Bài 17 trang 214 SBT đại số 10

Chứng minh rằng

LG a

\(\dfrac{{\sqrt {1 + \cos \alpha }  + \sqrt {1 - \cos \alpha } }}{{\sqrt {1 + \cos \alpha }  - \sqrt {1 - \cos \alpha } }} \) \(= \cot (\dfrac{\alpha }{2} + \dfrac{\pi }{4})\)     \((\pi  < \alpha  < 2\pi )\);

Lời giải chi tiết:

\(\sqrt {1 + \cos \alpha }  \) \(= \sqrt {1 + 2{{\cos }^2}\dfrac{\alpha }{2} - 1}  = \sqrt {2{{\cos }^2}\dfrac{\alpha }{2}} \) \(=  - \sqrt 2 \cos \dfrac{\alpha }{2}(do\dfrac{\pi }{2} < \dfrac{\alpha }{2} < \pi )\)

\(\sqrt {1 - \cos \alpha }  \) \( = \sqrt {1 - \left( {1 - 2{{\sin }^2}\dfrac{\alpha }{2}} \right)}  = \sqrt {2{{\sin }^2}\dfrac{\alpha }{2}} \) \(= \sqrt 2 \sin \dfrac{\alpha }{2}\)

Suy ra

\(\dfrac{{\sqrt {1 + \cos \alpha }  + \sqrt {1 - \cos \alpha } }}{{\sqrt {1 + \cos \alpha }  - \sqrt {1 - \cos \alpha } }} \) \(= \dfrac{{ - \sqrt 2 \cos \dfrac{\alpha }{2} + \sqrt 2 \sin \dfrac{\alpha }{2}}}{{ - \sqrt 2 \cos \dfrac{\alpha }{2} - \sqrt 2 \sin \dfrac{\alpha }{2}}}\)

\( = \dfrac{{\cos \dfrac{\alpha }{2} - \sin \dfrac{\alpha }{2}}}{{\cos \dfrac{\alpha }{2} + \sin \dfrac{\alpha }{2}}} = \dfrac{{1 - \tan \dfrac{\alpha }{2}}}{{1 + \tan \dfrac{\alpha }{2}}} \) \(= \dfrac{{\tan \dfrac{\pi }{4} - \tan \dfrac{\alpha }{2}}}{{1 + \tan \dfrac{\pi }{4}.\tan \dfrac{\alpha }{2}}}\) \(= \tan (\dfrac{\pi }{4} - \dfrac{\alpha }{2})\) \( = \tan \left[ {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)} \right]\)

\( = \cot (\dfrac{\alpha }{2} + \dfrac{\pi }{4})\)

LG b

\(\dfrac{{\cos 4a\tan 2a - \sin 4a}}{{\cos 4a\cot 2a + \sin 4a}} =  - {\tan ^2}2a\);

Lời giải chi tiết:

\( = \dfrac{{\cos 4a\tan 2a - \sin 4a}}{{\cos 4a\cot 2a + \sin 4a}} \)

\(\begin{array}{l} = \dfrac{{\cos 4a.\dfrac{{\sin 2a}}{{\cos 2a}} - \sin 4a}}{{\cos 4a.\dfrac{{\cos 2a}}{{\sin 2a}} + \sin 4a}}\\ = \dfrac{{\cos 4a\sin 2a - \sin 4a\cos 2a}}{{\cos 2a}}:\dfrac{{\cos 4a\cos 2a + \sin 4a\sin 2a}}{{\sin 2a}}\\ = \dfrac{{\cos 4a\sin 2a - \sin 4a\cos 2a}}{{\cos 2a}}.\dfrac{{\sin 2a}}{{\cos 4a\cos 2a + \sin 4a\sin 2a}}\end{array}\)

\(= \dfrac{{\cos 4a\sin 2a - \sin 4a\cos 2a}}{{\cos 4a\cos 2a + \sin 4a\sin 2a}}.\tan 2a\)

=\(\dfrac{{ - \sin 2a}}{{\cos 2a}}\tan 2a =  - {\tan ^2}2a\).

LG c

\(1 + 2\cos 7a = \dfrac{{\sin 10,5a}}{{\sin 3,5a}}\);

Lời giải chi tiết:

\(\dfrac{{\sin 10,5a}}{{\sin 3,5a}} = \dfrac{{\sin (7 + 3,5a)}}{{\sin 3,5a}} \) \(= \dfrac{{\sin 7a\cos 3,5a + \cos 7a\sin 3,5a}}{{\sin 3,5a}}\)

=\(\dfrac{{\sin 3,5a(2{{\cos }^2}3,5a + \cos 7a)}}{{\sin 3,5a}}\)

=\((2{\cos ^2}3,5a - 1) + 1 + cos7a\)

=\(2cos7a + 1.\)

LG d

\(\dfrac{{\tan 3a}}{{\tan a}} = \dfrac{{3 - {{\tan }^2}a}}{{1 - 3{{\tan }^2}a}}\).

Lời giải chi tiết:

\(\dfrac{{\tan (a + 2a)}}{{\tan a}} = \dfrac{{\tan a + \tan 2a}}{{\tan a(1 - {\mathop{\rm tanatan}\nolimits} 2a}} \) \(= \dfrac{{\tan a + \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}}}{{\tan a(1 - \dfrac{{2{{\tan }^2}a}}{{1 - {{\tan }^2}a}})}}\)

=\(\dfrac{{3 - {{\tan }^2}a}}{{1 - 3{{\tan }^2}a}}\)