Toán Lớp 11: mấy câu này khó lắm hả mn :(( hỏi lần 2 huhu
1) sin5x + cos5x = √2 cos13x
2) cos7x – sin5x – √3 ( cos5x – sin7x)
3) sin8x – cos6x = √3 ( sin6x + cos8x )
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Toán Lớp 11: mấy câu này khó lắm hả mn :(( hỏi lần 2 huhu
1) sin5x + cos5x = √2 cos13x
2) cos7x – sin5x – √3 ( cos5x – sin7x)
3) sin8x – cos6x = √3 ( sin6x + cos8x )
Comments ( 1 )
1,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{\pi }{{72}} + \dfrac{{k\pi }}{9}\\
x = – \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{6}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)
1,\\
\sin 5x + \cos 5x = \sqrt 2 \cos 13x\\
\Leftrightarrow \dfrac{1}{{\sqrt 2 }}.\sin 5x + \dfrac{1}{{\sqrt 2 }}.\cos 5x = \cos 13x\\
\Leftrightarrow \sin 5x.\cos \dfrac{\pi }{4} + \cos 5x.\sin \dfrac{\pi }{4} = \cos 13x\\
\Leftrightarrow \sin \left( {5x + \dfrac{\pi }{4}} \right) = \cos 13x\\
\Leftrightarrow \cos \left[ {\dfrac{\pi }{2} – \left( {5x + \dfrac{\pi }{4}} \right)} \right] = \cos 13x\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{4} – 5x} \right) = \cos 13x\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{4} – 5x = 13x + k2\pi \\
\dfrac{\pi }{4} – 5x = – 13x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
18x = \dfrac{\pi }{4} + k2\pi \\
8x = – \dfrac{\pi }{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{72}} + \dfrac{{k\pi }}{9}\\
x = – \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\cos 7x – \sin 5x = \sqrt 3 \left( {\cos 5x – \sin 7x} \right)\\
\Leftrightarrow \cos 7x + \sqrt 3 \sin 7x = \sqrt 3 \cos 5x + \sin 5x\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 7x + \dfrac{1}{2}\cos 7x = \dfrac{1}{2}\sin 5x + \dfrac{{\sqrt 3 }}{2}\cos 5x\\
\Leftrightarrow \sin 7x.\cos \dfrac{\pi }{6} + \cos 7x.\sin \dfrac{\pi }{6} = \sin 5x.\cos \dfrac{\pi }{3} + \cos 5x.\sin \dfrac{\pi }{3}\\
\Leftrightarrow \sin \left( {7x + \dfrac{\pi }{6}} \right) = \sin \left( {5x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{6} = 5x + \dfrac{\pi }{3} + k2\pi \\
7x + \dfrac{\pi }{6} = \pi – \left( {5x + \dfrac{\pi }{3}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{6} = 5x + \dfrac{\pi }{3} + k2\pi \\
7x + \dfrac{\pi }{6} = \dfrac{{2\pi }}{3} – 5x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
12x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{6}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\sin 8x – \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right)\\
\Leftrightarrow \sin 8x – \sqrt 3 \cos 8x = \sqrt 3 \sin 6x + \cos 6x\\
\Leftrightarrow \dfrac{1}{2}\sin 8x – \dfrac{{\sqrt 3 }}{2}\cos 8x = \dfrac{{\sqrt 3 }}{2}\sin 6x + \dfrac{1}{2}\cos 6x\\
\Leftrightarrow \sin 8x.\cos \dfrac{\pi }{3} – \cos 8x.\sin \dfrac{\pi }{3} = \sin 6x.\cos \dfrac{\pi }{6} + \cos 6x.\sin \dfrac{\pi }{6}\\
\Leftrightarrow \sin \left( {8x – \dfrac{\pi }{3}} \right) = \sin \left( {6x + \dfrac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
8x – \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
8x – \dfrac{\pi }{3} = \pi – \left( {6x + \dfrac{\pi }{6}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
8x – \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
8x – \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} – 6x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
14x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)