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If $A+B+C=\pi$,$$\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2\tag1$$$$\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos \dfrac C2\tag2$$$$\cos A+\cos B+\cos C=4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2+1\tag3$$$$\cos A+\cos B-\cos C=4\cos\dfrac A2\cos\dfrac B2\sin\dfrac C2-1\tag4$$$$\tan A+\tan B+\tan C=\tan A\tan B\tan C\tag5$$$$\cot\dfrac A2+\cot\dfrac B2+\cot\dfrac C2=\cot\dfrac A2\cot\dfrac B2\cot\dfrac C2\tag6$$

Formulae $(1)$ through $(6)$ were given with the condition that $A+B+C=180^{\circ}$. I'm not sure how to arrive at them.

Question: How do you arrive at $(1)$ through $(6)$?

I need a place to start. I am well aware that$$\sin A+\sin B=2\sin\dfrac {A+B}2\cos\dfrac {A-B}2$$And$$\cos A+\cos B=2\cos\dfrac {A+B}{2}\cos\dfrac {A-B}2$$ However, I'm not sure how to get $\sin A\pm\sin B\pm\sin C$. I'm guessing it has something to do with the expansion of $\sin(A+B+C)$.

Note: In your answer, give a hint on where I can begin, then hide the rest of your answer.

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hint for all of them: $\sin(C) = 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$, and also the little identities such as $\sin\left(\frac{A+B}{2}\right) = \cos\left(\frac{C}{2}\right)$. And the for $3)$, use $\cos(C) = 1- 2\sin^2\left(\frac{C}{2}\right)$.

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$$4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=2\cos\frac{A}{2}\left(\cos\frac{B+C}{2}+\cos\frac{B-C}{2}\right)=$$ $$=\cos\frac{A+B+C}{2}+\cos\frac{-A+B+C}{2}+\cos\frac{A+B-C}{2}+\cos\frac{A-B+C}{2}=$$ $$=\sin{A}+\sin{B}+\sin{C}$$

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