Bài 20 trang 217 SBT Toán Đại số 10: Chứng minh rằng...

Chứng minh rằng

a) ${{\sqrt {1 + \cos \alpha }  + \sqrt {1 - \cos \alpha } } \over {\sqrt {1 + \cos \alpha }  - \sqrt {1 - \cos \alpha } }} = \cot ({\alpha  \over 2} + {\pi  \over 4})$ $(\pi  < \alpha  < 2\pi )$

b) ${{\cos 4a\tan 2a - \sin 4a} \over {\cos 4a\cot 2a + \sin 4a}} =  - {\tan ^2}2a$

c) ${{{{\sin }^2}2a + 4{{\sin }^2}a - 4} \over {1 - 8{{\sin }^2}a - \cos 4a}} = {1 \over 2}{\cot ^4}a$

d) $1 + 2\cos 7a = {{\sin 10,5a} \over {\sin 3,5a}}$

e) ${{\tan 3a} \over {\tan a}} = {{3 - {{\tan }^2}a} \over {1 - 3{{\tan }^2}a}}$

Gợi ý làm bài

a) Vì $\sqrt {1 + \cos \alpha }  =  - \sqrt 2 \cos {\alpha  \over 2}(do{\pi  \over 2} < {\alpha  \over 2} < \pi )$

$\sqrt {1 - \cos \alpha }  = \sqrt 2 \sin {\alpha  \over 2}$ cho nên

${{\sqrt {1 + \cos \alpha }  + \sqrt {1 - \cos \alpha } } \over {\sqrt {1 + \cos \alpha }  - \sqrt {1 - \cos \alpha } }} = {{ - \sqrt 2 \cos {\alpha  \over 2} + \sqrt 2 \cos {\alpha  \over 2}} \over { - \sqrt 2 \cos {\alpha  \over 2} - \sqrt 2 \cos {\alpha  \over 2}}}$

$= {{\cos {\alpha  \over 2} - \sin {\alpha  \over 2}} \over {\cos {\alpha  \over 2} + \sin {\alpha  \over 2}}} = {{1 - \tan {\alpha  \over 2}} \over {1 + \tan {\alpha  \over 2}}} = \tan ({\pi  \over 4} - {\alpha  \over 2})$

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

b) 

$\eqalign{ & = {{\cos 4a\tan 2a - \sin 4a} \over {\cos 4a\cot 2a + \sin 4a}} \cr & = {{\cos 4a\sin 2a - \sin 4a\cos 2a} \over {\cos 4a\cos 2a + \sin 4a\sin 2a}}.\tan 2a \cr}$

$= {{ - \sin 2a} \over {\cos 2a}}\tan 2a =  - {\tan ^2}2a$$

c) 

$\eqalign{ & {{{{\sin }^2}2a + 4{{\sin }^2}4a} \over {1 - {{\sin }^2}a - \cos 4a}} \cr & = {{4{{\sin }^2}a{{\cos }^2}a + 4({{\sin }^2}a - 1)} \over {1 - 8{{\sin }^2}a - (1 - 2{{\sin }^2}2a)}} \cr}$

${{4{{\cos }^2}a({{\sin }^2}a - 1)} \over {8{{\sin }^2}a(co{s^2}a - 1)}} = {1 \over 2}{\cot ^4}a.$

d) 

$\eqalign{ & {{\sin 10,5a} \over {\sin 3,5a}} = {{\sin (7 + 3,5a)} \over {\sin 3,5a}} \cr & = {{\sin 7a\cos 3,5a + \cos 7a\sin 3,5a} \over {\sin 3,5a}} \cr}$

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

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

$= 2cos7a + 1.$

e) 

$\eqalign{ & {{\tan (a + 2a)} \over {\tan a}} = {{\tan a + \tan 2a} \over {\tan a(1 - {\mathop{\rm tanatan}\nolimits} 2a}} \cr & = {{\tan a + {{2\tan a} \over {1 - {{\tan }^2}a}}} \over {\tan a(1 - {{2{{\tan }^2}a} \over {1 - {{\tan }^2}a}})}} \cr}$

$= {{3 - {{\tan }^2}a} \over {1 - 3{{\tan }^2}a}}$$