Bài Ôn tập cuối năm SBT Toán lớp 10. Giải bài 21, 22, 23, 24 trang 218 Sách bài tập Toán Đại số 10. Câu 21: Rút gọn…
Bài 21: Rút gọn
a) \({{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha – 4{{\sin }^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha } \over {4 – {{\sin }^2}2\alpha – 4{{\sin }^2}\alpha }}\)
b) \(3 – 4\cos 2a + \cos 4a\)
c) \(\cos 4a – \sin 4a\cot 2a\)
d) \({{{\mathop{\rm cota}\nolimits} + \tan a} \over {1 + \tan 2a\tan a}}\)
a) \(\eqalign{
& {{{{\sin }^2}2\alpha + 4{{\sin }^2}4\alpha – 4{{\sin }^2}\alpha {{\cos }^2}\alpha } \over {4 – {{\sin }^2}2\alpha – 4{{\sin }^2}\alpha }} \cr
& = {{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha – {{\sin }^2}2\alpha } \over {4{{\cos }^2}a – 4{{\sin }^2}2\alpha {{\cos }^2}\alpha }} \cr} \)
\( = {{4{{\sin }^2}\alpha } \over {4co{s^2}\alpha (1 – {{\sin }^2}\alpha )}} = {\tan ^4}\alpha \)
b) \(\eqalign{
& 3 – 4\cos 2a + \cos 4a \cr
& = 3 – 4(1 – 2{\sin ^2}a) + (1 – 2{\sin ^2}2a) \cr} \)
\(\eqalign{
& = 8{\sin ^2}a – 8{\sin ^2}a{\cos ^2}a \cr
& = 8{\sin ^2}a(1 – {\cos ^2}a) \cr} \)
\( = 8{\sin ^4}a\)
c) \(\eqalign{
& \cos 4a – \sin 4a\cot 2a \cr
& = 2{\cos ^2}2a – 1 – 2\sin 2a\cos 2a{{\cos 2a} \over {\sin 2a}} = – 1 \cr} \)
d) \({{\cot a + \tan a} \over {1 + \tan 2a\tan a}} = {{{{\cos a} \over {\sin a}} + {{\sin a} \over {\cos a}}} \over {1 + {{\sin 2a\sin a} \over {\cos 2a\cos a}}}}\)
\( = {1 \over {\sin a\cos a}}.{{\cos acos2a} \over {\cos 2a\cos a + \sin 2a\sin a}}\)
\( = {2 \over {\sin 2a}}.{{\cos acos2a} \over {\cos (2a – a)}} = 2\cot 2a\)
Bài 22: Không dùng bảng số và máy tính, hãy tính
a) \(\cos {67^0}30’\) và \({\rm{cos7}}{{\rm{5}}^0}\)
b) \({{\cos {{15}^0} + 1} \over {2\cot {{15}^0}}}\)
c) \(\tan {20^0}\tan {40^0}\tan {80^0}\)
d) \(\cos {\pi \over 7}\cos {{4\pi } \over 7}\cos {{5\pi } \over 7}\)
a) \(\cos {67^0}30′ = \cos {{{{135}^0}} \over 2} = \sqrt {{{1 + \cos {{135}^0}} \over 2}} \)
\( = \sqrt {{{1 – {{\sqrt 2 } \over 2}} \over 2}} = {{\sqrt {2 – \sqrt 2 } } \over 2}\)
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\(\cos {75^0} = \cos ({45^0} + {30^0}) = {{\sqrt 2 } \over 4}(\sqrt 3 – 1)\)
b) \(\eqalign{
& \cos {30^0} = {1 \over {\tan {{2.15}^0}}} \cr
& = {{1 – {{\tan }^2}{{15}^0}} \over {2\tan {{15}^0}}} = {{{{\cot }^2}{{15}^0} – 1} \over {2\cot {{15}^0}}} \cr} \)
Đặt \(x = \cos {15^0}\) và chú ý rằng \(\cos {30^0} = \sqrt 3 \) ta có
\(\sqrt 3 = {{{x^2} – 1} \over {2x}} \Leftrightarrow {x^2} – 2\sqrt 3 – 1 = 0\)
