In a triangle. Abc prove by vector method cos2a +cos2b+cos2c
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Answer:
(Cos2A + Cos2B + Cos2C) > -3/2
Step-by-step explanation:
Let say O is Circumcenter of ΔABC
(OA + OB + OC)² > 0
|OA|² = |OB|² = |OC|² = R²
(OA + OB + OC)² > 0
=> |OA|² + |OB|² + |OC|² + 2| OA.OB + OB.OC + OA.OC| > 0
=> R² + R² + R² + 2R² (Cos2A + Cos2B + Cos2C) > 0
=> 3R² + 2R² (Cos2A + Cos2B + Cos2C) > 0
=> 2R² (Cos2A + Cos2B + Cos2C) > -3R²
Dividing by 2R² both sides
=> (Cos2A + Cos2B + Cos2C) > -3/2
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