Cho tam giác ABC, biết \(sin\dfrac{A}{2}.cos^3\dfrac{B}{2}=sin\dfrac{B}{2}.cos^3\dfrac{A}{2}\)
Chứng minh rằng tam giác ABC cân
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
M
1
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
\(\Leftrightarrow sinA=2sinB.cosC\)
\(\Leftrightarrow\dfrac{a}{2R}=2.\dfrac{b}{2R}.\dfrac{a^2+b^2-c^2}{2ab}\)
\(\Leftrightarrow a^2=a^2+b^2-c^2\)
\(\Leftrightarrow b^2=c^2\Leftrightarrow b=c\)
Vậy tam giác ABC cân tại A
HN
6 tháng 5 2017
Đặt \(f\left(A,B,C\right)=cosA+cosB+cosC+\dfrac{1}{sinA}+\dfrac{1}{sinB}+\dfrac{1}{sinC}-2\sqrt{3}-\dfrac{3}{2}\)
Ta có: \(f\left(A,B,C\right)-f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\)
\(=\left(cosB+cosC-2cos\left(\dfrac{B+C}{2}\right)\right)+\left(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\right)\)
\(=2cos\left(\dfrac{B+C}{2}\right)\left(cos\left(\dfrac{B-C}{2}\right)-1\right)+\left(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\right)\left(1\right)\)
Bên cạnh đó ta có:
\(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\ge\dfrac{4}{sinB+sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}=\dfrac{4\left(1-cos\left(\dfrac{B-C}{2}\right)\right)}{sinB+sinC}\)
Do đó \(\left(1\right)\ge2\left(1-cos\left(\dfrac{B-C}{2}\right)\right)\left(\dfrac{2}{sinB+sinC}-cos\left(\dfrac{B+C}{2}\right)\right)\)
\(=\left(1-cos\left(\dfrac{B-C}{2}\right)\right)\left(\dfrac{1-sin\left(\dfrac{B+C}{2}\right)cos\left(\dfrac{B+C}{2}\right)cos\left(\dfrac{B-C}{2}\right)}{sinB+sinC}\right)\ge0\)
\(\Rightarrow f\left(A,B,C\right)\ge f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\)
Giờ ta chỉ cần chứng minh bất đẳng thức đúng trong trường hợp tam giác cân.
Ta có: \(\left\{{}\begin{matrix}B=\dfrac{\pi}{2}-\dfrac{A}{2}\\cosB=cosC=\dfrac{sinA}{2}\\sinB=sinC=\dfrac{cosA}{2}\end{matrix}\right.\)
\(f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)=\left(cosA+2sin\left(\dfrac{A}{2}\right)-\dfrac{3}{2}\right)+\left(\dfrac{1}{sinA}+\dfrac{2}{cos\left(\dfrac{A}{2}\right)}-2\sqrt{3}\right)\)
\(=\dfrac{-2\left(sin\left(\dfrac{A}{2}\right)-1\right)^2}{2}+\dfrac{1+4sin\left(\dfrac{A}{2}\right)-2\sqrt{3}sinA}{sinA}\)
Mà ta có: \(1\ge sin\left(\dfrac{A}{2}+\dfrac{\pi}{3}\right)\)
\(\Rightarrow8sin\left(\dfrac{A}{2}\right)\ge2\sqrt{3}sinA+4sin^2\left(\dfrac{A}{2}\right)\)
\(\Rightarrow1+4sin\left(\dfrac{A}{2}\right)-2\sqrt{3}sinA\ge4sin^2\left(\dfrac{A}{2}\right)-4sin\left(\dfrac{A}{2}\right)+1=\left(2sin\left(\dfrac{A}{2}-1\right)\right)^2\)
Từ đó ta suy ra:
\(f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\ge\left(2sin-1\right)^2\left(\dfrac{1}{sinA}-\dfrac{1}{2}\right)\ge0\)
Vậy bài toán đã được chứng minh. Dấu = xảy ra khi \(A=B=C=\dfrac{\pi}{3}\)
6 tháng 5 2017
Hàm số \(f\left(x\right)=\cos\left(x\right)+\dfrac{1}{\sin\left(x\right)}\) là hàm lồi trên \(\left(0,\pi\right)\)
Do đó theo BĐT Jensen ( trường hợp của Karamata) có:
\(f\left(A\right)+f\left(B\right)+f\left(c\right)\ge3f\left(\dfrac{A+B+C}{3}\right)=3f\left(\dfrac{\pi}{3}\right)=2\sqrt{3}+\dfrac{3}{2}\)
P/s:Tính độ "lầy" của hàm số:
\(f''(x)=-\cos(x)-\frac{1}{\sin(x)}+\frac{2}{(\sin(x))^3}\)
Và cho \(x\in (0,\pi);f''(x)>0\) nếu \(2>(\sin(x))^2(\sin(x)\cos(x)+1)\) là xài dc Jensen :D
18 tháng 7 2022
Vì A+B+C=180^{\circ}A+B+C=180∘ nên V T=\dfrac{\sin ^{3} \dfrac{B}{2}}{\cos \left(\dfrac{180^{\circ}-B}{2}\right)}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\sin \left(\dfrac{180^{\circ}-B}{2}\right)}-\dfrac{\cos \left(180^{\circ}-B\right)}{\sin B} \cdot \tan BVT=cos(2180∘−B)sin32B+sin(2180∘−B)cos32B−sinBcos(180∘−B)⋅tanB.
V T=\dfrac{\sin ^{3} \dfrac{B}{2}}{\cos \left(\dfrac{180^{\circ}-B}{2}\right)}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\sin \left(\dfrac{180^{\circ}-B}{2}\right)}-\dfrac{\cos \left(180^{\circ}-B\right)}{\sin B} \cdot \tan BVT=cos(2180∘−B)sin32B+sin(2180∘−B)cos32B−sinBcos(180∘−B)⋅tanB =\dfrac{\sin ^{3} \dfrac{B}{2}}{\sin \dfrac{B}{2}}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\cos \dfrac{B}{2}}-\dfrac{-\cos B}{\sin B} \cdot \tan B=\sin ^{2} \dfrac{B}{2}+\cos ^{2} \dfrac{B}{2}+1=2=V P=sin2Bsin32B+cos2Bcos32B−sinB−cosB⋅tanB=sin22B+cos22B+1=2=VP
Suy ra điều phải chứng minh.
10 tháng 5 2021
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' = \(cos^2\dfrac{B-C}{4}-2\left(A-1\right)\ge0\)
\(\Rightarrow A-1\le\dfrac{1}{2}\Leftrightarrow A\le\dfrac{3}{2}\)
sinA/2.cos^3(B/2)=sinB/2.cos^3(A/2)
sinA/2.cos(B/2)[ 1 - sin^2B/2]=sinB/2.cos(A/2)[1 -sin^2A/2]
sinA/2.cosB/2 - sinB/2.cosA/2 = 1/2sinA/2.sinB/2[ sinB - sinA]
sin(A-B)/2 = sinA/2.sinB/2 cos(A+B)/2.sin(A-B)/2
sin(A-B)/2[ 1 - sinA/2.sinB/2 cos(A+B)/2] = 0
Vì [1 - sinA/2.sinB/2 cos(A+B)/2] >0
=> sin(A-B)/2 =0
=> A = B