Bài tập 46 trang 215 SGK Toán 10 NC
Bài tập 46 trang 215 SGK Toán 10 NC
Chứng minh rằng:
\(\begin{array}{*{20}{l}}
\begin{array}{l}
a)\sin 3\alpha = 3\sin \alpha - 4{\sin ^3}\alpha ;\\
\cos 3\alpha = 4{\cos ^3}\alpha - 3\cos \alpha
\end{array}\\
\begin{array}{l}
b)\sin \alpha \sin \left( {\frac{\pi }{3} - \alpha } \right)\sin \left( {\frac{\pi }{3} + \alpha } \right)\\
= \frac{1}{4}\sin 3\alpha
\end{array}\\
\begin{array}{l}
{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \cos \alpha \cos \left( {\frac{\pi }{3} - \alpha } \right)\cos \left( {\frac{\pi }{3} + \alpha } \right)\\
= \frac{1}{4}\cos 3\alpha
\end{array}
\end{array}\)
Ứng dụng: Tính sin 200 sin 400 sin 800 và tan 200 tan 400 tan 800
a)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin 3\alpha = \sin \left( {2\alpha + \alpha } \right)\\
= \sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha
\end{array}\\
{ = 2\sin \alpha {{\cos }^2}\alpha + \sin \alpha \left( {1 - 2{{\sin }^2}\alpha {\mkern 1mu} } \right)}\\
{ = 2\sin \alpha \left( {1 - {{\sin }^2}\alpha } \right) + \sin \alpha \left( {1 - {{\sin }^2}\alpha } \right)}\\
{ = 3\sin \alpha - 4{{\sin }^3}\alpha }\\
\begin{array}{l}
\cos 3\alpha = \cos \left( {2\alpha + \alpha } \right)\\
= \cos 2\alpha \cos \alpha - \sin 2\alpha \sin \alpha
\end{array}\\
{ = \left( {2{{\cos }^2}\alpha - 1} \right)\cos \alpha - 2{{\sin }^2}\alpha \cos \alpha }\\
{ = 2{{\cos }^3}\alpha - \cos \alpha - 2\cos \alpha \left( {1 - {{\cos }^2}\alpha } \right)}\\
{ = 4{{\cos }^3}\alpha - 3\cos \alpha }
\end{array}\)
b)
\(\begin{array}{*{20}{l}}
{\sin \alpha \sin \left( {\frac{\pi }{3} - \alpha } \right)\sin \left( {\frac{\pi }{3} + \alpha } \right)}\\
{ = \sin \alpha .\frac{1}{2}\left( {\cos 2\alpha - \cos \frac{{2\pi }}{3}} \right)}\\
{ = \frac{1}{2}\sin \alpha \left( {1 - 2{{\sin }^2}\alpha + \frac{1}{2}} \right)}\\
{ = \frac{1}{4}\sin \alpha \left( {3 - 4{{\sin }^2}\alpha } \right) = \frac{1}{4}\sin 3\alpha }\\
{\cos \alpha \cos \left( {\frac{\pi }{3} - \alpha } \right)\cos \left( {\frac{\pi }{3} + \alpha } \right)}\\
{ = \cos \alpha .\frac{1}{2}\cos \left( {\cos \alpha + \cos \frac{{2\pi }}{3}} \right)}\\
{ = \frac{1}{2}\cos \alpha \left( {2{{\cos }^2}\alpha - 1 - \frac{1}{2}} \right)}\\
{ = \frac{1}{2}\cos \alpha \left( {4{{\cos }^2}\alpha - 3} \right) = \frac{1}{4}\cos 3\alpha }
\end{array}\)
Ứng dụng:
\(\begin{array}{l}
\sin {20^0}\sin {40^0}\sin {80^0}\\
= \sin {20^0}.\sin \left( {{{69}^0} - {{20}^0}} \right)\sin \left( {{{60}^0} + {{20}^0}} \right)\\
= \frac{1}{4}\sin \left( {{{3.20}^0}} \right) = \frac{1}{4}\sin {60^0} = \frac{{\sqrt 3 }}{8}\\
\cos {20^0}\cos {40^0}\cos {80^0} = \frac{1}{4}\cos {60^0} = \frac{1}{8}\\
\Rightarrow \tan {20^0}\tan {40^0}\tan {80^0} = \sqrt 3
\end{array}\)
-- Mod Toán 10
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