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 Jul 24th 2017, 08:59 PM #1 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 510 This text book solution looks wrong This is a solution given in a text book but as far as I can tell: $\displaystyle E - V_0 \tan^2(\frac{\pi x}{2a}) \neq E - (E+V_0)\frac{\sin^2(\frac{\pi x}{2a}) }{\cos^2(\frac{\pi x}{2a}) }$ Can anyone see anything I am missing? Attached Thumbnails
 Jul 25th 2017, 05:20 AM #2 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,268 You have miswritten what they wrote - it should be: $\displaystyle E - V_0 \tan^2(\frac{\pi x}{2a}) = \frac {E - (E+V_0)\sin^2(\frac{\pi x}{2a})}{\cos^2(\frac{\pi x}{2a}) }$ The steps are: $\displaystyle E = \frac {E\cos^2( \frac {\pi x}{ 2a})}{\cos^2 (\frac {\pi x}{ 2a})} = \frac {E(1 - \sin^2 (\frac {\pi x}{ 2a}))}{\cos^2 (\frac {\pi x}{ 2a})} = \frac{ E - E\sin^2 ( \frac {\pi x}{ 2a})}{cos^2 (\frac {\pi x}{ 2a})}$
 Jul 25th 2017, 01:15 PM #3 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 510 Ok, thanks. I missed the trig identity substitution and was trying to figure out how he distributed the sin^2 term alone. I knew I had seen the tan^2 term in QM but hadn't previously seen the pedagogical algebraic advantage of the expression.

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