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Old Jul 24th 2017, 09:59 PM   #1
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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?
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Old Jul 25th 2017, 06:20 AM   #2
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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:

$\displaystyle E = \frac {E\cos^2( \frac {\pi x}{ 2a})}{\cos^2 (\frac {\pi x}{ 2a})} = \frac {E(1 - \sin^2 (\frac {\pi x}{ 2a}))}{\cos^2 (\frac {\pi x}{ 2a})} = \frac{ E - E\sin^2 ( \frac {\pi x}{ 2a})}{cos^2 (\frac {\pi x}{ 2a})}$
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Old Jul 25th 2017, 02:15 PM   #3
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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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