Câu 45 trang 47 SGK Đại số và Giải tích 11 Nâng cao

LG a

$\sin x + \tan {\pi \over 7}\cos x$

Lời giải chi tiết:

Ta có:

$\begin{array}{l} \sin x + \tan \frac{\pi }{7}\cos x\\ = \sin x + \frac{{\sin \frac{\pi }{7}}}{{\cos \frac{\pi }{7}}}.\cos x\\ = \sin x + \frac{{\sin \frac{\pi }{7}\cos x}}{{\cos \frac{\pi }{7}}}\\ = \frac{{\sin x\cos \frac{\pi }{7} + \sin \frac{\pi }{7}\cos x}}{{\cos \frac{\pi }{7}}}\\ = \frac{{\sin \left( {x + \frac{\pi }{7}} \right)}}{{\cos \frac{\pi }{7}}} \end{array}$

$= \frac{1}{{\cos \frac{\pi }{7}}}\sin \left( {x + \frac{\pi }{7}} \right)$

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LG b

$\tan {\pi \over 7}\sin x + \cos x$

Lời giải chi tiết:

$\begin{array}{l} \tan \frac{\pi }{7}\sin x + \cos x\\ = \frac{{\sin \frac{\pi }{7}}}{{\cos \frac{\pi }{7}}}.\sin x + \cos x\\ = \frac{{\sin \frac{\pi }{7}\sin x}}{{\cos \frac{\pi }{7}}} + \cos x\\ = \frac{{\sin \frac{\pi }{7}\sin x + \cos x\cos \frac{\pi }{7}}}{{\cos \frac{\pi }{7}}}\\ = \frac{{\cos \left( {x - \frac{\pi }{7}} \right)}}{{\cos \frac{\pi }{7}}} = \frac{{\cos \left( {\frac{\pi }{7} - x} \right)}}{{\cos \frac{\pi }{7}}}\\ = \frac{{\sin \left( {\frac{\pi }{2} - \frac{\pi }{7} + x} \right)}}{{\cos \frac{\pi }{7}}}\\ = \frac{{\sin \left( {\frac{{5\pi }}{{14}} + x} \right)}}{{\cos \frac{\pi }{7}}}\\ = \frac{1}{{\cos \frac{\pi }{7}}}\sin \left( {x + \frac{{5\pi }}{{14}}} \right) \end{array}$

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