Cho tam giác ABC thỏa mãn: $\frac{sin2B+sin2C}{sin2A}=\frac{sin B+sinC}{sinA}$
Chứng minh rằng: $cosB+cosC=1$
Cho tam giác ABC thỏa mãn: $\frac{sin2B+sin2C}{sin2A}=\frac{sin B+sinC}{sinA}$
Chứng minh rằng: $cosB+cosC=1$
Sống là cho, đâu chỉ nhận riêng mình
Cho tam giác ABC thỏa mãn: $\frac{sin2B+sin2C}{sin2A}=\frac{sin B+sinC}{sinA}$
Chứng minh rằng: $cosB+cosC=1$
Ta có: $sin2B+sin2C=2.sin(B+C).cos(B-C)$
$=4.sin\frac{B+C}{2}.cos\frac{B+C}{2}.cos(B-C)$
$sin2A=2.sinA.cosA$
$sinB+sinC=2.sin\frac{B+C}{2}.cos\frac{B-C}{2}$
Do đó $\frac{sin2B+sin2C}{sin2A}=\frac{sin B+sinC}{sinA}$
$\Leftrightarrow cos\frac{B+C}{2}.cos(B-C)=cosA.cos\frac{B-C}{2}$
$\Leftrightarrow cos\frac{B+C}{2}\left ( cos^2\frac{B-C}{2}-sin^2\frac{B-C}{2} \right )=-\left ( cos^2\frac{B+C}{2}-sin^2\frac{B+C}{2} \right ).cos\frac{B-C}{2}$
$\Leftrightarrow cos\frac{B+C}{2}.cos\frac{B-C}{2}\left ( cos\frac{B+C}{2}+cos\frac{B-C}{2} \right )=sin^2\frac{B+C}{2}.cos\frac{B-C}{2}+sin^2\frac{B-C}{2}.cos\frac{B+C}{2}$
$\Leftrightarrow cos\frac{B+C}{2}.cos\frac{B-C}{2}\left ( cos\frac{B+C}{2}+cos\frac{B-C}{2} \right )=\left ( 1-cos^2\frac{B+C}{2} \right ).cos\frac{B-C}{2}+\left ( 1-cos^2\frac{B-C}{2} \right ).cos\frac{B+C}{2}$
$\Leftrightarrow 2cos\frac{B+C}{2}.cos\frac{B-C}{2}\left ( cos\frac{B+C}{2}+cos\frac{B-C}{2} \right )=cos\frac{B+C}{2}+cos\frac{B-C}{2}$
$\Leftrightarrow cosB+cosC=1$
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