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The process of the Heaviside Cover-up method works because the substitution replicates the results of Drsoccerball's generic substitution method. Heaviside allows you to skip the step of common denominating

They multiplied the 1/2 inside the big brackets.

3. \frac{1- \cos x}{\sin x} = \frac{2 \sin^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \tan \frac{x}{2} Beware that: 1- \cos x = (\sin \frac{x}{2})^2 + (\cos \frac{x}{2})^2 - \left((\cos \frac{x}{2})^2 - (\sin \frac{x}{2})^2 \right ) = 2(\sin \frac{x}{2})^2 and 1- \cos 2x = 2...

2. \cos (4x)= 2(\cos2x)^2 -1 = 2(2 \cos^2x - 1)^2 - 1 = \dots

1. \frac{\sin 2x + 1}{\cos 2x} = \frac{2 \sin x \cos x + \sin^2x + \cos^2x}{\cos^2x - \sin^2x} = \frac{(\sin x + \cos x)^2}{(\cos x + \sin x)(\cos x - \sin x)} \\= \frac{\sin x + \cos x}{\cos x - \sin x}

Re: HSC 2015 4U Marathon Which one? How is that not elementary methods?

Re: HSC 2015 4U Marathon Let x = \frac{1}{u} \lim_{u \rightarrow \infty} \frac{1}{u} \text{ln}\left ( \frac{1}{u} \right ) = \lim_{u \rightarrow \infty} 0 - \frac{\text{ln}(u)}{u} = 0 As u approaches infinity, u increases more rapidly than ln(u), so the fraction ln(u)/u will tend to 0...