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Bài tập 51 trang 216 SGK Toán 10 NC

Bài tập 51 trang 216 SGK Toán 10 NC

Chứng minh rằng nếu \(\alpha  + \beta  + \gamma  = \pi \) thì:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
a)\sin \alpha  + \sin \beta  + \sin \gamma \\
 = 4\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}
\end{array}\\
\begin{array}{l}
b)\cos \alpha  + \cos \beta  + \cos \gamma \\
 = 1 + 4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}
\end{array}\\
\begin{array}{l}
c)\sin 2\alpha  + \sin 2\beta  + \sin 2\gamma \\
 = 4\sin \alpha \sin \beta \sin \gamma 
\end{array}\\
\begin{array}{l}
d){\cos ^2}\alpha  + {\cos ^2}\beta  + {\cos ^2}\gamma \\
 = 1 - 2\cos \alpha \cos \beta \cos \gamma 
\end{array}
\end{array}\)

ADSENSE

Hướng dẫn giải chi tiết

a)

\(\begin{array}{l}
\sin \alpha  + \sin \beta  + \sin \gamma \\
 = \sin \alpha  + 2\sin \frac{{\beta  + \gamma }}{2}\cos \frac{{\beta  - \gamma }}{2}\\
 = \sin \alpha  + 2\sin \frac{{\pi  - \alpha }}{2}\cos \frac{{\beta  - \gamma }}{2}\\
 = 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} + 2\cos \frac{\alpha }{2}\cos \frac{{\beta  - \gamma }}{2}\\
 = 2\cos \frac{\alpha }{2}\left( {\sin \frac{\alpha }{2} + \cos \frac{{\beta  - \gamma }}{2}} \right)\\
 = 2\cos \frac{\alpha }{2}\left[ {\sin \frac{{\pi  - \left( {\beta  + \gamma } \right)}}{2} + \cos \frac{{\beta  - \gamma }}{2}} \right]\\
 = 2\cos \frac{\alpha }{2}\left( {\cos \frac{{\beta  + \gamma }}{2} + \cos \frac{{\beta  - \gamma }}{2}} \right)\\
 = 4\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}
\end{array}\)

b)

\(\begin{array}{l}
\cos \alpha  + \cos \beta  + \cos \gamma \\
 = 2\cos \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2} + 1 - 2{\sin ^2}\frac{\gamma }{2}\\
 = 2\cos \left( {\frac{\pi }{2} - \frac{\gamma }{2}} \right)\cos \frac{{\alpha  - \beta }}{2} + 1 - 2{\sin ^2}\frac{\gamma }{2}\\
 = 1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha  - \beta }}{2} - \sin \frac{\gamma }{2}} \right)\\
 = 1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha  - \beta }}{2} - \cos \frac{{\alpha  + \beta }}{2}} \right)\\
 = 1 + 4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}
\end{array}\)

c)

\(\begin{array}{l}
\sin 2\alpha  + \sin 2\beta  + \sin 2\gamma \\
 = 2\sin \left( {\alpha  + \beta } \right)\cos \left( {\alpha  - \beta } \right) + 2\sin \gamma \cos \gamma \\
 = 2\sin \gamma \left[ {\cos \left( {\alpha  - \beta } \right) - \cos \left( {\alpha  + \beta } \right)} \right]\\
 = 4\sin \alpha \sin \beta \sin \gamma 
\end{array}\)

d)

\(\begin{array}{l}
{\cos ^2}\alpha  + {\cos ^2}\beta  + {\cos ^2}\gamma \\
 = \frac{{1 + \cos 2\alpha }}{2} + \frac{{1 + \cos 2\beta }}{2} + {\cos ^2}\gamma \\
 = 1 + \frac{1}{2}\left( {\cos 2\alpha  + \cos 2\beta } \right) + {\cos ^2}\gamma \\
 = 1 + \cos \left( {\alpha  + \beta } \right)\cos \left( {\alpha  - \beta } \right) + {\cos ^2}\gamma \\
 = 1 + \cos \gamma \left[ {\cos \gamma  - \cos \left( {\alpha  - \beta } \right)} \right]\\
 = 1 - \cos \gamma \left[ {\cos \left( {\alpha  + \beta } \right) + \cos \left( {\alpha  - \beta } \right)} \right]\\
 = 1 - 2\cos \alpha \cos \beta \cos \gamma 
\end{array}\)

-- Mod Toán 10 HỌC247

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