Question

Cho A, B, C là 3 góc 1 tam giác. Chứng minh

a) \(cos2A+cos2B+cos2C=-1-4cosA.cosB.cosC\)

b) \(sin2A+sin2B+sin2C=4.sinA.sinB.sinC\)

Answer 1

\(cos2A+cos2B+cos2C=2cos\left(A+B\right).cos\left(A-B\right)+2cos^2C-1\)

\(=-2cosC.cos\left(A-B\right)+2cos^2C-1\)

\(=-2cosC\left[cos\left(A-B\right)-cosC\right]-1\)

\(=-2cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]-1\)

\(=-4cosC.cosA.cosB-1\)

\(sin2A+sin2B+sin2C=2sin\left(A+B\right)cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)

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

\(=-4sinC.sinA.sin\left(-B\right)=4sinA.sinB.sinC\)