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Lời giải:
Ta có:
$\sin 2A+\sin 2B=2\sin \frac{2A+2B}{2}\cos \frac{2A-2B}{2}=2\sin (A+B)\cos (A-B)$
$=2\sin (\pi -C)\cos (A-B)=2\sin C\cos (A-B) $
Do đó:
$\sin 2A+\sin 2B+\sin 2C=\sin 2C+2\sin C\cos (A-B)=2\sin C\cos C+2\sin C\cos (A-B)$
$=2\sin C[\cos C+\cos (A-B)]=2\sin C[\cos (\pi -A-B)+\cos (A-B)]$
$=2\sin C[\cos (A-B)-\cos (A+B)]=-2.\sin C[\cos (A+B)-\cos (A-B)]$
$=-2\sin C. (-2).\sin \frac{(A+B)+(A-B)}{2}.\sin \frac{(A+B)-(A-B)}{2}=4\sin C.\sin A.\sin B$
Ta có đpcm.
\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC=2sinC\left[cos\left(A-B\right)+cosC\right]\)
\(=4sinC.cos\left(\frac{A+C-B}{2}\right).cos\left(\frac{A-B-C}{2}\right)\)
\(=4sinC.cos\left(\frac{\pi-2B}{2}\right).cos\left(\frac{2A-\pi}{2}\right)=4sinC.cos\left(\frac{\pi-2B}{2}\right).cos\left(\frac{\pi-2A}{2}\right)\)
\(=4sinC.cos\left(\frac{\pi}{2}-B\right).cos\left(\frac{\pi}{2}-A\right)\)
\(=4sinA.sinB.sinC\)
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/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)
\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(=1-cosC.cos\left(A-B\right)+cos^2C\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=1-2cosC.cosA.cosB\)
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/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
a)\(VT=sinA+sinB+sinC=2sin\frac{A+B}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)(đpcm)
Ta có \(A+B+C=\pi\)
\(\Rightarrow A+B=\pi-C\)
\(\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\)
\(\Rightarrow tanA+tanB=-tanC\left(1-tanA.tanB\right)\)
\(\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\) ( đpcm )
\(cos2A+cos2B+cos2C=2cos\left(A+B\right).cos\left(A-B\right)+2cos^2C-1\)
\(=-2cosC.cos\left(A-B\right)+2cos^2C-1\)
\(=-2cosC\left[cos\left(A-B\right)-cosC\right]-1\)
\(=-2cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]-1\)
\(=-4cosC.cosA.cosB-1\)
\(sin2A+sin2B+sin2C=2sin\left(A+B\right)cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC\left[cos\left(A-B\right)+cosC\right]=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)
\(=-4sinC.sinA.sin\left(-B\right)=4sinA.sinB.sinC\)
Vì A, B, C là ba góc của tam giác nên ta có : A + B + C = π.
⇒ C = π - (A + B); A + B = π - C
a) Ta có: tan A + tan B + tan C = (tan A + tan B) + tan C
= tan (A + B). (1 – tan A.tan B) + tan C
= tan (π – C).(1 – tan A. tan B) + tan C
= -tan C.(1 – tan A. tan B) + tan C
= -tan C + tan A. tan B. tan C + tan C
= tan A. tan B. tan C
b) sin 2A + sin 2B + sin 2C
= 2. sin (A + B). cos (A – B) + 2.sin C. cos C
= 2. sin (π – C). cos (A – B) + 2.sin C. cos (π – (A + B))
= 2.sin C. cos (A – B) - 2.sin C. cos (A + B)
= 2.sin C.[cos (A – B) - cos (A + B)]
= 2.sin C.[-2sinA. sin(- B)]
= 2.sin C. 2.sin A. sin B ( vì sin(- B)= - sinB )
= 4. sin A. sin B. sin C
Tại sao câu b) cái phần sin2A + sin2B lại bằng 2sin(A+B).cos(A-B) vậy ạ