Bài 19: Chứng minh rằng các biểu thức sau là những hằng số không phụ thuộc \(\alpha ,\beta \)
a) \(\sin 6\alpha \cot 3\alpha – c{\rm{os6}}\alpha \)
b) \({{\rm{[}}\tan ({90^0} – \alpha ) – \cot ({90^0} + \alpha ){\rm{]}}^2} – {{\rm{[}}c{\rm{ot(18}}{{\rm{0}}^0} + \alpha ) + \cot ({270^0} + \alpha ){\rm{]}}^2}\)
c) \((\tan \alpha – \tan \beta )cot(\alpha – \beta ) – \tan \alpha \tan \beta \)
d) \((\cot {\alpha \over 3} – \tan {\alpha \over 3})\tan {{2\alpha } \over 3}\)
a) \(\eqalign{
& \sin 6\alpha \cot 3\alpha – c{\rm{os6}}\alpha \cr
& = 2\sin 3\alpha \cos 3\alpha .{{\cos 3\alpha } \over {\sin 3\alpha }} – (2{\cos ^2}3\alpha – 1) \cr} \)
= \(2{\cos ^2}3\alpha – 2{\cos ^2}3\alpha + 1 = 1\)
b) \({{\rm{[}}\tan ({90^0} – \alpha ) – \cot ({90^0} + \alpha ){\rm{]}}^2} – {{\rm{[}}c{\rm{ot(18}}{{\rm{0}}^0} + \alpha ) + \cot ({270^0} + \alpha ){\rm{]}}^2}\)
= \({(\cot \alpha + \tan \alpha )^2} – {(\cot \alpha – \tan \alpha )^2}\)
= \({\cot ^2}\alpha + 2 + {\tan ^2}\alpha – {\cot ^2}\alpha + 2 – {\tan ^2}\alpha = 4\)
c) \(\eqalign{
& (\tan \alpha – \tan \beta )cot(\alpha – \beta ) – \tan \alpha \tan \beta \cr
& = {{\tan \alpha – \tan \beta } \over {\tan (\alpha – \beta )}} – \tan \alpha \tan \beta \cr} \)
=\(1 + \tan \alpha \tan \beta – \tan \alpha \tan \beta = 1\)
d) \(\eqalign{
& (\cot {\alpha \over 3} – \tan {\alpha \over 3})\tan {{2\alpha } \over 3} \cr
& = ({{\cos {\alpha \over 3}} \over {\sin {\alpha \over 3}}} – {{\sin {\alpha \over 3}} \over {\cos {\alpha \over 3}}}){{\sin {{2\alpha } \over 3}} \over {\cos {{2\alpha } \over 3}}} \cr} \)
= \(\eqalign{
& {{{{\cos }^2}{\alpha \over 3} – {{\sin }^2}{\alpha \over 3}} \over {\sin {\alpha \over 3}\cos {\alpha \over 3}}}.{{\sin {{2\alpha } \over 3}} \over {\cos {{2\alpha } \over 3}}} \cr
& = {{\cos {{2\alpha } \over 3}} \over {{1 \over 2}\sin {{2\alpha } \over 3}}}.{{\sin {{2\alpha } \over 3}} \over {\cos {{2\alpha } \over 3}}} = 2 \cr} \)
Bài 20: Không sử dụng bảng số và máy tính, hãy tính
a) \({\sin ^4}{\pi \over {16}} + {\sin ^4}{{3\pi } \over {16}} + {\sin ^4}{{5\pi } \over {16}} + {\sin ^4}{{7\pi } \over {16}}\)
b) \(\cot 7,{5^0} + \tan 67,{5^0} – \tan 7,{5^0} – \cot 67,{5^0}\)
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a) \({\sin ^4}{\pi \over {16}} + {\sin ^4}{{3\pi } \over {16}} + {\sin ^4}{{5\pi } \over {16}} + {\sin ^4}{{7\pi } \over {16}}\)
