Advertisements (Quảng cáo)
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)
\(\eqalign{
& {{\tan (a + 2a)} \over {\tan a}} = {{\tan a + \tan 2a} \over {\tan a(1 – {\mathop{\rm tanatan}\nolimits} 2a}} \cr
& = {{\tan a + {{2\tan a} \over {1 – {{\tan }^2}a}}} \over {\tan a(1 – {{2{{\tan }^2}a} \over {1 – {{\tan }^2}a}})}} \cr} \)
\( = {{3 – {{\tan }^2}a} \over {1 – 3{{\tan }^2}a}}$\)