin x) = (\cos x-\sin x)(1+ 3\sin x\cos x) = \\ = \cos x-\sin x +3\sin x\cos^2 x -3\sin^2 x\cos x = \\ 1) \sqrt3\sin x -\sin x + 3\sin^3 x -3\sqrt3\sin^3x = \\ = -\sqrt3\sin x(3\sin^2 x -1) +\sin x(3\sin^2x-1) =\\ (3\sin^2x-1)(\sin x-\sqrt3\sin x) \ne1 \\ 2) -\sqrt3\sin x -\sin x + 3\sin^3 x +3\sqrt3\sin^3x = \\ = \sqrt3\sin x(3\sin^2 x -1) +\sin x(3\sin^2x-1) =\\ (3\sin^2x-1)(\sin x+\sqrt3\sin x) \ne1 \\ \ \\ answer: \sin3x+\cos3x \ne 1


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