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b + c = 2a
⇔ \(\dfrac{b+c}{2R}=\dfrac{2a}{2R}\) (1) với R là bán kính đường tròn ngoại tiếp
Theo định lí sin \(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=2R\)
nên (1) ⇔ sinB + sinC = 2sinA
Chọn B
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)\(VT=sinA+sinB+sinC=2sin\frac{A+B}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)(đpcm)
Sai đề:
Thử với \(A=B=C=60^0\) thay vào ta được:
\(-\dfrac{3}{2}=-1+\dfrac{1}{8}\) (vô lí)
TL:
sinA+sinB+sinC=1-cosA+cosB+cosC => Tam giác ABC Vuông tại A
Vế trái = sinA + sinB + sinC
= 2sin(A + B)/2.cos(A - B)/2 + 2sinC/2.cosC/2
= 2cosC/2.cos(A - B)/2 + 2sinC/2.cosC/2
= 2cosC/2[cos(A - B)/2 + sinC/2]
=2.cosC/2.[cos(A - B)/2 + cos(A + B)/2]
= 4.cosC/2.cosB/2.cosA/2
Vế phải = 1 - cosA + cosB + cosC
= 2sin²A/2 + 2cos(B + C)/2.cos(B - C)/2
= 2.sinA/2[sinA/2 + cos(B - C)/2] (vì cos(B + C)/2 = sinA/2)
= 2.sinA/2[cos(B + C)/2 + cos(B - C)/2
= 4.sinA/2.cosB/2.cosC/2
Vậy sinA + sinB + sinC = 1 - cosA + cosB + cosC
<=> cosA/2.cosB/2.cosC/2 = sinA/2.cosB/2.cosC/2
<=> cosB/2.cosC/2(sinA/2 - cosA/2) = 0
mà cosB/2 ≠ 0 và cosC/2 ≠ 0
=> sinA/2 = cosA/2
<=> A/2 = 45o
<=> A = 90o
tam giác ABC vuông tại A
\(\cos 4A+\cos 4B+\cos 4C=(\cos 4A+\cos 4B)+\cos 4C\)
\(=2\cos (2A+2B)\cos (2A-2B)+2\cos ^22C-1\)
\(=2\cos (2\pi -2C)\cos (2A-2B)+2\cos ^22C-1\)
\(=2\cos 2C\cos (2A-2B)+2\cos ^2C-1\)
\(=2\cos 2C[\cos (2A-2B)+\cos 2C]-1\)
\(=2\cos 2C[\cos (2A-2B)+\cos (2A+2B)]-1\)
\(=2\cos 2C.2\cos 2A\cos (-2B)-1\)
\(=-4\cos 2A\cos 2B\cos 2C-1\)
Theo đl sin có:
\(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\Rightarrow b=a\dfrac{sinB}{sinA};c=\dfrac{sinC}{sinA}.a\)
Mà `b+c=2a`
\(\Rightarrow a\dfrac{sinB}{sinA}+a\dfrac{sinC}{sinA}=2a\\ \Rightarrow\dfrac{sinB}{sinA}+\dfrac{sinC}{sinA}=2\\ \Leftrightarrow sinB+sinC=2sinA\)
Chọn B
3/
\(cos4A+cos4B+cos4C=2cos\left(2A+2B\right).cos\left(2A-2B\right)+2cos^22C-1\)
\(=2cos2C.cos\left(2A-2B\right)+2cos^22C-1\)
\(=2cos2C\left(cos\left(2A-2B\right)+cos2C\right)-1\)
\(=2cos2C\left(cos\left(2A-2B\right)+cos\left(2A+2B\right)\right)-1\)
\(=4cos2A.cos2B.cos2C-1\Rightarrow\left\{{}\begin{matrix}a=-1\\b=4\end{matrix}\right.\)
4/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+\frac{1}{2}+\frac{1}{2}cos2C\)
\(=\frac{3}{2}+\frac{1}{2}\left(cos2A+cos2B+cos2C\right)\)
\(=\frac{3}{2}+\frac{1}{2}\left[2cos\left(A+B\right).cos\left(A-B\right)+2cos^2C-1\right]\)
\(=1+\frac{1}{2}\left(-2cosC.cos\left(A-B\right)+2cos^2C\right)\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left(cos\left(A-B\right)+cos\left(A+B\right)\right)\)
\(=1-2cosA.cosB.cosC\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-2\end{matrix}\right.\)
1/ \(sinA+sinB+sin2\frac{C}{2}=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2cos\frac{A+B}{2}.cos\frac{C}{2}=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)
\(=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\Rightarrow\left\{{}\begin{matrix}a=0\\b=4\end{matrix}\right.\)
2/ \(sin4A+sin4B+sin4C=2sin\left(2A+2B\right)cos\left(2A-2B\right)+2sin2C.cos2C\)
\(=-2sin2C.cos\left(2A-2B\right)+2sin2C.cos2C\)
\(\)\(=2sin2C\left(cos2C-cos\left(2A-2B\right)\right)\)
\(=-4sin2C.sin\left(C+A-B\right)sin\left(C-A+B\right)\)
\(=-4sin2A.sin2B.sin2C\Rightarrow\left\{{}\begin{matrix}a=0\\b=-4\end{matrix}\right.\)