Bài 42: Chứng minh rằng:
a) \(\sin {{11\pi } \over {12}}\cos {{5\pi } \over {12}} = {1 \over 4}(2 – \sqrt 3 )\)
b) \(\cos {\pi \over 7}\cos {{3\pi } \over 7}\cos {{5\pi } \over 7} = – {1 \over 8}\)
c) \(\sin {6^0}\sin {42^0}\sin {66^0}\sin {78^0} = {1 \over 6}\) (
Nhân hai vế với cos 60)
Đáp án
a) Ta có:
\(\eqalign{
& \sin {{11\pi } \over {12}}\cos {{5\pi } \over {12}} = \sin (\pi – {\pi \over {12}})cos({\pi \over 2} – {\pi \over {12}}) \cr
& = {\sin ^2}{\pi \over {12}} = {1 \over 2}(1 – \cos {\pi \over 6}) = {1 \over 2}(1 – {{\sqrt 3 } \over 2})\cr& = {1 \over 4}(2 – \sqrt 3 ) \cr} \)
b) Ta có:
\(\eqalign{
& \cos {{3\pi } \over 7} = \cos (\pi – {{4\pi } \over 7}) = – \cos {{4\pi } \over 7} \cr
& \cos {{5\pi } \over 7} = \cos (\pi – {{2\pi } \over 7}) = – \cos {{2\pi } \over 7} \cr} \)
Nên:
\(\eqalign{
& \cos {\pi \over 7}\cos {{3\pi } \over 7}\cos {{5\pi } \over 7} = \cos {\pi \over 7}\cos {{2\pi } \over 7}\cos {{4\pi } \over 7} \cr
& = {1 \over {\sin {\pi \over 7}}}(\sin {\pi \over 7}\cos {\pi \over 7})\cos {{2\pi } \over 7}\cos {{4\pi } \over 7}\cr& = {1 \over {\sin {\pi \over 7}}}.{1 \over 2}(\sin {{2\pi } \over 7}\cos {{2\pi } \over 7}).cos{{4\pi } \over 7} \cr
& = {1 \over {\sin {\pi \over 7}}}.{1 \over 4}\sin {{4\pi } \over 7}.cos{{4\pi } \over 7} = {1 \over {8\sin {\pi \over 7}}}.\sin {{8\pi } \over 7} \cr
& = {{ – \sin {\pi \over 7}} \over {8\sin {\pi \over 7}}} = – {1 \over 8} \cr} \)
c) Ta có:
\(\eqalign{
& \sin {6^0}\sin {42^0}\sin {66^0}\sin {78^0} \cr&= \sin {6^0}\cos {48^0}\cos {24^0}\cos {12^0} \cr
& = {1 \over {\cos {6^0}}}(\sin {6^0}\cos {6^0})cos{12^0}\cos {24^0}\cos {48^0} \cr
& ={1 \over {\cos {6^0}}}\left( {{1 \over 2}\sin {{12}^0}\cos {{12}^0}} \right)cos{24^0}.cos{48^0}\cr&={1 \over {\cos {6^0}}}.{1 \over 4}\sin{24^0}\cos{24^0}.\cos{48^0}\cr&= {1 \over {\cos {6^0}}}.{1 \over 8}\sin {48^0}\cos {48^0} \cr
& = {1 \over {\cos {6^0}}}.{1 \over {16}}.\sin {96^0} = {{\cos {6^0}} \over {16\cos {6^0}}} = {1 \over {16}} \cr} \)
Bài 43: Dùng công thức biến đổi tích thành tổng, chứng minh:
a) \(\cos {75^0}\cos {15^0} = \sin {75^0}\sin {15^0} = {1 \over 4}\)
b) \(\cos {75^0}\sin {15^0} = {{2 – \sqrt 3 } \over 4}\)
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c) \(\sin {75^0}\cos {15^0} = {{2 + \sqrt 3 } \over 4}\)
d) \(\cos \alpha \sin (\beta – \gamma ) + \cos \beta \sin (\gamma – \alpha ) \)
\(+ \cos \gamma \sin (\alpha – \beta ) = 0\,\,\,\,\,\forall \alpha ,\beta ,\gamma \)
Đáp án
a) Ta có:
\(\eqalign{
& \cos {75^0}\cos {15^0} = {1 \over 2}(\cos {90^0} + \cos {60^0}) = {1 \over 4} \cr
& \sin {75^0}\sin {15^0} = {1 \over 2}(cos{60^0} – \cos {90^0}) = {1 \over 4} \cr} \)
