Bài tập 43 trang 214 SGK Toán 10 NC
Bài tập 43 trang 214 SGK Toán 10 NC
Dùng công thức biến đổi tích thành tổng, chứng minh:
\(\begin{array}{*{20}{l}}
{a)\cos {{75}^0}\cos {{15}^0} = \sin {{75}^0}\sin {{15}^0} = \frac{1}{4}}\\
{b)\cos {{75}^0}\sin {{15}^0} = \frac{{2 - \sqrt 3 }}{4}}\\
{c)\sin {{75}^0}\cos {{15}^0} = \frac{{2 + \sqrt 3 }}{4}}\\
\begin{array}{l}
d)\cos \alpha \sin \left( {\beta - \gamma } \right) + \cos \beta \sin \left( {\gamma - \alpha } \right)\\
+ \cos \gamma \sin \left( {\alpha - \beta } \right) = 0,\forall \alpha ,\beta ,\gamma
\end{array}
\end{array}\)
a)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos {75^0}\cos {15^0}\\
= \frac{1}{2}\left( {\cos \left( {{{75}^0} - {{15}^0}} \right) + \cos \left( {{{75}^0} + {{15}^0}} \right)} \right)\\
= \frac{1}{2}\left( {\cos {{60}^0} + \cos {{90}^0}} \right) = \frac{1}{4}
\end{array}\\
\begin{array}{l}
\sin {75^0}\sin {15^0}\\
= \frac{1}{2}\left( {\cos \left( {{{75}^0} - {{15}^0}} \right) - \cos \left( {{{75}^0} + {{15}^0}} \right)} \right)\\
= \frac{1}{2}\left( {\cos {{60}^0} - \cos {{90}^0}} \right) = \frac{1}{4}
\end{array}
\end{array}\)
Vậy \(\cos {75^0}\cos {15^0} = \sin {75^0}\sin {15^0} = \frac{1}{4}\)
b)
\(\begin{array}{l}
\cos {75^0}\sin {15^0} = \sin {15^0}\cos {75^0}\\
= \frac{1}{2}\left[ {\sin \left( {{{15}^0} - {{75}^0}} \right) + \sin \left( {{{15}^0} + {{75}^0}} \right)} \right]\\
= \frac{1}{2}\left[ {\sin \left( { - {{60}^0}} \right) + \sin {{90}^0}} \right]\\
= \frac{{2 - \sqrt 3 }}{4}
\end{array}\)
c)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin {75^0}\cos {15^0}\\
= \frac{1}{2}\left[ {\sin \left( {{{75}^0} - {{15}^0}} \right) + \sin \left( {{{75}^0} + {{15}^0}} \right)} \right]
\end{array}\\
{ = \frac{1}{2}\left[ {\sin {{60}^0} + \sin {{90}^0}} \right]}\\
{ = \frac{{2 + \sqrt 3 }}{4}}
\end{array}\)
d)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos \alpha \sin \left( {\beta - \gamma } \right)\\
= \frac{1}{2}\left[ {\sin \left( {\alpha + \beta - \gamma } \right) - \sin \left( {\alpha - \beta + \gamma } \right)} \right]
\end{array}\\
\begin{array}{l}
\cos \beta \sin \left( {\gamma - \alpha } \right)\\
= \frac{1}{2}\left[ {\sin \left( {\beta + \gamma - \alpha } \right) - \sin \left( {\beta - \gamma + \alpha } \right)} \right]
\end{array}\\
\begin{array}{l}
\cos \gamma \sin \left( {\alpha - \beta } \right)\\
= \frac{1}{2}\left[ {\sin \left( {\gamma + \alpha - \beta } \right) - \sin \left( {\gamma - \alpha + \beta } \right)} \right]
\end{array}\\
\begin{array}{l}
\Rightarrow \cos \alpha \sin \left( {\beta - \gamma } \right) + \cos \beta \sin \left( {\gamma - \alpha } \right)\\
+ \cos \gamma \sin \left( {\alpha - \beta } \right) = 0,\forall \alpha ,\beta ,\gamma
\end{array}
\end{array}\)
-- Mod Toán 10
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