Giải phương trình trên ta được \(x = 2 + \sqrt 3 \) (nghiệm \(x = \sqrt 3 – 2\) loại vì \(\cot {15^0} > 0\)). Do đó
\(\eqalign{
& {{{{\cot }^2}{{15}^0} + 1} \over {2\cot {{15}^0}}} = {{2 + \sqrt 3 + 1} \over {2(2 + \sqrt 3 )}} \cr
& = {{3 + \sqrt 3 } \over {2(2 + \sqrt 3 )}} = {{3 – \sqrt 3 } \over 2} \cr} \)
c) Ta có:
\(\tan {20^0}\tan {40^0}\tan {80^0} = – \tan {20^0}\tan {40^0}\tan {100^0}\)
\( = – \tan ({60^0} – {40^0})\tan {40^0}\tan ({60^0} + {40^0})\)
\( = – {{\tan {{60}^0} – \tan {{40}^0}} \over {1 + \tan {{60}^0}\tan {{40}^0}}}\tan {40^0}{{\tan {{60}^0} + \tan {{40}^0}} \over {1 – \tan {{60}^0}\tan {{40}^0}}}\)
\( = – {{3 – {{\tan }^2}{{40}^0}} \over {1 – 3{{\tan }^2}{{40}^0}}}\tan {40^0} = – \tan {120^0} = \sqrt 3 \)
d) Nhân thêm \(\sin {\pi \over 7}\)
Đáp số: \({1 \over 8}\)
Bài 23: Chứng minh rằng
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a) \({{1 – \cos 2a + \sin 2a} \over {1 + \cos 2a + \sin 2a}} = \tan a\)
b) \({{\cot a + {\mathop{\rm tana}\nolimits} } \over {1 + \tan 2\tan a}} = 2\cot 2a\)
c) \({{\sqrt 2 – {\mathop{\rm sina}\nolimits} – \cos a} \over {\sin a – \cos a}} = – \tan \left( {{a \over 2} – {\pi \over 8}} \right)\)
d) \(\cos 2a – \cos 3a – \cos 4a + \cos 5a = – 4\sin {a \over 2}\sin a\cos {{7a} \over 2}\)
a) \(\eqalign{
& {{1 – \cos 2a + \sin 2a} \over {1 + \cos 2a + \sin 2a}} \cr
& = {{2{{\sin }^2}a + 2\sin a\cos a} \over {1 + 2{{\cos }^2}a – 1 + 2\sin a\cos a}} \cr} \)
\( = {{2\sin a(\sin a + {\mathop{\rm cosa}\nolimits} )} \over {2\cos a(\sin a + \cos a)}} = \tan a\)
b) \({{\cot a + \tan a} \over {1 + \tan 2a\tan a}} = {{{1 \over {\tan a}} + \tan a} \over {1 + {{2tana} \over {1 – {{\tan }^2}a}}}}\)
\( = {{1 + {{\tan }^2}a} \over {\tan a}}:{{1 – {{\tan }^2}a + 2{{\tan }^2}a} \over {1 – {{\tan }^2}a}}\)
\( = {{1 – {{\tan }^2}a} \over {\tan a}} = 2\cot 2a\)
c) \({{\sqrt 2 – \sin a – \cos a} \over {\sin a – \cos a}} = {{\sqrt 2 – \sqrt 2 sin(a + {\pi \over 4})} \over {\sqrt 2 sin(a – {\pi \over 4})}}\)
\( = {{1 – \sin (a + {\pi \over 4})} \over {sin(a – {\pi \over 4})}} = {{sin{\pi \over 2} – sin(a + {\pi \over 4})} \over {sin(a – {\pi \over 4})}}\)
\(\eqalign{
& = {{\cos \left( {{a \over 2} + {{3\pi } \over 8}} \right)sin\left( {{\pi \over 8} – {a \over 2}} \right)} \over {2sin\left( {{a \over 2} – {\pi \over 8}} \right)\cos \left( {{a \over 2} – {\pi \over 8}} \right)}} \cr