\( = {\left( {{{1 – \cos {\pi \over 8}} \over 2}} \right)^2} + {\left( {{{1 – \cos {{3\pi } \over 8}} \over 2}} \right)^2} + {\left( {{{1 – \cos {{5\pi } \over 8}} \over 2}} \right)^2} + {\left( {{{1 – \cos {{7\pi } \over 8}} \over 2}} \right)^2}\)
\( = {1 \over 4}\left( {1 – 2\cos {\pi \over 8} + {{\cos }^2}{\pi \over 8} + 1 – 2\cos {{3\pi } \over 8} + {{\cos }^2}{{3\pi } \over 8} + 1 – 2\cos {{5\pi } \over 8} + {{\cos }^2}{{5\pi } \over 8} + 1 – 2\cos {{7\pi } \over 8} + {{\cos }^2}{{7\pi } \over 8}} \right)\)
\( = 1 – {1 \over 2}\left( {\cos {\pi \over 8} + \cos {{3\pi } \over 8} + \cos {{5\pi } \over 8} + \cos {{7\pi } \over 8}} \right) + {1 \over 4}\left( {{{1 + \cos {\pi \over 4}} \over 2} + {{1 + \cos {{3\pi } \over 4}} \over 2} + {{1 + \cos {{5\pi } \over 4}} \over 2} + {{1 + \cos {{7\pi } \over 4}} \over 2}} \right)$\)
=\(1 – {1 \over 2}\left( {\cos {\pi \over 8} + \cos {{3\pi } \over 8} – \cos {{3\pi } \over 8} – \cos {\pi \over 8}} \right) + {1 \over 8}\left( {4 + {{\sqrt 2 } \over 2} – {{\sqrt 2 } \over 2} – {{\sqrt 2 } \over 2} + {{\sqrt 2 } \over 2}} \right)\)
= \({3 \over 2}\)
b) \(\cot 7,{5^0} + \tan 67,{5^0} – \tan 7,{5^0} – \cot 67,{5^0}\)
= \({{\cos 7,{5^0}} \over {\sin 7,{5^0}}} – {{\sin 7,{5^0}} \over {\cos 7,{5^0}}} + {{\sin 67,{5^0}} \over {\cos 67,{5^0}}} – {{\cos 67,{5^0}} \over {\sin 67,{5^0}}}\)
= \({{{{\cos }^2}7,{5^0} – {{\sin }^2}7,{5^0}} \over {\sin 7,{5^0}\cos 7,{5^0}}} + {{{{\sin }^2}67,{5^0} – {{\cos }^2}67,{5^0}} \over {sin67,{5^0}\cos 67,{5^0}}}\)
= \(\eqalign{
& {{\cos {{15}^0}} \over {{1 \over 2}\sin {{15}^0}}} – {{\cos {{135}^0}} \over {{1 \over 2}\sin {{135}^0}}} \cr
& = {{2(\sin {{135}^0}\cos {{15}^0} – \cos {{135}^0}\sin {{15}^0})} \over {\sin {{15}^0}\sin {{135}^0}}} \cr} \)
= \({{\sin ({{135}^0} – {{15}^0})} \over {\sin ({{45}^0} – {{30}^0})\sin ({{180}^0} – {{45}^0})}}\)
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= \({{2\sin {{120}^0}} \over {(\sin {{45}^0}\cos {{30}^0} – \cos {{45}^0}\sin {{30}^0})sin{{45}^0}}}\)
\(\eqalign{
& = {{\sqrt 3 } \over {{{\sqrt 2 } \over 2}({{\sqrt 3 } \over 2} – {1 \over 2}).{{\sqrt 2 } \over 2}}} \cr
& = {{4\sqrt 3 } \over {\sqrt 3 – 1}} = 6 + 2\sqrt 3 \cr} $\)
Bài 21: Rút gọn các biểu thức
a) \({{\sin 2\alpha + \sin \alpha } \over {1 + c{\rm{os2}}\alpha {\rm{ + cos}}\alpha }}\)
b) \({{4{{\sin }^2}\alpha } \over {1 – c{\rm{o}}{{\rm{s}}^2}{\alpha \over 2}}}\)
c) \({{1 + c{\rm{os}}\alpha – \sin \alpha } \over {1 – c{\rm{os}}\alpha – {\rm{sin}}\alpha }}\)
