<s>[INST] Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees I most humbly beseech help for this question.
If $A+B+C=180$ degrees, then prove
$$
\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)
$$
I am not sure what trig identity I should use to begin this problem. 
 [/INST] $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin(B+C)\sin(B-C)$
 either using the identity $\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$
or $\sin^2B-\sin^2C=\frac{1}{2}(2\sin^2B-2\sin^2C)=\frac{1}{2}(1-\cos2B-(1-\cos2C))=\frac{1}{2}(\cos2C-\cos2B)=-\frac{1}{2}2\sin(B+C)\sin(C-B)=\sin(B+C)\sin(B-C)$
as $\sin(-x)=-\sin(x)$
Now, $\sin(B+C)=\sin(180^\circ-A)=\sin{A}$
So, $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin{A} \sin(B-C)$
$=\sin{A}(\sin{A}+\sin(B-C))$
$=\sin{A}(\sin(B+C)+\sin(B-C))$ replacing $\sin{A}$ with $\sin(B+C)$
$=\sin{A}(2\sin{B}\cos{C})$
$=2\sin{A}\sin{B}\cos{C}$
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