Sin^2A+sin^2B+sin^2C =2(1+cosAcosBcosC) giải thích các bước giải giúp mình nhé

\( \sin^2A + \sin^2B + \sin^2C\\ = \dfrac{1 - \cos 2A}{2} + \dfrac{1 - \cos 2B}{2} + \dfrac{1 - \cos 2C}{2}\\ = \dfrac{1}{2}[3 - ( {\cos 2A + \cos 2B + \cos 2C}) ] \\ = \dfrac{1}{2}[ 3-( 2.\cos ( A + B ).\cos (A - B) + 2\cos^2C - 1 ]\\ = \dfrac{1}{2}.[4 - 2( - \cos ( 180^o -A-B )).\cos ( A - B ) - 2\cos ^2C )]\\ = \dfrac{1}{2}[ 4 + 2.cos C .\cos ( A - B ) - 2\cos^2C ]\\ = 2 + \cos C( \cos (A - B )-\cos C)\\ = 2 - 2\cos C\cos\left( 90^o - \dfrac{A - B + C}{2} \right)\cos\left(90^o - \dfrac{A-B-C}{2} \right)\\ = 2 - 2\cos C\cos \dfrac{2B}{2}\cos \left( {B + C} \right)\\= 2 - 2.\cos C\cos B(- \cos A)\\ = 2(1+\cos A\cos B\cos C)\)