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Ta có: A = \(sin\dfrac{A}{2}+sin\dfrac{B}{2}+sin\dfrac{C}{2}=cos\dfrac{B+C}{2}+2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}\)
\(\Leftrightarrow A-2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}-cos^2\dfrac{B+C}{4}+sin^2\dfrac{B+C}{4}=0\)\(\Leftrightarrow A-2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}+2sin^2\dfrac{B+C}{4}-1=0\)
Δ' = \(cos^2\dfrac{B-C}{4}-2\left(A-1\right)\ge0\)
\(\Rightarrow A-1\le\dfrac{1}{2}\Leftrightarrow A\le\dfrac{3}{2}\)
cho tam giác ABC, chứng minh rằng: \(sinA+sinB-sinC=4.sin\frac{A}{2}.sin\frac{B}{2}.cos\frac{C}{2}\)
\(sinA+sinB-sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}-sinC\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}-2sin\frac{C}{2}cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=4cos\frac{C}{2}sin\frac{A}{2}sin\frac{B}{2}\)
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
a) ta có : \(cos^2\left(a-b\right)-sin^2\left(a+b\right)\)
\(=\left(cosa.cosb+sina.sinb\right)^2-\left(sina.cosb+sinb.cosa\right)^2\)
\(=cos^2a.cos^2b+sin^2a.sin^2b-sin^2a.cos^2b-sin^2b.cos^2a\)
\(=cos^2a.cos^2b-sin^2a.cos^2b+sin^2a.sin^2b-sin^2b.cos^2a\)\(=cos^2b\left(cos^2a-sin^2a\right)-sin^2b\left(cos^2a-sin^2a\right)\)
\(=\left(cos^2b-sin^2b\right)\left(cos^2a-sin^2a\right)=cos2a.cos2b\left(đpcm\right)\)
\(0< A;B;C< 180^0\Rightarrow\left\{{}\begin{matrix}sinA>0\\sinB>0\\sinC>0\end{matrix}\right.\)
\(\Rightarrow A=sinA+sinB+sinC>0\)
\(B=sinA.sinB.sinC>0\)
Riêng 2 câu c;d đâu biết \(\alpha\) là góc nào mà xét dấu?
\(sin^4x+cos^4x=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-\frac{1}{2}\left(2sinx.cosx\right)^2\)
\(=1-\frac{1}{2}sin^22x\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=2\end{matrix}\right.\) \(\Rightarrow a+3b+c=?\)
\(\frac{sin\left(A-B\right)}{sinC}=\frac{sin\left(A-B\right).sinC}{sin^2C}=\frac{sin\left(A-B\right).sin\left(A+B\right)}{sin^2C}=\frac{-\frac{1}{2}\left(cos2A-cos2B\right)}{sin^2C}\)
\(=\frac{-\frac{1}{2}\left(1-2sin^2A-1+2sin^2B\right)}{sin^2C}=\frac{sin^2A-sin^2B}{sin^2C}=\frac{\left(\frac{a}{2R}\right)^2-\left(\frac{b}{2R}\right)^2}{\left(\frac{c}{2R}\right)^2}=\frac{a^2-b^2}{c^2}\)
Câu 3:
a/ Đề dị dị, là \(\frac{cosA+cosB}{sinB+sinC}\) hay \(\frac{cosB+cosC}{sinB+sinC}\) bạn?
b/ \(cos\left(B-C\right)-cos\left(B+C\right)=1+cosA\)
\(\Leftrightarrow cos\left(B-C\right)+cosA=1+cosA\)
\(\Leftrightarrow cos\left(B-C\right)=1\)
\(\Rightarrow B=C\Rightarrow\Delta ABC\) cân tại A
Theo định lí sin trong tam giác ta có:
a sin A = 2 R ⇒ a = 2 R . sin A
Tương tự, b = 2 R . sin B ; c = 2 R . sin C
Theo bất đẳng thức tam giác ta có: a + b > c
Do đó, 2Rsin A + 2Rsin B > 2Rsin C ⇒ sin A + sin B > sin C
Tương tự, sin A + sin C > sin B và sin B + sin C > sin A
Vậy D sai.
ĐÁP ÁN D