Vậy \(\cos {75^0}\cos {15^0} = \sin {75^0}\sin {15^0} = {1 \over 4}\)
b) Ta có:
\(\eqalign{
& \cos {75^0}\sin {15^0} = {1 \over 2}(\sin {90^0} – \sin {60^0}) \cr
& = {1 \over 2}(1 – {{\sqrt 3 } \over 2}) = {{2 – \sqrt 3 } \over 4} \cr} \)
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c) Ta có:
\(\eqalign{
& \sin {75^0}\cos {15^0} = {1 \over 2}(\sin {90^0} + \sin {60^0}) \cr
& = {1 \over 2}(1 + {{\sqrt 3 } \over 2}) = {{2 + \sqrt 3 } \over 4} \cr} \)
d) Ta có:
\(\eqalign{
& \cos \alpha \sin (\beta – \gamma )\cr& = {1 \over 2}{\rm{[sin(}}\alpha {\rm{ + }}\beta – \gamma {\rm{)}}\,{\rm{ – }}\,{\rm{sin(}}\alpha {\rm{ – }}\beta {\rm{ + }}\gamma {\rm{)]}} \cr
& \cos \beta \sin (\gamma – \alpha ) \cr&= {1 \over 2}{\rm{[}}\sin (\beta + \gamma – \alpha {\rm{)}}\,{\rm{ – }}\,{\rm{sin(}}\beta – \gamma + \alpha ){\rm{]}} \cr
& \cos \gamma \sin (\alpha – \beta ) \cr&= {1 \over 2}{\rm{[sin(}}\gamma {\rm{ + }}\alpha {\rm{ – }}\beta {\rm{)}}\,{\rm{ – }}\,{\rm{sin(}}\gamma {\rm{ – }}\alpha {\rm{ + }}\beta {\rm{)]}} \cr} \)
Cộng các vế của ba đẳng thức, ta có:
\(\cos \alpha \sin (\beta – \gamma ) + \cos \beta \sin (\gamma – \alpha ) \)
\(+ \cos \gamma \sin (\alpha – \beta ) = 0\,\,\,\,\,\forall \alpha ,\beta ,\gamma \)
Bài 44: Đơn giản các biểu thức sau:
a) \(\sin ({\pi \over 3} + \alpha ) – \sin ({\pi \over 3} – \alpha )\)
b) \({\cos ^2}({\pi \over 4} + \alpha ) – {\cos ^2}({\pi \over 4} – \alpha )\)
Đáp án
a) Ta có:
\(\sin ({\pi \over 3} + \alpha ) – \sin ({\pi \over 3} – \alpha ) = 2\cos {\pi \over 3}\sin \alpha = \sin \alpha \)
b) Áp dụng: \({\cos ^2}a = {{1 + \cos 2a} \over 2}\) , ta có:
\(\eqalign{
& {\cos ^2}({\pi \over 4} + \alpha ) – {\cos ^2}({\pi \over 4} – \alpha ) \cr&= {{1 + \cos ({\pi \over 2} + 2\alpha )} \over 2} – {{1 + \cos ({\pi \over 2} – 2\alpha )} \over 2} \cr
& = {1 \over 2}( – \sin 2\alpha – \sin 2\alpha ) = – \sin 2\alpha \cr} \)
Bài 45: Chứng minh rằng:
a) \({{\sin \alpha – \sin \beta } \over {\cos \alpha – \cos \beta }} = – \sqrt 3 \) nếu
\(\left\{ \matrix{
\alpha + \beta = {\pi \over 3} \hfill \cr
\cos \alpha \ne \cos \beta \hfill \cr} \right.\)
b) \({{\cos \alpha – \cos 7\alpha } \over {\sin 7\alpha – sin\alpha }} = \tan 4\alpha \) (khi các biểu thức có nghĩa)
Đáp án
a)
\(\eqalign{
& {{\sin \alpha – \sin \beta } \over {\cos \alpha – \cos \beta }} = {{2\cos {{\alpha + \beta } \over 2}\sin {{\alpha – \beta } \over 2}} \over { – 2\sin {{\alpha + \beta } \over 2}\sin {{\alpha – \beta } \over 2}}} \cr
& = – \cot {{\alpha + \beta } \over 2} = – \cot {\pi \over 6} = – \sqrt 3 \cr} \)
b)
\({{\cos \alpha – \cos 7\alpha } \over {\sin 7\alpha – sin\alpha }} = {{2\sin 4\alpha \sin 3\alpha } \over {2\cos 4\alpha \sin 3\alpha }} = \tan 4\alpha \)
