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Câu hỏi: Rút gọn biểu thức
Phương pháp giải:
Áp dụng các công thức:
\(\begin{array}{l}
+ )\cos2\alpha = 1 - 2{\sin ^2}\alpha = 2{\cos ^2}\alpha - 1.\\
+ )\tan\alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}.\\
+ )\tan\alpha .\cot\alpha = 1.
\end{array}\)
Lời giải chi tiết:
\(\eqalign{ & {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }} \cr& = \frac{{2\sin 2\alpha - \sin \left( {2.2\alpha } \right)}}{{2\sin 2\alpha + \sin \left({2.2\alpha } \right)}}\cr&= {{2\sin 2\alpha - 2\sin 2\alpha. Cos2\alpha } \over {2\sin 2\alpha + 2\sin 2\alpha. Cos2\alpha }} \cr & = \frac{{2\sin 2\alpha \left({1 - \cos 2\alpha } \right)}}{{2\sin 2\alpha \left({1 + \cos 2\alpha } \right)}}\cr &= {{1 - \cos 2\alpha } \over {1 + \cos 2\alpha }} = \frac{{1 - \left({1 - 2{{\sin }^2}\alpha } \right)}}{{1 + \left({2{{\cos }^2}\alpha - 1} \right)}}\cr&= {{2{{\sin }^2}\alpha } \over {2{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}\cr&={\left({\frac{{\sin \alpha }}{{\cos \alpha }}} \right)^2}=\tan^2\alpha.\cr} \)
Lời giải chi tiết:
\(\eqalign{& \tan \alpha \left({{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha\right) \cr&= {{\sin \alpha } \over {\cos \alpha }}\left({{1 + {{\cos }^2}\alpha - {{\sin }^2}\alpha } \over {\sin \alpha }}\right) \cr & = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha + {{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\sin \alpha }}\cr &= {{\sin \alpha } \over {\cos \alpha }}.{{2{{\cos }^2}\alpha } \over {\sin \alpha }} = 2\cos \alpha. \cr} \)
Lời giải chi tiết:
\(\displaystyle {{\sin ({\pi \over 4} - \alpha) + \cos ({\pi \over 4} - \alpha)} \over {\sin ({\pi \over 4} - \alpha) - \cos ({\pi \over 4} - \alpha)}}\)
\(\begin{array}{l}
= \dfrac{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\frac{{\sin \left({\frac{\pi }{4} - \alpha } \right)}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)}} + 1} \right]}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\frac{{\sin \left({\frac{\pi }{4} - \alpha } \right)}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)}} - 1} \right]}}\\
= \dfrac{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) + 1} \right]}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) - 1} \right]}}\\
= \dfrac{{\tan \left({\frac{\pi }{4} - \alpha } \right) + 1}}{{\tan \left({\frac{\pi }{4} - \alpha } \right) - 1}}\\
= \left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) + 1} \right]:\left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) - 1} \right]\\
= \left({\frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }} + 1} \right)
:\left({\frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }} - 1} \right)\\
= \left({\frac{{1 - \tan \alpha }}{{1 + \tan \alpha }} + 1} \right):\left({\frac{{1 - \tan \alpha }}{{1 + \tan \alpha }} - 1} \right)\\
= \frac{{1 - \tan \alpha + 1 + \tan \alpha }}{{1 + \tan \alpha }}:\frac{{1 - \tan \alpha - 1 - \tan \alpha }}{{1 + \tan \alpha }}\\
= \frac{2}{{1 + \tan \alpha }}:\frac{{ - 2\tan \alpha }}{{1 + \tan \alpha }}\\
= \frac{2}{{1 + \tan \alpha }}.\frac{{1 + \tan \alpha }}{{ - 2\tan \alpha }}\\
= - \frac{1}{{\tan \alpha }} = - \cot \alpha
\end{array}\)
Cách khác:
c) Cách $1: \frac{\sin \left(\frac{\pi}{4}-\alpha\right)+\cos \left(\frac{\pi}{4}-\alpha\right)}{\sin \left(\frac{\pi}{4}-\alpha\right)-\cos \left(\frac{\pi}{4}-\alpha\right)}$
