Bài 16: Cho \(\cos \alpha = {1 \over 3}\) tính \(sin(\alpha + {\pi \over 6}) – \cos (\alpha – {{2\pi } \over 3})\)
Ta có: \(sin(\alpha + {\pi \over 6}) – \cos (\alpha – {{2\pi } \over 3})\)
= \(sin\alpha c{\rm{os}}{\pi \over 6} + \cos \alpha \sin {\pi \over 6} – \cos \alpha \cos {{2\pi } \over 3} – \sin \alpha \sin {{2\pi } \over 3}\)
\( = {{\sqrt 3 } \over 2}sin\alpha + {1 \over 2}\cos \alpha + {1 \over 2}\cos \alpha – {{\sqrt 3 } \over 2}\sin \alpha \)
\( = \cos \alpha = {1 \over 3}\)
Bài 17: Cho \(\sin \alpha = {8 \over {17}},\sin \beta = {{15} \over {17}}\) với \(0 < \alpha < {\pi \over 3},0 < \beta < {\pi \over 2}\). Chứng minh rằng \(\alpha + \beta = {\pi \over 2}\)
Ta có: \(\eqalign{
& \cos \alpha = \sqrt {1 – {{64} \over {289}}} = \sqrt {{{225} \over {289}}} = {{15} \over {17}}; \cr
& \cos \beta = \sqrt {1 – {{225} \over {289}}} = \sqrt {{{64} \over {289}}} = {8 \over {17}} \cr} \)
Do đó: \(\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
\({8 \over {17}}.{8 \over {17}} + {{15} \over {17}}.{{15} \over {17}} = {{289} \over {289}} = 1\)
Vì \(0 < \alpha < {\pi \over 3},0 < \beta < {\pi \over 2}\) nên từ đó suy ra \(\alpha + \beta = {\pi \over 2}\)
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Bài 18: Không dùng bảng số và máy tính, chứng minh rằng
a) \(\sin {20^0} + 2\sin {40^0} – \sin {100^0} = \sin {40^0}\)
b) \({{\sin ({{45}^0} + \alpha ) – c{\rm{os(}}{{45}^0} + \alpha )} \over {\sin ({{45}^0} + \alpha ) + c{\rm{os(}}{{45}^0} + \alpha )}} = \tan \alpha \)
c) \({{3{{\cot }^2}{{15}^0} – 1} \over {3 – c{\rm{o}}{{\rm{t}}^2}{{15}^0}}} = – \cot {15^0}\)
d) \(\sin {200^0}\sin {310^0} + c{\rm{os34}}{{\rm{0}}^0}{\rm{cos5}}{{\rm{0}}^0}{\rm{ = }}{{\sqrt 3 } \over 2}\)
a) \(\eqalign{
& \sin {20^0} + 2\sin {40^0} – \sin {100^0} \cr
& = (\sin {20^0} – \sin {100^0}) + 2\sin {40^0} \cr} \)
=\(2\cos {60^0}\sin ( – {40^0}) + 2\sin {40^0}\)
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=\( – \sin {40^0} + 2\sin {40^0} = \sin {40^0}\)
b) \(\eqalign{
& {{\sin ({{45}^0} + \alpha ) – c{\rm{os(}}{{45}^0} + \alpha )} \over {\sin ({{45}^0} + \alpha ) + c{\rm{os(}}{{45}^0} + \alpha )}} \cr
& = {{\sin ({{45}^0} + \alpha ) – \sin {\rm{(}}{{45}^0} – \alpha )} \over {\sin ({{45}^0} + \alpha ) + \sin {\rm{(}}{{45}^0} – \alpha )}} \cr} \)
=\({{2\cos {{45}^0}\sin \alpha } \over {2\sin {{45}^0}\cos \alpha }} = {{\sqrt 2 \sin \alpha } \over {\sqrt 2 \cos \alpha }} = \tan \alpha \)
c) \({{3{{\cot }^2}{{15}^0} – 1} \over {3 – c{\rm{o}}{{\rm{t}}^2}{{15}^0}}} = {{{{\cot }^2}{{30}^0}{{\cot }^2}{{15}^0} – 1} \over {c{\rm{o}}{{\rm{t}}^2}{{30}^0} – {{\cot }^2}{{15}^0}}}\)
=\({{\cot {{30}^0}\cot {{15}^0} + 1} \over {c{\rm{ot}}{{30}^0} – \cot {{15}^0}}}.{{\cot {{30}^0}\cot {{15}^0} – 1} \over {c{\rm{ot}}{{30}^0} + \cot {{15}^0}}}\)
Mặt khác ta có
\(\cot (\alpha + \beta ) = {{\cos (\alpha + \beta )} \over {\sin (\alpha + \beta )}} = {{\cos \alpha \cos \beta – \sin \alpha \sin \beta } \over {\sin \alpha \cos \beta + \cos \alpha \sin \beta }}\)
Chia cả tử và mẫu của biểu thức cho \(\sin \alpha \sin \beta \) ta được
\(\cot (\alpha + \beta ) = {{\cot \alpha \cot \beta – 1} \over {\cot \alpha + \cot \beta }}\)
Tương tự
\(\cot (\alpha – \beta ) = {{\cot \alpha \cot \beta + 1} \over {\cot \beta – \cot \alpha }}\)
Do đó
\(A = \cot ({15^0} – {30^0})\cot ({15^0} + {30^0}) = – \cot {15^0}\)
d) \(\sin {200^0}\sin {310^0} + c{\rm{os34}}{{\rm{0}}^0}{\rm{cos5}}{{\rm{0}}^0}\)
= \(\sin ({180^0} + {20^0})\sin ({360^0} – {50^0}) + c{\rm{os(36}}{{\rm{0}}^0}{\rm{ – 2}}{{\rm{0}}^0}{\rm{)cos5}}{{\rm{0}}^0}\)
\( = ( – \sin {20^0})( – \sin {50^0}) + \cos {20^0}\cos {50^0}\)
\( = \cos {50^0}\cos {20^0} + \sin {50^0}\sin {20^0}\)
= \(\cos ({50^0} – {20^0}) = {{\sqrt 3 } \over 2}\)
