<s>[INST] Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
 [/INST] $$\cos^2B+\cos^2C+2\cos A\cos B\cos C =1+\cos^2B-\sin^2C+2\cos A\cos B\cos C$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $$\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
Again, $\cos(B+C)=\cos(\pi-A)=-\cos A$
Using Werner Formula,
$$2\cos A\cos B\cos C=\cos A[2\cos B\cos C]$$
$$=\cos A[\cos(B-C)+\cos(B+C)]=\cos A[\cos(B-C)-\cos A]$$
I hope the rest should not be too tough to deal with
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