# A “known” tangent half-angle formula?

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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I mangled the identity somehow; I'll figure out what it should have said and edit accordingly. – Michael Hardy Dec 2 '12 at 0:07
Haste makes waste: The question now says what it should have said. – Michael Hardy Dec 2 '12 at 1:58