$=\frac{\left(\sin \frac{\pi}{4} \cdot \cos \alpha-\cos \frac{\pi}{4} \cdot \sin \alpha\right)+\left(\cos \frac{\pi}{4} \cdot \cos \alpha+\sin \frac{\pi}{4} \cdot \sin \alpha\right)}{\left(\sin \frac{\pi}{4} \cdot \cos \alpha-\cos \frac{\pi}{4} \cdot \sin \alpha\right)-\left(\cos \frac{\pi}{4} \cdot \cos \alpha+\sin \frac{\pi}{4} \cdot \sin \alpha\right)}$
$=\frac{\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha-\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)+\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha+\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)}{\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha-\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)-\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha+\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)}$
$=\frac{\sqrt{2} \cdot \cos \alpha}{-\sqrt{2} \cdot \sin \alpha}=-\cot \alpha$
Cách 2
$\frac{\sin \left(\frac{\pi}{4}-\alpha\right)+\cos \left(\frac{\pi}{4}-\alpha\right)}{\sin \left(\frac{\pi}{4}-\alpha\right)-\cos \left(\frac{\pi}{4}-\alpha\right)}=\frac{\tan \left(\frac{\pi}{4}-\alpha\right)+1}{\tan \left(\frac{\pi}{4}-\alpha\right)-1}$
$=-\frac{\tan \left(\frac{\pi}{4}-\alpha\right)+1}{1-1 \cdot \tan \left(\frac{\pi}{4}-\alpha\right)}$
$=-\frac{\tan \left(\frac{\pi}{4}-\alpha\right)+\tan \frac{\pi}{4}}{1-\tan \left(\frac{\pi}{4}-\alpha\right) \cdot \tan \frac{\pi}{4}}$
$=-\tan \left(\frac{\pi}{4}-\alpha+\frac{\pi}{4}\right)=-\tan \left(\frac{\pi}{2}-\alpha\right)=-\cot \alpha$
$$
\begin{array}{l}
\text { Cách 3: } \frac{\sin \left(\frac{\pi}{4}-\alpha\right)+\cos \left(\frac{\pi}{4}-\alpha\right)}{\sin \left(\frac{\pi}{4}-\alpha\right)-\cos \left(\frac{\pi}{4}-\alpha\right)} \\
=\frac{\frac{1}{\sqrt{2}} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)+\frac{1}{\sqrt{2}} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)}{\frac{1}{\sqrt{2}} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)-\frac{1}{\sqrt{2}} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)} \\
\end{array}
$$
(Nhân cả tử và mẫu với $\frac{1}{\sqrt{2}}$).
$$
=\frac{\sin \frac{\pi}{4} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)+\cos \frac{\pi}{4} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)}{\cos \frac{\pi}{4} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)-\sin \frac{\pi}{4} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)}
$$
$=\frac{\cos \left(\frac{\pi}{4}-\alpha-\frac{\pi}{4}\right)}{\sin \left(\frac{\pi}{4}-\alpha-\frac{\pi}{4}\right)}=\frac{\cos (-\alpha)}{\sin (-\alpha)}$
$=\frac{\cos \alpha}{-\sin \alpha}=-\cot \alpha$
Lời giải chi tiết:
\(\displaystyle {{\sin 5\alpha - \sin 3\alpha } \over {2\cos 4\alpha }} \) \(\displaystyle = {{2\cos {{5\alpha + 3\alpha } \over 2}\sin {{5\alpha - 3\alpha } \over 2}} \over {2\cos 4\alpha }} \) \(\displaystyle = \frac{{2\cos 4\alpha \sin \alpha }}{{2\cos 4\alpha }}\)
\(\displaystyle = \sin \alpha \)
Câu a
\(\displaystyle {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }}\)Phương pháp giải:
Áp dụng các công thức:
\(\begin{array}{l}
+ )\cos2\alpha = 1 - 2{\sin ^2}\alpha = 2{\cos ^2}\alpha - 1.\\
+ )\tan\alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}.\\
+ )\tan\alpha .\cot\alpha = 1.