& = {{sin\left( { – {a \over 2} + {\pi \over 8}} \right)sin\left( {{\pi \over 8} – {a \over 2}} \right)} \over {sin\left( {{a \over 2} – {\pi \over 8}} \right)sin\left( {{a \over 2} – {\pi \over 8}} \right)}} \cr} \)
\( = {{ – sin\left( {{a \over 2} – {\pi \over 8}} \right)} \over {\cos \left( {{a \over 2} – {\pi \over 8}} \right)}} = – \tan \left( {{a \over 2} – {\pi \over 8}} \right)\)
d) \(\eqalign{
& \cos 2a – \cos 3a – \cos 4a + \cos 5a \cr
& = (\cos 2a – \cos 4a) + (\cos 5a – \cos 3a) \cr} \)
\(\eqalign{
& = – 2\sin 3a\sin ( – a) – 2\sin 4a\sin a \cr
& = 2\sin a(\sin 3a – \sin 4a) \cr} \)
\(\eqalign{
& = 4\sin a\cos {{7a} \over 2}\sin \left( { – {a \over 2}} \right) \cr
& = – 4\sin {a \over 2}\sin a\cos {{7a} \over 2} \cr} \)
Bài 24: Rút gọn
a) \({{1 + \cos a} \over {1 – \cos a}}{\tan ^2}{a \over 2} – {\cos ^2}a\)
b) \(4{\cos ^4}a – 2\cos 2a – {1 \over 2}\cos 4a\)
c) \({\sin ^2}a\left( {1 + {1 \over {\sin a}} + \cot a} \right)\left( {1 – {1 \over {\sin a}} + \cot a} \right)\)
d) \({{\cos 2a} \over {{{\cos }^4}a – {{\sin }^4}a}} – {{{{\cos }^4}a + {{\sin }^4}a} \over {1 – {1 \over 2}{{\sin }^2}2a}}\)
a) \(\eqalign{
& {{1 + \cos a} \over {1 – \cos a}}{\tan ^2}{a \over 2} – {\cos ^2}a \cr
& = {{2{{\cos }^2}{a \over 2}} \over {2{{\sin }^2}{a \over 2}}}{\tan ^2}{a \over 2} – {\cos ^2}a = {\sin ^2}a \cr} \)
b) \(4{\cos ^4}a – 2\cos 2a – {1 \over 2}\cos 4a\)
\( = 4{\cos ^4}a – 2(2{\cos ^2}a – 1) – {1 \over 2}(2{\cos ^2}2a – 1)\)
\( = 4{\cos ^4}a – 4{\cos ^2}a + 2 – {(2{\cos ^2}a – 1)^2} + {1 \over 2}\)
\( = 4{\cos ^4}a – 4{\cos ^2}a + {5 \over 2} – 4{\cos ^4}a + 4{\cos ^2}a – 1 = {3 \over 2}\)
c) \({\sin ^2}a(1 + {1 \over {\sin a}} + \cot a)(1 – {1 \over {\sin a}} + \cot a)\)
\(\eqalign{
& = {\sin ^2}a\left[ {{{(1 + cota)}^2} – {1 \over {{{\sin }^2}a}}} \right] \cr
& = {\sin ^2}a(1 + {\cot ^2}a + 2\cot a) – 1 \cr} \)
\(\eqalign{
& = {\sin ^2}a + {\cos ^2}a + 2{\sin ^2}a{{\cos a} \over {\sin a}} – 1 \cr
& = \sin 2a \cr} \)
d) \({{\cos 2a} \over {{{\cos }^4}a – {{\sin }^4}a}} – {{{{\cos }^4}a + {{\sin }^4}a} \over {1 – {1 \over 2}{{\sin }^2}2a}}\)
\(= {{{{\cos }^2}a – {{\sin }^2}a} \over {({{\cos }^2}a + {{\sin }^2}a)({{\cos }^2}a – {{\sin }^2}a)}} – {{{{\cos }^4}a + {{\sin }^4}a} \over {1 – {1 \over 2}{{(2\sin a\cos a)}^2}}}\)
\( = 1 – {{{{\cos }^4}a + {{\sin }^4}a} \over {{{\sin }^2}a – si{n^2}aco{s^2}a + {{\cos }^2}a – {{\sin }^2}a{{\cos }^2}a}}\)
\( = 1 – {{{{\cos }^4}a + {{\sin }^4}a} \over {{{\sin }^2}a(1 – co{s^2}a) + {{\cos }^2}a(1 – {{\sin }^2}a)}} = 0\).
