Cho ΔABC, CMR: \(sin\dfrac{A}{2}+sin\dfrac{B}{2}+sin\dfrac{C}{2}\le\dfrac{3}{2}\).
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
Cho tam giác ABC; AB = c; AC = b; BC = a; đường phân giác AD. Chứng minh:
1) \(\sin\dfrac{A}{2}\le\dfrac{a}{b+c}\)
2) \(\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{S}< 2\)
3) \(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\ge6\)
4) \(\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le\dfrac{1}{8}\)
5) \(\dfrac{1}{\sin^2\dfrac{A}{2}}+\dfrac{1}{\sin^2\dfrac{B}{2}}+\dfrac{1}{\sin^2\dfrac{C}{2}}\ge12\)
1)
Kẻ phân giác AD,BK vuông góc với AD
sin A/2=sinBAD
xét tam giác AKB vuông tại K,có:
sinBAD=BK/AB (1)
xét tam giác BKD vuông tại K,có
BK<=BD thay vào (1):
sinBAD<=BD/AB(2)
lại có:BD/CD=AB/AC
=>BD/(BD+CD)=AB/(AB+AC)
=>BD/BC=AB/(AB+AC)
=>BD=(AB*BC)/(AB+AC) thay vào (2)
sinBAD<=[(AB*BC)/(AB+AC)]/AB
= BC/(AB + AC)
=>ĐPCM
Cho tam giác ABC, chứng minh rằng:
a) \(Sin\dfrac{A}{2}+Sin\dfrac{B}{2}+Sin\dfrac{C}{2}\le\dfrac{3}{2}\)
b) \(SinA+SinB+SinC\le\dfrac{3\sqrt{3}}{2}\)
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 có 3 cạnh có độ dài là a, b, c.
Chứng minh rằng: a) \(\sin\dfrac{a}{2}\le\dfrac{a}{\sqrt{bc}}\)
b) \(\sin\dfrac{a}{2}\cdot\sin\dfrac{b}{2}\cdot\sin\dfrac{c}{2}\le\dfrac{1}{8}\)
c) \(\sin\dfrac{a}{2}\cdot\sin\dfrac{b}{2}\cdot\sin\dfrac{c}{2}=\dfrac{1}{8}\) khi tam giác đã cho là tam giác đều.
Chứng minh rằng với mọi tam giác ABC ta có:
a) \(SinA+SinB+SinC\le Cos\dfrac{A}{2}+Cos\dfrac{B}{2}+Cos\dfrac{C}{2}\)
b) \(CosA.CosB.CosC\le Sin\dfrac{A}{2}.Sin\dfrac{B}{2}.Sin\dfrac{C}{2}\)
Cho tam giác ABC có BC = a ; CA = b ; AB = c. Chứng minh rằng:
a) \(sin\dfrac{A}{2}\)≤\(\dfrac{a}{b+c}\)
b) \(\sin\dfrac{A}{2}.\sin\dfrac{B}{2}.\sin\dfrac{C}{2}\) ≤ \(\dfrac{1}{8}\)
a, Vẽ phân giác AD của góc BAC
Kẻ BH\(\perp\)AD tại H ; CK\(\perp AD\) tại K
Dễ thấy \(sin\widehat{A_1}=sin\widehat{A_2}=sin\dfrac{A}{2}=\dfrac{BH}{AB}=\dfrac{CK}{AC}=\dfrac{BH+CK}{AB+AC}\le\)\(\le\dfrac{BD+CD}{b+c}=\dfrac{a}{b+c}\)
b, Tượng tự \(sin\dfrac{B}{2}\le\dfrac{b}{a+c};sin\dfrac{C}{2}\le\dfrac{c}{a+b}\)
Mặt khác \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\)
\(\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\le\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)
2. CM:
a) \(-1\le\dfrac{2\sin x+\cos x}{\sin x-\cos x+3}\le\dfrac{5}{7}\)
b) \(\dfrac{2}{11}\le\dfrac{2\sin x+\cos x+2}{2\cos x-\sin x+4}\le2\)
a.
Đặt \(y=\dfrac{2sinx+cosx}{sinx-cosx+3}\)
\(\Leftrightarrow y.sinx-y.cosx+3y=2sinx+cosx\)
\(\Leftrightarrow\left(2-y\right)sinx+\left(y+1\right)cosx=3y\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(2-y\right)^2+\left(y+1\right)^2\ge9y^2\)
\(\Leftrightarrow7y^2+2y-5\le0\)
\(\Leftrightarrow-1\le y\le\dfrac{5}{7}\) (đpcm)
b.
Hoàn toàn tương tự câu a:
Đặt \(y=\dfrac{2sinx+cosx+2}{2cosx-sinx+4}\)
\(\Leftrightarrow2y.cosx-y.sinx+4y=2sinx+cosx+2\)
\(\Leftrightarrow\left(y+2\right)sinx+\left(1-2y\right)cosx=4y-2\)
Theo đk có nghiệm pt lượng giác bậc nhất:
\(\left(y+2\right)^2+\left(1-2y\right)^2\ge\left(4y-2\right)^2\)
\(\Leftrightarrow11y^2-16y-1\le0\)
\(\Leftrightarrow\dfrac{8-5\sqrt{3}}{11}\le y\le\dfrac{8+5\sqrt{3}}{11}\)
Đề bài chắc sai, em kiểm tra lại số liệu đề câu b nhé
1. CM:
\(\dfrac{1}{2}\le\dfrac{\sin x+2\cos x+3}{2\sin x\cos x+3}\le2\)
2. Giải PT:
a) \(\dfrac{1}{\cos x}=4\sin x+6\cos x\)
b) \(\sin^3\left(x-\dfrac{\pi}{4}\right)=\sqrt{2}\sin x\)
c) \(\dfrac{1}{\cos x}+\dfrac{1}{\sin2x}=\dfrac{2}{\sin4x}\)
1.
Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng
2.a
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)
\(\Leftrightarrow1+tan^2x=4tanx+6\)
\(\Leftrightarrow tan^2x-4tanx-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)
2b.
Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)
\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow sin^3t=sint+cost\)
\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)
\(\Leftrightarrow sint.cos^2t+cost=0\)
\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
2c.
ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{4}\)
\(\dfrac{4sinx.cos2x}{sin4x}+\dfrac{2cos2x}{sin4x}=\dfrac{2}{sin4x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)
\(\Leftrightarrow sinx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\cos2x-sinx=0\end{matrix}\right.\)
\(\Leftrightarrow1-2sin^2x-sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)
\(\sin^2\dfrac{A}{2}+\sin^2\dfrac{B}{2}+\sin^2\dfrac{C}{2}\)\(=\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\)
Sai đề: \(sin^2\dfrac{A}{2}+sin^2\dfrac{B}{2}+sin^2\dfrac{C}{2}=1-2sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\)
\(sin^2\dfrac{A}{2}+sin^2\dfrac{B}{2}+sin^2\dfrac{C}{2}\)
\(=1-\dfrac{cosA+cosB}{2}+sin^2\dfrac{C}{2}\)
\(=1-cos\dfrac{A+B}{2}.cos\dfrac{A-B}{2}+sin\dfrac{C}{2}.cos\dfrac{A+B}{2}\)
\(=1-sin\dfrac{C}{2}.cos\dfrac{A-B}{2}+sin\dfrac{C}{2}.cos\dfrac{A+B}{2}\)
\(=1+sin\dfrac{C}{2}\left(cos\dfrac{A+B}{2}-cos\dfrac{A-B}{2}\right)\)
\(=1-2sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\)