Bài 20 trang 214 SBT đại số 10

Chứng minh rằng

LG a

 \(\dfrac{{\cot a + {\mathop{\rm tana}\nolimits} }}{{1 + \tan 2a\tan a}} = 2\cot 2a\);

Lời giải chi tiết:

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

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

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

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

LG b

\(\dfrac{{\sqrt 2  - {\mathop{\rm sina}\nolimits}  - \cos a}}{{\sin a - \cos a}} =  - \tan \left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\);

Lời giải chi tiết:

\(\dfrac{{\sqrt 2  - \sin a - \cos a}}{{\sin a - \cos a}} \)\( = \dfrac{{\sqrt 2  - \left( {\sin a + \cos a} \right)}}{{\sin a - \cos a}}\)

\(= \dfrac{{\sqrt 2  - \sqrt 2 \sin(a + \dfrac{\pi }{4})}}{{\sqrt 2 \sin(a - \dfrac{\pi }{4})}}\)

=\(\dfrac{{1 - \sin (a + \dfrac{\pi }{4})}}{{\sin(a - \dfrac{\pi }{4})}} \) \(= \dfrac{{\sin\dfrac{\pi }{2} - \sin(a + \dfrac{\pi }{4})}}{{\sin(a - \dfrac{\pi }{4})}}\)

=\(\dfrac{{2\cos \left( {\dfrac{a}{2} + \dfrac{{3\pi }}{8}} \right)\sin\left( {\dfrac{\pi }{8} - \dfrac{a}{2}} \right)}}{{2\sin\left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\cos \left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}}\) \( = \dfrac{{\sin\left( { - \dfrac{a}{2} + \dfrac{\pi }{8}} \right)\sin\left( {\dfrac{\pi }{8} - \dfrac{a}{2}} \right)}}{{\sin\left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\sin\left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}}\).

=\(\dfrac{{ - \sin\left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}}{{\cos \left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)}} =  - \tan \left( {\dfrac{a}{2} - \dfrac{\pi }{8}} \right)\)

LG c

\(\cos 2a - \cos 3a - \cos 4a + \cos 5a\) \(= - 4\sin \dfrac{a}{2}\sin a\cos \dfrac{{7a}}{2}\).

Lời giải chi tiết:

\(\cos 2a - \cos 3a - \cos 4a + \cos 5a \) \(= (\cos 2a - \cos 4a) + (\cos 5a - \cos 3a)\)

=\( - 2\sin 3a\sin ( - a) - 2\sin 4a\sin a \) \(= 2\sin a(\sin 3a - \sin 4a)\)

=\(4\sin a\cos \dfrac{{7a}}{2}\sin \left( { - \dfrac{a}{2}} \right)\) \( =  - 4\sin \dfrac{a}{2}\sin a\cos \dfrac{{7a}}{2}\)