Question

Cho A, B, C là 3 góc của tam giác. CMR:

sin ( A + 2B + C) = -sinBcos A = sin B sin C - cos B cos Ccos A + cos B + cos C = 1 + 4 sin \(\frac{A}{2}\)sin \(\frac{B}{2}\)sin \(\frac{C}{2}\)sin²A + sin²B + sin²C = 2 cos A cos B cos C

Answer 1

1) \(sin\left(A+2B+C\right)=sin\left(\pi-B+2B\right)\)

=\(sin\left(\pi+B\right)=sin\left(-B\right)=-sinB\)

2) \(sinBsinC-cosBcosC=-cos\left(B+C\right)\)

\(=-cos\left(\pi-A\right)=cosA\)

Answer 2

4) bạn ơi +2 vào vế phải mới đúng nhé

2+ \(2cosAcosBcosC=\left[cos\left(A+B\right)+cos\left(A-B\right)\right]cosC+2\)

\(=cos\left(\pi-C\right)cosC+cos\left(A-B\right)cos\left(\pi-\left(A+B\right)\right)+2\)

=\(-cos^2C-cos\left(A-B\right)cos\left(A+B\right)+2\)

\(=-cos^2C-\frac{1}{2}\left(cos2A+cos2B\right)+2\)

\(=-cos^2C-\frac{1}{2}\left(2cos^2A-1\right)-\frac{1}{2}\left(2cos^2B-1\right)+2\)

\(=-cos^2C-cos^2A+\frac{1}{2}-cos^2C+\frac{1}{2}+2\)

= sin²C - 1 + sin²A - 1 + sin²C - 1 + 3

= sin²A + sin²B + sin²C