Cho A, B, C là 3 góc trong tam giác. Chứng minh rằng:1, sin A + sin B - sin C = 4sin\(\dfrac{A}{2}\) sin \(\dfrac{B}{2}\)sin \(\dfrac{C}{2}\)2, \(\dfrac{sinA+sinB-sinC}{cosA+cosB-cosC+1}=tan\dfrac{A}{2}tan\dfrac{B}{2}tan\dfrac{C}{2}\) (ΔABC nhọn)3, \(\dfrac{cosA+cosB+cosC+3}{sinA+sinB+sinC}=tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}\)GIÚP MÌNH...
Đọc tiếp
Cho A, B, C là 3 góc trong tam giác. Chứng minh rằng:
1, sin A + sin B - sin C = 4sin\(\dfrac{A}{2}\) sin \(\dfrac{B}{2}\)sin \(\dfrac{C}{2}\)
2, \(\dfrac{sinA+sinB-sinC}{cosA+cosB-cosC+1}=tan\dfrac{A}{2}tan\dfrac{B}{2}tan\dfrac{C}{2}\) (ΔABC nhọn)
3, \(\dfrac{cosA+cosB+cosC+3}{sinA+sinB+sinC}=tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}\)
GIÚP MÌNH VỚI!!!
1.
\(sinA+sinB-sinC=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-sin\left(A+B\right)\)
\(=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-2sin\dfrac{A+B}{2}.cos\dfrac{A+B}{2}\)
\(=2sin\dfrac{A+B}{2}.\left(cos\dfrac{A-B}{2}-cos\dfrac{A+B}{2}\right)\)
\(=2sin\dfrac{A+B}{2}.2sin\dfrac{A}{2}.sin\dfrac{B}{2}\)
\(=4sin\dfrac{A}{2}.sin\dfrac{B}{2}.cos\dfrac{C}{2}\)
Sao t lại đc như này v, ai check hộ phát