Bài 20 trang 213 SBT đại số 10

Câu hỏi: Chứng minh rằng

Câu a​

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

Câu b​

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

Câu c​

\(\dfrac{{\sqrt 2 - {\mathop{\rm sina}\nolimits} - \cos a}}{{\sin a - \cos a}} = - \tan \left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\);
Lời giải chi tiết:
\(\dfrac{{\sqrt 2 - \sin a - \cos a}}{{\sin a - \cos a}} \)\( = \dfrac{{\sqrt 2 - \left( {\sin a + \cos a} \right)}}{{\sin a - \cos a}}\)
\(= \dfrac{{\sqrt 2 - \sqrt 2 \sin(a + \dfrac{\pi }{4})}}{{\sqrt 2 \sin(a - \dfrac{\pi }{4})}}\)
=\(\dfrac{{1 - \sin (a + \dfrac{\pi }{4})}}{{\sin(a - \dfrac{\pi }{4})}} \) \(= \dfrac{{\sin\dfrac{\pi }{2} - \sin(a + \dfrac{\pi }{4})}}{{\sin(a - \dfrac{\pi }{4})}}\)
=\(\dfrac{{2\cos \left( {\dfrac{a}{2} + \dfrac{{3\pi }}{8}} \right)\sin\left({\dfrac{\pi }{8} - \dfrac{a}{2}} \right)}}{{2\sin\left({\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\cos \left({\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}}\) \(= \dfrac{{\sin\left( { - \dfrac{a}{2} + \dfrac{\pi }{8}} \right)\sin\left({\dfrac{\pi }{8} - \dfrac{a}{2}} \right)}}{{\sin\left({\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\sin\left({\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}}\).
=\(\dfrac{{ - \sin\left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}}{{\cos \left({\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}} = - \tan \left({\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\)

Câu d​

\(\cos 2a - \cos 3a - \cos 4a + \cos 5a\) \(= - 4\sin \dfrac{a}{2}\sin a\cos \dfrac{{7a}}{2}\).
Lời giải chi tiết:
\(\cos 2a - \cos 3a - \cos 4a + \cos 5a \) \(= (\cos 2a - \cos 4a) + (\cos 5a - \cos 3a)\)
=\(- 2\sin 3a\sin ( - a) - 2\sin 4a\sin a \) \(= 2\sin a(\sin 3a - \sin 4a)\)
=\(4\sin a\cos \dfrac{{7a}}{2}\sin \left( { - \dfrac{a}{2}} \right)\) \(= - 4\sin \dfrac{a}{2}\sin a\cos \dfrac{{7a}}{2}\)