1. CMR cos^2.(a-b) - sin^2.(a+b) = cos2a.cos2b
2. CMR nếu tam giá ABC tm sinA=\(\dfrac{c\text{os}B+c\text{os}C}{sinB+sinC}\) thì tg ABC vuông
Những câu hỏi liên quan
cho tam giác abc. cmr sin^3a*cos(b-c)+sin^3b*cos(c-a)+sin^3c*cos(a-b)=sina*sinb*sinc
cho tam giác abc. cmr sin^3a*cos(b-c0+sin^3b*cos(c-a)+sin^3c*cos(a-b)=sina*sinb*sinc
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. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
Cho tam giác ABC. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
Cho tam giác ABC. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
Cho tam giác ABC. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
chứng minh tam giác ABC đều
a) sin2A+sin2B+sin2C=sinA+sinB+sinC
b) sin6A + sin6B + sin 6C = 0
c) sin A + sinB + sinC = \(cos\frac{A}{2}+cos\frac{B}{2}+cos\frac{C}{2}\)
d) \(sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}=\frac{1}{8}\)
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' = \(cos^2\dfrac{B-C}{4}-2\left(A-1\right)\ge0\)
\(\Rightarrow A-1\le\dfrac{1}{2}\Leftrightarrow A\le\dfrac{3}{2}\)
Đúng 2
Bình luận (0)