Cho A, B, C là 3 góc của tam giác. CMR:
sin ( A + 2B + C) = -sinBcos A = sin B sin C - cos B cos Ccos A + cos B + cos C = 1 + 4 sin \(\frac{A}{2}\)sin \(\frac{B}{2}\)sin \(\frac{C}{2}\)sin2A + sin2B + sin2C = 2 cos A cos B cos C
Những câu hỏi liên quan
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) \(sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}=\frac{1}{8}\)
30 tháng 4 2016 lúc 12:57
cho A , B , C là 3 góc của tam giác ABC . chứng minh rằng : a) sin2A + sin2B + sin2C = 4sinAsinBsinC ; b) cosA + cosB + cosC = 1 = 4sin\(\frac{A}{2}\)sin\(\frac{B}{2}\)sin\(\frac{C}{2}\) ; c) cos2A + cos2B + cos2C = 1 - 2cosAcosBcosC
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
Đọc tiếp
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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cho tam giác ABC .chứng minh
\(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}+sin\frac{B}{2}cos\frac{C}{2}cos\frac{A}{2}+sin\frac{C}{2}cos\frac{A}{2}cos\frac{B}{2}=sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}+tan\frac{A}{2}tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}\)
Tự chứng minh từng cái này rồi suy ra cái đó nhé b.
Ta có: \(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}-sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}\)
Tương tự ta suy ra:
\(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}sin\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+3sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\left(1\right)\)
Tiếp theo chứng minh:
\(2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{cosA+cosB+cosC-1}{2}\left(2\right)\)
\(sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}\left(3\right)\)
\(tan\frac{A}{2}tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}=1\left(4\right)\)
Từ (1), (2), (3), (4) suy được điều phải chứng minh
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trinh le na
cho bạn 4 năm nữa cũng chưa hiểu đâu
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27 tháng 4 2022 lúc 1:49
1. cos 2a + cos 2b - 2 cos(a+b) cos( a-b)
2. cos2a + sin2b 1
3. cos a2 + sin b2 1
4. cos2 a + sin2 a 1
5. cos 2a cos2 a - 2 sin 2a
6. sin 2a - 2 sin a. cos a.
7. sin 2a cos2 a - sin2 a
8. sin 2a - sin 2b 2 sin ( a+b) cos ( a - b)
9. sin 2a - sin 2b 2 cos( a+b) sin ( a - b)
10. cos a2 + sin a2 1
Câu số mấy đúng?
Đọc tiếp
1. cos 2a + cos 2b = - 2 cos(a+b) cos( a-b)
2. cos2a + sin2b = 1
3. cos a2 + sin b2= 1
4. cos2 a + sin2 a = 1
5. cos 2a = cos2 a - 2 sin 2a
6. sin 2a = - 2 sin a. cos a.
7. sin 2a = cos2 a - sin2 a
8. sin 2a - sin 2b= 2 sin ( a+b) cos ( a - b)
9. sin 2a - sin 2b= 2 cos( a+b) sin ( a - b)
10. cos a2 + sin a2 = 1
Câu số mấy đúng?
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:
a) \(\frac{cos\left(a-b\right)}{sin\left(a+b\right)}=\frac{cota.cotb+1}{cota.cotb-1}\)
b) sin(a+b).sin(a-b)=\(sin^2a-sin^2b=cos^2a-cos^2b\)
c) cos(a+b).cos(a-b)=\(cos^2a-sin^2b=cos^2b-sin^2a\)
\(\frac{cos\left(a-b\right)}{sin\left(a+b\right)}=\frac{cosa.cosb+sina.sinb}{sina.cosb+cosa.sinb}=\frac{\frac{cosa.cosb}{sina.sinb}+1}{\frac{sina.cosb}{sina.sinb}+\frac{cosa.sinb}{sina.sinb}}=\frac{cota.cotb+1}{cota+cotb}\)
Bạn ghi đề ko đúng
\(sin\left(a+b\right)sin\left(a-b\right)=\frac{1}{2}\left[cos2b-cos2a\right]\)
\(=\frac{1}{2}\left[1-2sin^2b-1+2sin^2a\right]\)
\(=sin^2a-sin^2b\)
\(=1-cos^2a-1+cos^2b=cos^2b-cos^2a\)
Câu này bạn cũng ghi đề ko đúng
\(cos\left(a+b\right)cos\left(a-b\right)=\frac{1}{2}\left[cos2a+cos2b\right]\)
\(=\frac{1}{2}\left[2cos^2a-1+1-2sin^2b\right]=cos^2a-sin^2b\)
\(=1-sin^2a-1+cos^2b=cos^2b-sin^2a\)
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