Toán Trong ΔABC, CMR : cos2A + cos2B + cos2C + 1 = -4.cosA.cosB.cosC 06/08/2021 By Remi Trong ΔABC, CMR : cos2A + cos2B + cos2C + 1 = -4.cosA.cosB.cosC
$\begin{array}{l}\cos2A + \cos2B + \cos2C + 1\\ =2\cos(A + B)\cos(A – B) +2\cos^2C – 1 +1\\ = – 2\cos C.\cos(A – B) + 2\cos^2C\\ = -2\cos C[\cos(A – B) – \cos C]\\ = -2\cos C[\cos(A – B) + \cos(A + B)]\\ = -4\cos C.\cos A.\cos B\end{array}$ Trả lời
Giải thích các bước giải: \(\cos 2A+\cos 2B+\cos 2C+1\) \(=2.\cos (A+B).\cos (A-B)+2\cos^{2} C-1+1\) \(=-2\cos C.\cos (A-B)+2\cos^{2} C\) \(=-2\cos C[\cos (A-B)+\cos (A+B)]\) \(=-2\cos C.2\cos A \cos B=-4.\cos A.\cos B.\cos C\) Trả lời
$\begin{array}{l}\cos2A + \cos2B + \cos2C + 1\\ =2\cos(A + B)\cos(A – B) +2\cos^2C – 1 +1\\ = – 2\cos C.\cos(A – B) + 2\cos^2C\\ = -2\cos C[\cos(A – B) – \cos C]\\ = -2\cos C[\cos(A – B) + \cos(A + B)]\\ = -4\cos C.\cos A.\cos B\end{array}$
Giải thích các bước giải:
\(\cos 2A+\cos 2B+\cos 2C+1\)
\(=2.\cos (A+B).\cos (A-B)+2\cos^{2} C-1+1\)
\(=-2\cos C.\cos (A-B)+2\cos^{2} C\)
\(=-2\cos C[\cos (A-B)+\cos (A+B)]\)
\(=-2\cos C.2\cos A \cos B=-4.\cos A.\cos B.\cos C\)