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Given the relation

$$ \tan x = \tan y \;\cos i, $$

what is the simplest expression (not involving inverse trigonometric functions) for

$$ \sin2(x-z). $$

in terms of $y$, $i$ and $z$ only?

So far, I obtained rather lengthy expressions from the doubling and difference formulas for the sine function in conjunction with \begin{align} \cos x = \frac{\cos y}{\sqrt{\cos^2\!y + \sin^2\!y\cos^2\!i}} ,\\ \sin x = \frac{\sin y\;\cos i}{\sqrt{\cos^2\!y + \sin^2\!y\cos^2\!i}}, \end{align}
but I'm hoping for something better. In particular, I would like a formula that trivially and obviously obtains $\sin2(y-z)$ in case of $\cos i=1$ and $\sin2(-y-z)$ for $\cos i=-1$

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Apply $\sin(A-B)$ in $$\sin(2x-2z)$$

Now use https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution to find $$\sin2x,\cos2x$$ in terms of $y,i$

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  • $\begingroup$ That is essentially what I've got already. (only that I used cos and sin instead of tan). $\endgroup$
    – Walter
    Oct 17 '19 at 18:11
  • $\begingroup$ @Walter, then simplify $\endgroup$ Oct 17 '19 at 18:12
  • $\begingroup$ Exactly. But I couldn't. At least not in such a way that the result meets my requirement (see edited post). $\endgroup$
    – Walter
    Oct 17 '19 at 18:14

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