Bài 111131

Question

Chứng minh rằng trong tam giác $ABC$, ta có:
a)   $\sin A+ \sin B+ \sin C = 4\cos \frac{A}{2} \cos \frac{B}{2}\cos \frac{C}{2}   $
b)   $\cos A + \cos B + \cos C=1+ 4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}   $
c)   $\sin 2A+ \sin 2B+ \sin 2C = 4 \sin A \sin B \sin C$
d)   $\cos^2 A+\cos^2 B+ \cos^2 C+2\cos A \cos B \cos C =1$
e)   $\sin^2 A +\sin^2 B +\sin^2 C = 2(1+\cos A \cos B \cos C) $

score 0 · Answer 1

Trong $\Delta ABC$ ta có $\frac{B+C}{2}=\frac{\pi}{2}-\frac{A}{2}$ nên
                              $\sin \frac{B+C}{2}=\cos \frac{A}{2}, \cos \frac{B+C}{2}=\sin \frac{A}{2} .    $
a) $\sin A+ \sin B+ \sin C    = 2\sin \frac{A}{2} \cos \frac{A}{2} + 2\sin \frac{B+C}{2} \cos \frac{B-C}{2}    $
                                            $=2\cos \frac{A}{2} (\cos \frac{B+C}{2}=\cos \frac{B-C}{2} ) $
                                                $=4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}   $ (đpcm)
b) $\cos A+ \cos B +\cos C   =1-2\sin^2 \frac{A}{2}+ 2\cos \frac{B+C}{2} \cos \frac{B-C}{2}    $
      $= 1-2\sin \frac{A}{2} \sin \frac{A}{2}+2\sin \frac{A}{2}\cos \frac{B-C}{2} $
                                                $= 1+ 2\sin \frac{A}{2} (\cos \frac{B-C}{2} -\cos \frac{B+C}{2} ) $
                                                $=1+4\sin \frac{A}{2}\sin \frac{B}{2} \sin \frac{c}{2}    (đpcm)$
c) $\sin 2A +\sin 2B+ \sin 2C = 2\sin A\cos A+2\sin (B+C) \cos (B-C)$
                                               $= 2\sin A [ \cos (B-C)-\cos (B+C) ]$
                                                   $=4\sin A \sin B \sin C$

d) Dùng công thức hạ bậc, thay vào vế trái ta được:
     $\cos^2 A+\frac{1}{2}(1+\cos 2B)+ \frac{1}{2}(1+\cos 2C)= $
        $=1+\cos ^2 A +\frac{1}{2} (\cos 2B+ \cos 2C)=1+\cos ^2 A+ \cos (B+C) \cos(B-C)$
            $=1+\cos A \cos A-\cos A \cos(B-C)=1-\cos A[\cos (B+C)+\cos(B-C)] $
   Từ đó suy ra đpcm.
e) Tương tự như câu d).

Bài 111131

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Bài 111131

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