Answer to Question #146110 in Trigonometry for PRITAM

Question

TanA × Tan2A = 1

Prove sin3A + cos3A = 1

Answer 1

$\tan x * \tan 2x = 1\\ \dfrac{\sin x}{\cos x} * \dfrac{\sin 2x}{\cos 2x} = 1\\ \ \\ \dfrac{\sin x}{\cos x} * \dfrac{2\sin x *cosx}{\cos^2 x-\sin^2x} = 1\\ \ \\ \ \dfrac{2\sin^2 x}{\cos^2 x-\sin^2x} = 1\\ 2\sin^2 x = \cos^2 x - \sin^2 x\\ therefore \ \cos^2 x = 3\sin^2 x\\ \cos x = \pm \sqrt3\sin x\\ \ \\ \ \\ \sin 3x +\cos 3x = \\ = \sin x*\cos 2x +\sin 2x * \cos x + \\ \cos x *\cos 2x -\sin x *sin2x = \\ = \sin x *\cos^2x -\sin^3x + 2\sin x*\cos^2x + \\ +\cos^3 x -\sin x*\cos^2x - 2 \sin^2x* \cos x = \\ = \cos^3x-\sin^3x+2\sin x\cos^2x-2\sin^2 x\cos x = \\ = (\cos x-\sin x)(\cos^2x+\sin x\cos x +\sin^2 x) +\\ +2\sin x\cos x(\cos x-\sin x) = (\cos x-\sin x)(1+ 3\sin x\cos x) = \\ = \cos x-\sin x +3\sin x\cos^2 x -3\sin^2 x\cos x = \\ 1) \sqrt3\sin x -\sin x + 3\sin^3 x -3\sqrt3\sin^3x = \\ = -\sqrt3\sin x(3\sin^2 x -1) +\sin x(3\sin^2x-1) =\\ (3\sin^2x-1)(\sin x-\sqrt3\sin x) \ne1 \\ 2) -\sqrt3\sin x -\sin x + 3\sin^3 x +3\sqrt3\sin^3x = \\ = \sqrt3\sin x(3\sin^2 x -1) +\sin x(3\sin^2x-1) =\\ (3\sin^2x-1)(\sin x+\sqrt3\sin x) \ne1 \\ \ \\ answer: \sin3x+\cos3x \ne 1$