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I am having difficulty in proving the following trigonometric equality. It represents the conservation of energy in my physics context. The two variable $\theta_i$ and $\theta_t$ depends on each other by Snell's Law of refraction. But I am assure (by 3Dplot made by Mathematica) the equality should hold even when the two variables are considered independent. $$\frac{tan^2(\theta_i-\theta_t)}{tan^2(\theta_i+\theta_t)} + \frac{sin(2\theta_i)sin(2\theta_t)}{sin^2(\theta_i+\theta_t)cos^2(\theta_i-\theta_t)} \equiv 1$$

I have tried expending all terms into basic $sins$ and $coses$, but it turns out to be extremely messy and error prone. May anyone suggest how to do this?

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Start with the left-hand side (clearly...). You have $$ \frac{\tan^2{(\theta_i-\theta_t)}}{\tan^2{(\theta_i+\theta_t)}} = \frac{\sin^2{(\theta_i-\theta_t)}\cos^2{(\theta_i+\theta_t)}}{\cos^2{(\theta_i-\theta_t)}\sin^2{(\theta_i+\theta_t)}}, $$ by expanding the tangents as sine/cosine. Now the denominators are the same, so we only need to work with the numerators, and want to show that $$\sin^2{(\theta_i-\theta_t)}\cos^2{(\theta_i+\theta_t)}+\sin{2\theta_1}\sin{2\theta_2}=\cos^2{(\theta_i-\theta_t)}\sin^2{(\theta_i+\theta_t)}$$ At this point the sensible thing to do is to use the difference of two squares factorisation on $$ \cos^2{(\theta_i-\theta_t)}\sin^2{(\theta_i+\theta_t)} - \sin^2{(\theta_i-\theta_t)}\cos^2{(\theta_i+\theta_t)} \\ = \left(\cos{(\theta_i-\theta_t)}\sin{(\theta_i+\theta_t)} - \sin{(\theta_i-\theta_t)}\cos{(\theta_i+\theta_t)} \right) \times \\ \left( \cos{(\theta_i-\theta_t)}\sin{(\theta_i+\theta_t)} + \sin{(\theta_i-\theta_t)}\cos{(\theta_i+\theta_t)} \right)$$ At this point you should go "Oh, hang on, I recognise the sine addition formulae there!" Indeed, the first bracket is $$ \cos{(\theta_i-\theta_t)}\sin{(\theta_i+\theta_t)} - \sin{(\theta_i-\theta_t)}\cos{(\theta_i+\theta_t)} = \sin{((\theta_i+\theta_t)-(\theta_i-\theta_t))} = \sin{2\theta_t}, $$ and similarly the second bracket is $\sin{((\theta_i+\theta_t)+(\theta_i-\theta_t))}=\sin{2\theta_i}$, which gives the result.

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  • $\begingroup$ Oh good! I am really doing it the bad way. I should have notice that square. $\endgroup$ – taper Apr 6 '15 at 13:18
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$$1-\frac{\tan^2(A-B)}{\tan^2(A+B)}$$ $$=1-\frac{\sin^2(A-B)\cos^2(A+B)}{\sin^2(A+B)\cos^2(A-B)}$$

$$=\frac{\sin^2(A+B)\cos^2(A-B)-\sin^2(A-B)\cos^2(A+B)}{\sin^2(A+B)\cos^2(A-B)}$$

Now, $\sin^2(A+B)\cos^2(A-B)-\sin^2(A-B)\cos^2(A+B)$

$=\{\sin(A+B)\cos(A-B)-\sin(A-B)\cos(A+B)\}\cdot\{\sin(A+B)\cos(A-B)+\sin(A-B)\cos(A+B)\}$

$=\sin\{(A+B)-(A-B)\}\cdot\sin\{(A+B)+(A-B)\}$

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  • $\begingroup$ Thanks for your answer too! $\endgroup$ – taper Apr 6 '15 at 13:20

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