Several trigonometric identities in three variables \alpha \beta \gamma   that satisfy \alpha + \beta + \gamma = \pi   and thus can be thought of as the angles of a triangle, are either outright equivalent or have very similar proofs.

  1. \displaystyle \sin\alpha + \sin\beta + \sin\gamma = 4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}
  2. \sin{2}\alpha + \sin{2}\beta + \sin{2}\gamma = 4\sin\alpha\sin\beta\sin\gamma
  3. \sin{4}\alpha + \sin{4}\beta + \sin{4}\gamma = -4\sin{2}\alpha\sin{2}\beta\sin{2}\gamma
  4. \displaystyle \cos\alpha + \cos\beta + \cos\gamma - 1 = 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}
  5. \cos{2}\alpha + \cos{2}\beta + \cos{2}\gamma + 1 = -4\cos\alpha\cos\beta\cos\gamma
  6. \displaystyle \cos{4}\alpha + \cos{4}\beta + \cos{4}\gamma + 1 = 4\cos{2}\alpha\cos{2}\beta\cos{2}\gamma
  7. \displaystyle \tan\frac{\alpha}{2}\tan\frac{\beta}{2}+\tan\frac{\beta}{2}\tan\frac{\gamma}{2}+\tan\frac{\gamma}{2}\tan\frac{\alpha}{2}=1
  8. \displaystyle \cot\alpha\cot\beta+\cot\beta\cot\gamma+\cot\gamma\cot\alpha=1
  9. \displaystyle \sin^{2}\frac{\alpha}{2} + \sin^{2}\frac{\beta}{2} + \sin^{2}\frac{\gamma}{2} + 2\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2} = 1
  10. \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma + 2\cos\alpha\cos\beta\cos\gamma = 1
  11. \tan\alpha + \tan\beta +\tan\gamma = \tan\alpha\tan\beta\tan\gamma, if none of the angles is right

Let's show the dependencies.

1 \Rightarrow 2

Indeed, if \alpha + \beta + \gamma = \pi   then also

(\pi - 2\alpha) + (\pi - 2\beta) + (\pi - 2\gamma) = \pi

and the assertion follows because

\sin(\pi - x) = \sin(x)  and \cos(\frac{\pi}{2} - x) = \sin(x).

2 \Rightarrow 3

This follows with the same substitution using

\sin(2\pi - x) = -\sin(x)  and \sin(\pi - x) = \sin(x).

4 \Rightarrow 5

This is because

\cos(\pi - x) = -\cos(x)  and \sin(\frac{\pi}{2} - x) = \cos(x).

5 \Rightarrow 6

\cos(2\pi - x) = \cos(x)  and \cos(\pi - x) = -\cos(x).

Now, the first identity is an easy consequence of the following

(*)    \displaystyle \sin\alpha + \sin\beta + \sin\gamma - \sin(\alpha + \beta + \gamma) = 4\sin\frac{(\alpha + \beta)}{2}\sin\frac{(\beta + \gamma)}{2}\sin\frac{(\gamma + \alpha)}{2}

which holds for any \alpha \beta \gamma.   For the proof see One Trigonometric Formula and Its Consequences

The substitution \displaystyle \frac{\pi}{2}-\frac{\alpha}{2}\rightarrow\alpha,  etc. reduces (*) to

(**)  \displaystyle \cos\alpha + \cos\beta + \cos\gamma + \cos(\alpha + \beta + \gamma) = 4\cos\frac{(\alpha + \beta)}{2}\cos\frac{(\beta + \gamma)}{2}\cos\frac{(\gamma + \alpha)}{2}

from which 4 follows.

4 \Leftrightarrow 9

One is transformed into the other with \displaystyle\cos\nu =1-\sin^{2}\frac{\nu}{2}.

5 \Leftrightarrow 10

Use the identity \cos 2\nu =2\cos^{2}\nu - 1.

9 \Leftrightarrow 10

One is transformed into the other with \nu\leftrightarrow \pi -2\nu.

The #7 stands somewhat apart but its proof is rather easy. Note that \displaystyle\tan\frac{\gamma}{2}=\cot\frac{\alpha+\beta}{2}. Given that, rewrite the left-hand side of #7 as

\displaystyle\tan\frac{\alpha}{2}\tan\frac{\beta}{2}+\left(\tan\frac{\beta}{2}+\tan\frac{\alpha}{2}\right)\cot\frac{\alpha+\beta}{2}=\tan\frac{\alpha}{2}\tan\frac{\beta}{2}+\left(\tan\frac{\beta}{2}+\tan\frac{\alpha}{2}\right)\frac{1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}=1.

Since (\pi - 2\alpha) + (\pi - 2\beta) + (\pi - 2\gamma) = \pi and \tan (\frac{\pi}{2}-\delta)=\cot\delta, #8 is equivalent to #7.

7 \Leftrightarrow 8 \Leftrightarrow 11