Cho tam giác ABC. Tính P = sinA. cos(B + C) + cos A.sin(B + C).
A. P = 0
A. P = 0
B. P = 1
B. P = 1
C. P = - 1
C. P = - 1
D. P = 2
D. P = 2
Những câu hỏi liên quan
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:
1) Sin2A+Sin2B+Sin2C=4SinA.SinB.SinC
2) Cos 2A + Cos 2B + Cos 2C = 4.CosA.CosB.CosC
3) 4R + r = P.( Tan \(\frac{a}{2}\) + Tan \(\frac{b}{2}\) + Tan \(\frac{c}{2}\) )
4) a.Sin(b-c) +b.Sin(c-a) +c.Sin(a-b)=0
Help me
cho tam giác ABC . chứng minh:
a, sin(A+B)sinC. ; cos (A+B)cos-C; tan ( A+B) -tan C
b, sinfrac{A+B}{2}cosfrac{C}{2} ; cosfrac{A+B}{2}sinfrac{C}{2} ; tanfrac{A+B}{2}cotfrac{C}{2}
c, tan A+tanB+tanC tanA.tanB.tanc( tam giác không vuông)
d, sinA+sinB+sinC 4cosfrac{A}{2}cosfrac{B}{2}cosfrac{C}{2}
e, cos A+cosB+cosC 1+4sinfrac{A}{2}sinfrac{B}{2}sinfrac{C}{2}
f, sin2A+sin2B+sin2C 4sinAsinBsinC
g, cos 2A+cos2B+cos2C1-2cosAcosBcosC
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cho tam giác ABC . chứng minh:
a, sin(A+B)=sinC. ; cos (A+B)=cos-C; tan ( A+B)= -tan C
b, \(sin\frac{A+B}{2}=cos\frac{C}{2}\) ; \(cos\frac{A+B}{2}=sin\frac{C}{2}\) ; tan\(\frac{A+B}{2}=cot\frac{C}{2}\)
c, tan A+tanB+tanC= tanA.tanB.tanc( tam giác không vuông)
d, sinA+sinB+sinC= \(4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
e, cos A+cosB+cosC= \(1+4sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\)
f, sin2A+sin2B+sin2C= 4sinAsinBsinC
g, cos 2A+cos2B+cos2C=1-2cosAcosBcosC
\(A+B+C=180^0\Rightarrow A+B=180^0-C\)
\(\Rightarrow sin\left(A+B\right)=sin\left(180^0-C\right)=sinC\)
\(cos\left(A+B\right)=cos\left(180^0-C\right)=-cosC\)
\(tan\left(A+B\right)=tan\left(180^0-C\right)=-tanC\)
b/ \(\frac{A+B+C}{2}=90^0\Rightarrow\frac{A+B}{2}=90^0-\frac{C}{2}\)
\(\Rightarrow sin\frac{A+B}{2}=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)
\(cos\frac{A+B}{2}=cos\left(90^0-\frac{C}{2}\right)=sin\frac{C}{2}\)
\(tan\frac{A+B}{2}=tan\left(90-\frac{C}{2}\right)=cot\frac{C}{2}\)
c/ \(A+B=180^0-C\Rightarrow tan\left(A+B\right)=-tanC\)
\(\Leftrightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\)
\(\Leftrightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
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d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)
\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)
e/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
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f/
\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)
\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)
\(=4sinC.sinA.sinB\)
g/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)
\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(=1-cosC.cos\left(A-B\right)+cos^2C\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=1-2cosC.cosA.cosB\)
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Rut gon bieu thuc:
A=(sin4a+cos4a-1)/(sin6a+cos6a-1)
B=1-sin2a+cos2a
C= 1-sina.cosa.tana
D= cos4a+cos2a.sin2a+sin2a
E=cansin2a.(1+cotx)+cos2a(1+tana)
F= 1+sina/cosa.[1-(1-sina/cosa)2]
Ai giup minh voiii. Minh cam on nhieuu!
B=1-sin2a+cos2a
\(=\sin^2a+\cos^2a-\sin^2a+\cos^2a=2\cos^2a\)
C= 1-sina.cosa.tana
\(=1-\sin a.\cos a.\frac{\sin a}{\cos a}=1-\sin^2a=\cos^2a\)
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Cho tam giác ABC. Chứng minh:
\(\frac{a^2-b^2}{\cos A+\cos B}+\frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}=0\)