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