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Question

Chứng minh trong mọi

$\triangle ABC$ ta luôn có:

$\cos^2A+\cos^2 B+\cos^2 C=1-2 \cos A. \cos B. \cos C$

score 4 · Answer 1

Có:

$cosC=cos(180-A-B)=-cos(A+B)=-(cosA.cosB-sinA.sinB)=sinA.sinB-cosA.cosB$

$VT=cos^2A+cos^2B+cos^2C$

$=cos^2A+cos^2B+(sinA.sinB-cosA.cosB)^2$

$=cos^2A+cos^2B+sin^2A.sin^2B+cos^2A.cos^2B-2sinA.sinB.cosA.cosB$

$=cos^2A+cos^2B+(1-cos^2A)(1-cos^2B)+cos^2A.cos^2B-2sinA.sinB.cosA.cosB$

$=cos^2A+cos^2B+1-cos^2A-cos^2B+2cos^2A.cos^2B-2inA.sinB.cosA.cosB$

$=1-2.cosA.cosB(sinA.sinB-cosA.cosB)$

$=1-2.cosA.cosB.cosC$ (đpcm)

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