\end{array}\)
Lời giải chi tiết:
\(\eqalign{ & {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }} \cr& = \frac{{2\sin 2\alpha - \sin \left( {2.2\alpha } \right)}}{{2\sin 2\alpha + \sin \left({2.2\alpha } \right)}}\cr&= {{2\sin 2\alpha - 2\sin 2\alpha. Cos2\alpha } \over {2\sin 2\alpha + 2\sin 2\alpha. Cos2\alpha }} \cr & = \frac{{2\sin 2\alpha \left({1 - \cos 2\alpha } \right)}}{{2\sin 2\alpha \left({1 + \cos 2\alpha } \right)}}\cr &= {{1 - \cos 2\alpha } \over {1 + \cos 2\alpha }} = \frac{{1 - \left({1 - 2{{\sin }^2}\alpha } \right)}}{{1 + \left({2{{\cos }^2}\alpha - 1} \right)}}\cr&= {{2{{\sin }^2}\alpha } \over {2{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}\cr&={\left({\frac{{\sin \alpha }}{{\cos \alpha }}} \right)^2}=\tan^2\alpha.\cr} \)
Câu b
\(\tan \alpha ({{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha)\)Lời giải chi tiết:
\(\eqalign{& \tan \alpha \left({{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha\right) \cr&= {{\sin \alpha } \over {\cos \alpha }}\left({{1 + {{\cos }^2}\alpha - {{\sin }^2}\alpha } \over {\sin \alpha }}\right) \cr & = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha + {{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\sin \alpha }}\cr &= {{\sin \alpha } \over {\cos \alpha }}.{{2{{\cos }^2}\alpha } \over {\sin \alpha }} = 2\cos \alpha. \cr} \)
Câu c
\(\displaystyle {{\sin ({\pi \over 4} - \alpha) + \cos ({\pi \over 4} - \alpha)} \over {\sin ({\pi \over 4} - \alpha) - \cos ({\pi \over 4} - \alpha)}}\)Lời giải chi tiết:
\(\displaystyle {{\sin ({\pi \over 4} - \alpha) + \cos ({\pi \over 4} - \alpha)} \over {\sin ({\pi \over 4} - \alpha) - \cos ({\pi \over 4} - \alpha)}}\)
\(\begin{array}{l}
= \dfrac{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\frac{{\sin \left({\frac{\pi }{4} - \alpha } \right)}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)}} + 1} \right]}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\frac{{\sin \left({\frac{\pi }{4} - \alpha } \right)}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)}} - 1} \right]}}\\
= \dfrac{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) + 1} \right]}}{{\cos \left({\frac{\pi }{4} - \alpha } \right)\left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) - 1} \right]}}\\
= \dfrac{{\tan \left({\frac{\pi }{4} - \alpha } \right) + 1}}{{\tan \left({\frac{\pi }{4} - \alpha } \right) - 1}}\\
= \left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) + 1} \right]:\left[ {\tan \left({\frac{\pi }{4} - \alpha } \right) - 1} \right]\\
= \left({\frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }} + 1} \right)
:\left({\frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }} - 1} \right)\\
= \left({\frac{{1 - \tan \alpha }}{{1 + \tan \alpha }} + 1} \right):\left({\frac{{1 - \tan \alpha }}{{1 + \tan \alpha }} - 1} \right)\\
= \frac{{1 - \tan \alpha + 1 + \tan \alpha }}{{1 + \tan \alpha }}:\frac{{1 - \tan \alpha - 1 - \tan \alpha }}{{1 + \tan \alpha }}\\
= \frac{2}{{1 + \tan \alpha }}:\frac{{ - 2\tan \alpha }}{{1 + \tan \alpha }}\\
= \frac{2}{{1 + \tan \alpha }}.\frac{{1 + \tan \alpha }}{{ - 2\tan \alpha }}\\
= - \frac{1}{{\tan \alpha }} = - \cot \alpha
\end{array}\)
Cách khác:
c) Cách $1: \frac{\sin \left(\frac{\pi}{4}-\alpha\right)+\cos \left(\frac{\pi}{4}-\alpha\right)}{\sin \left(\frac{\pi}{4}-\alpha\right)-\cos \left(\frac{\pi}{4}-\alpha\right)}$
$=\frac{\left(\sin \frac{\pi}{4} \cdot \cos \alpha-\cos \frac{\pi}{4} \cdot \sin \alpha\right)+\left(\cos \frac{\pi}{4} \cdot \cos \alpha+\sin \frac{\pi}{4} \cdot \sin \alpha\right)}{\left(\sin \frac{\pi}{4} \cdot \cos \alpha-\cos \frac{\pi}{4} \cdot \sin \alpha\right)-\left(\cos \frac{\pi}{4} \cdot \cos \alpha+\sin \frac{\pi}{4} \cdot \sin \alpha\right)}$
$=\frac{\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha-\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)+\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha+\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)}{\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha-\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)-\left(\frac{1}{\sqrt{2}} \cdot \cos \alpha+\frac{1}{\sqrt{2}} \cdot \sin \alpha\right)}$
$=\frac{\sqrt{2} \cdot \cos \alpha}{-\sqrt{2} \cdot \sin \alpha}=-\cot \alpha$
Cách 2
$\frac{\sin \left(\frac{\pi}{4}-\alpha\right)+\cos \left(\frac{\pi}{4}-\alpha\right)}{\sin \left(\frac{\pi}{4}-\alpha\right)-\cos \left(\frac{\pi}{4}-\alpha\right)}=\frac{\tan \left(\frac{\pi}{4}-\alpha\right)+1}{\tan \left(\frac{\pi}{4}-\alpha\right)-1}$
$=-\frac{\tan \left(\frac{\pi}{4}-\alpha\right)+1}{1-1 \cdot \tan \left(\frac{\pi}{4}-\alpha\right)}$
$=-\frac{\tan \left(\frac{\pi}{4}-\alpha\right)+\tan \frac{\pi}{4}}{1-\tan \left(\frac{\pi}{4}-\alpha\right) \cdot \tan \frac{\pi}{4}}$
$=-\tan \left(\frac{\pi}{4}-\alpha+\frac{\pi}{4}\right)=-\tan \left(\frac{\pi}{2}-\alpha\right)=-\cot \alpha$
$$
\begin{array}{l}
\text { Cách 3: } \frac{\sin \left(\frac{\pi}{4}-\alpha\right)+\cos \left(\frac{\pi}{4}-\alpha\right)}{\sin \left(\frac{\pi}{4}-\alpha\right)-\cos \left(\frac{\pi}{4}-\alpha\right)} \\
=\frac{\frac{1}{\sqrt{2}} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)+\frac{1}{\sqrt{2}} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)}{\frac{1}{\sqrt{2}} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)-\frac{1}{\sqrt{2}} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)} \\
\end{array}
$$
(Nhân cả tử và mẫu với $\frac{1}{\sqrt{2}}$).
$$
=\frac{\sin \frac{\pi}{4} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)+\cos \frac{\pi}{4} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)}{\cos \frac{\pi}{4} \cdot \sin \left(\frac{\pi}{4}-\alpha\right)-\sin \frac{\pi}{4} \cdot \cos \left(\frac{\pi}{4}-\alpha\right)}
$$
$=\frac{\cos \left(\frac{\pi}{4}-\alpha-\frac{\pi}{4}\right)}{\sin \left(\frac{\pi}{4}-\alpha-\frac{\pi}{4}\right)}=\frac{\cos (-\alpha)}{\sin (-\alpha)}$
$=\frac{\cos \alpha}{-\sin \alpha}=-\cot \alpha$
Câu d
\(\displaystyle {{\sin 5\alpha - \sin 3\alpha } \over {2\cos 4\alpha }}\)Lời giải chi tiết:
\(\displaystyle {{\sin 5\alpha - \sin 3\alpha } \over {2\cos 4\alpha }} \) \(\displaystyle = {{2\cos {{5\alpha + 3\alpha } \over 2}\sin {{5\alpha - 3\alpha } \over 2}} \over {2\cos 4\alpha }} \) \(\displaystyle = \frac{{2\cos 4\alpha \sin \alpha }}{{2\cos 4\alpha }}\)
\(\displaystyle = \sin \alpha \)
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