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The equality $ \cos\eta+i\sin\eta = \frac{1+\cos\theta+c\cdot i\sin\theta}{1+\cos\theta-c\cdot i\sin\theta} \tag{1} $ or equivalently $ \frac{1+\cos\eta+\frac1c\cdot i\sin\eta}{1+\cos\eta-\frac1c\cdot i\sin\eta} = \cos\theta+i\sin\theta \tag{1} $ holds precisely if $\tan\frac\eta2 = c\cdot\tan\frac\theta2.\tag{2}$

Does this appear in any published tabulation of trigonometric identities, or is it otherwise "known"?

PS: My original posting mangled $(1)$ so that it said something that didn't make sense and was simpler than what should have been there.

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I don't get it: two complex numbers are equal iff their respective real and imaginary parts are equal , so:

$\cos\theta+i\sin\theta=\cos\eta+ci\sin\eta\Longleftrightarrow$

$ \cos\theta=\cos\eta\Longleftrightarrow \theta=\pm\eta+2k\pi\,\,,\,\,k\in\Bbb Z$

and

$\sin\theta=c\sin\eta$

and choosing $\,k=0\,$ above and the plus sign we get something close to what you want, though I don't understand why you have to take the halved angles.

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    Haste makes waste: The question now says what it _should_ have said.2012-12-02