Bài 17 trang 213 SBT đại số 10

Câu hỏi: Chứng minh rằng

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

Câu b​

\(\dfrac{{\cos 4a\tan 2a - \sin 4a}}{{\cos 4a\cot 2a + \sin 4a}} = - {\tan ^2}2a\);
Lời giải chi tiết:
\(= \dfrac{{\cos 4a\tan 2a - \sin 4a}}{{\cos 4a\cot 2a + \sin 4a}} \)
\(\begin{array}{l} = \dfrac{{\cos 4a.\dfrac{{\sin 2a}}{{\cos 2a}} - \sin 4a}}{{\cos 4a.\dfrac{{\cos 2a}}{{\sin 2a}} + \sin 4a}}\\ = \dfrac{{\cos 4a\sin 2a - \sin 4a\cos 2a}}{{\cos 2a}}:\dfrac{{\cos 4a\cos 2a + \sin 4a\sin 2a}}{{\sin 2a}}\\ = \dfrac{{\cos 4a\sin 2a - \sin 4a\cos 2a}}{{\cos 2a}}.\dfrac{{\sin 2a}}{{\cos 4a\cos 2a + \sin 4a\sin 2a}}\end{array}\)
\(= \dfrac{{\cos 4a\sin 2a - \sin 4a\cos 2a}}{{\cos 4a\cos 2a + \sin 4a\sin 2a}}.\tan 2a\)
=\(\dfrac{{ - \sin 2a}}{{\cos 2a}}\tan 2a = - {\tan ^2}2a\).

Câu c​

$\frac{\sin ^{2} 2 a+4 \sin ^{2} a-4}{1-8 \sin ^{2} a-\cos 4 a}=\frac{1}{2} \cot ^{4} a$
Lời giải chi tiết:
$\begin{aligned} \frac{\sin ^{2} 2 a+4 \sin ^{2} a-4}{1-8 \sin ^{2} a-\cos 4 a} &=\frac{4 \sin ^{2} a \cos ^{2} a+4\left(\sin ^{2} a-1\right)}{1-8 \sin ^{2} a-\left(1-2 \sin ^{2} 2 a\right)} \\ &=\frac{4 \cos ^{2} a\left(\sin ^{2} a-1\right)}{8 \sin ^{2} a\left(\cos ^{2} a-1\right)}=\frac{1}{2} \cot ^{4} a \end{aligned}$

Câu d​

\(1 + 2\cos 7a = \dfrac{{\sin 10,5a}}{{\sin 3,5a}}\);
Lời giải chi tiết:
\(\dfrac{{\sin 10,5a}}{{\sin 3,5a}} = \dfrac{{\sin (7 + 3,5a)}}{{\sin 3,5a}} \) \(= \dfrac{{\sin 7a\cos 3,5a + \cos 7a\sin 3,5a}}{{\sin 3,5a}}\)
=\(\dfrac{{\sin 3,5a(2{{\cos }^2}3,5a + \cos 7a)}}{{\sin 3,5a}}\)
=\((2{\cos ^2}3,5a - 1) + 1 + cos7a\)
=\(2cos7a + 1.\)

Câu e​

\(\dfrac{{\tan 3a}}{{\tan a}} = \dfrac{{3 - {{\tan }^2}a}}{{1 - 3{{\tan }^2}a}}\).
Lời giải chi tiết:
\(\dfrac{{\tan (a + 2a)}}{{\tan a}} = \dfrac{{\tan a + \tan 2a}}{{\tan a(1 - {\mathop{\rm tanatan}\nolimits} 2a}} \) \(= \dfrac{{\tan a + \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}}}{{\tan a(1 - \dfrac{{2{{\tan }^2}a}}{{1 - {{\tan }^2}a}})}}\)
=\(\dfrac{{3 - {{\tan }^2}a}}{{1 - 3{{\tan }^2}a}}\)