Proving $\cos2A+\cos2B-\cos2C=1-4\sin A\sin B\cos C$ with identities for $\sin2A+\sin2B+\sin2C$ and $\cos2A+\cos2B+\cos2C$ I know the solution to this question, which is very long.

Prove the trigonometric identity $$\cos2A+\cos2B-\cos2C=1-4\sin A\sin B\cos C \tag{$\star$}$$ 
  where the angles are part of $\triangle ABC$

I would like to know:

Is it possible to prove $(\star)$ with the following trigonometric identities?
  $$\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$$
$$\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C$$

My teacher told me such questions can be solved very quickly using these identities instead of the transformation formulas. However I just can figure out how to use them.
 A: Using $$\cos(2x)=2\cos^2(x)-1$$ we have to prove
$$\cos^2(A)+\cos^2(B)+\cos^2(C)=1-2\cos(A)\cos(B)\cos(C)$$ now use the theorem of cosines.
A: I don't agree with you. It's very quickly! 
The first identity:
$$\cos2\alpha+\cos2\beta-\cos2\gamma-1=2\cos(\alpha+\beta)\cos(\alpha-\beta)-2\cos^2\gamma=$$
$$=-2\cos\gamma(\cos(\alpha-\beta)-\cos(\alpha+\beta))=-4\cos\gamma\sin\alpha\sin\beta.$$
The second identity:
$$\sin2\alpha+\sin2\beta+\sin2\gamma=2\sin(\alpha+\beta)\cos(\alpha-\beta)+2\sin\gamma\cos\gamma=$$
$$=2\sin\gamma(\cos(\alpha-\beta)-\cos(\alpha+\beta))=4\sin\gamma\sin\alpha\sin\beta.$$
The third identity:
$$\cos2\alpha+\cos2\beta+\cos2\gamma+1=2\cos(\alpha+\beta)\cos(\alpha-\beta)+2\cos^2\gamma=$$
$$=-2\cos\gamma(\cos(\alpha-\beta)+\cos(\alpha+\beta))=4\cos\gamma\cos\alpha\cos\beta.$$
A: We can derive the identities even without using the other two identities.
$$F=\cos2A+\cos2B-(1+\cos2C) =2\cos(A+B)\cos(A-B)-2\cos^2C$$
Now $\cos(A+B)=\cos(\pi-C)=-\cos C$
$$F=2(-\cos C)\cos(A-B)-2\cos C[-\cos(A+B)]$$
$$=-2\cos C[\cos(A-B)-\cos(A+B)]$$
$$=-2\cos C[?]$$
