Bài 3 trang 231 SBT đại số và giải tích 11

Giả sử A, B, C là ba góc của tam giác ABC, chứng minh rằng:

LG a

\(\frac{{\sin C}}{{\cos A\cos B}} = \tan A + \tan B\)

Lời giải chi tiết:

\(\begin{array}{l}
VT = \frac{{\sin C}}{{\cos A\cos B}}\\
= \frac{{\sin \left( {{{180}^0} - \left( {A + B} \right)} \right)}}{{\cos A\cos B}}\\
= \frac{{\sin \left( {A + B} \right)}}{{\cos A\cos B}}\\
= \frac{{\sin A\cos B + \sin B\cos A}}{{\cos A\cos B}}\\
= \frac{{\sin A\cos B}}{{\cos A\cos B}} + \frac{{\sin B\cos A}}{{\cos A\cos B}}\\
= \frac{{\sin A}}{{\cos A}} + \frac{{\sin B}}{{\cos B}}\\
= \tan A + \tan B = VP
\end{array}\)

LG b

\(\sin A + \sin B + \sin C \) \(= 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\)

Lời giải chi tiết:

\(\begin{array}{l}
VT = \sin A + \sin B + \sin C\\
= 2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} + 2\sin \frac{C}{2}\cos \frac{C}{2}\\
= 2\cos \frac{C}{2}\cos \frac{{A - B}}{2} + 2\sin \frac{C}{2}\cos \frac{C}{2}\\
= 2\cos \frac{C}{2}\left( {\cos \frac{{A - B}}{2} + \sin \frac{C}{2}} \right)\\
= 2\cos \frac{C}{2}\left( {\cos \frac{{A - B}}{2} + \cos \frac{{A + B}}{2}} \right)\\
= 2\cos \frac{C}{2}.2\cos \frac{A}{2}\cos \left( { - \frac{B}{2}} \right)\\
= 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}
\end{array}\)

LG c

\(\frac{{\sin A + \sin B + \sin C}}{{\sin A + \sin B - \sin C}} = \cot \frac{A}{2}\cot \frac{B}{2}\)

Lời giải chi tiết:

Từ câu b ta có:

\(\sin A + \sin B + \sin C \) \(= 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\)

Lại có:

\(\begin{array}{l}
\sin A + \sin B - \sin C\\
= 2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} - 2\sin \frac{C}{2}\cos \frac{C}{2}\\
= 2\cos \frac{C}{2}\cos \frac{{A - B}}{2} - 2\sin \frac{C}{2}\cos \frac{C}{2}\\
= 2\cos \frac{C}{2}\left( {\cos \frac{{A - B}}{2} - \sin \frac{C}{2}} \right)\\
= 2\cos \frac{C}{2}\left( {\cos \frac{{A - B}}{2} - \cos \frac{{A + B}}{2}} \right)\\
= 2\cos \frac{C}{2}.\left[ { - 2\sin \frac{A}{2}\sin \left( { - \frac{B}{2}} \right)} \right]\\
= 4\sin \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}
\end{array}\)

Do đó,

\(\begin{array}{l}
\frac{{\sin A + \sin B + \sin C}}{{\sin A + \sin B - \sin C}} = \frac{{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}}{{4\sin \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}}}\\
= \frac{{\cos \frac{A}{2}\cos \frac{B}{2}}}{{\sin \frac{A}{2}\sin \frac{B}{2}}} = \cot \frac{A}{2}\cot \frac{B}{2}
\end{array}\)