d) \({{1 + \sin \alpha – 2{{\sin }^2}({{45}^0} – {\alpha \over 2})} \over {4c{\rm{os}}{\alpha \over 2}}}\)
a) \({{\sin 2\alpha + \sin \alpha } \over {1 + c{\rm{os2}}\alpha {\rm{ + cos}}\alpha }} = {{\sin \alpha (2\cos \alpha + 1)} \over {2c{\rm{o}}{{\rm{s}}^2}\alpha {\rm{ + cos}}\alpha }}\)
= \({{\sin \alpha (2\cos \alpha + 1)} \over {c{\rm{os}}\alpha (2{\rm{cos}}\alpha + 1)}} = \tan \alpha \)
b) \({{4{{\sin }^2}\alpha } \over {1 – c{\rm{o}}{{\rm{s}}^2}{\alpha \over 2}}} = {{16{{\sin }^2}{\alpha \over 2}{{\cos }^2}{\alpha \over 2}} \over {{{\sin }^2}{\alpha \over 2}}} = 16{\cos ^2}{\alpha \over 2}\)
c) \({{1 + c{\rm{os}}\alpha – \sin \alpha } \over {1 – c{\rm{os}}\alpha – {\rm{sin}}\alpha }} = {{2{{\cos }^2}{\alpha \over 2} – 2\sin {\alpha \over 2}\cos {\alpha \over 2}} \over {2si{n^2}{\alpha \over 2} – 2\sin {\alpha \over 2}\cos {\alpha \over 2}}}\)
= \({{2\cos {\alpha \over 2}(\cos {\alpha \over 2} – \sin {\alpha \over 2})} \over {2\sin {\alpha \over 2}(sin{\alpha \over 2} – \cos {\alpha \over 2})}} = – \cot {\alpha \over 2}\)
d) \({{1 + \sin \alpha – 2{{\sin }^2}({{45}^0} – {\alpha \over 2})} \over {4c{\rm{os}}{\alpha \over 2}}} = {{\sin \alpha + \cos ({{90}^0} – \alpha )} \over {4\cos {\alpha \over 2}}}\)
=\({{\sin \alpha + \sin \alpha } \over {4\cos {\alpha \over 2}}} = {{4\sin {\alpha \over 2}\cos {\alpha \over 2}} \over {4\cos {\alpha \over 2}}} = \sin {\alpha \over 2}\)
Bài 22: Cho hình thang cân ABCD có đáy nhỏ AB = AD. Biết \(\tan \widehat {BDC} = {3 \over 4}\) tính các giá trị lượng giác của \(\widehat {BAD}\)
Ta có (h.64)
\(\widehat {ABD} = \widehat {ADB}\)
\(\widehat {ABD} = \widehat {BDC}\)
=> \(\widehat {BDC} = \widehat {ADB}\)
Suy ra \(\widehat {BAD} = \pi – 2\widehat {BDC}\)
Từ đó ta có:
\(\eqalign{
& \tan \widehat {BAD} = – \tan 2\widehat {BDC} = – {{2\tan \widehat {BDC}} \over {1 – {{\tan }^2}\widehat {BDC}}} \cr
& = – {{2.{3 \over 4}} \over {1 – {9 \over {16}}}} = – {3 \over 2}.{{16} \over 7} = – {{24} \over 7} \cr} \)
Vì \({\pi \over 2} < \widehat {BAD} < \pi \) nên \(\cos \widehat {BAD} < 0\). Do đó
\(\eqalign{
& \cos \widehat {BAD} = – {1 \over {\sqrt {1 + {{\tan }^2}\widehat {BAD}} }} \cr
& = – {1 \over {\sqrt {1 + {{576} \over {49}}} }} = – {7 \over {25}} \cr} \)
\(\eqalign{
& \sin \widehat {BAD} = \cos \widehat {BAD}.\tan \widehat {BAD} \cr
& = {{ – 7} \over {25}}.{{ – 24} \over 7} = {{24} \over {25}} \cr} \).
