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) \(\cos {75^0}\sin {15^0} = {{2 - \sqrt 3 } \over 4}\)
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 \(\cos {75^0}\cos {15^0} = \sin {75^0}\sin {15^0} = {1 \over 4}\)
Advertisements (Quảng cáo)
b) Ta có:
\(\eqalign{
& \cos {75^0}\sin {15^0} = {1 \over 2}(\sin {90^0} - \sin {60^0}) \cr
& = {1 \over 2}(1 - {{\sqrt 3 } \over 2}) = {{2 - \sqrt 3 } \over 4} \cr} \)
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:
\(\cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) \)
\(+ \cos \gamma \sin (\alpha - \beta ) = 0\,\,\,\,\,\forall \alpha ,\beta ,\gamma \)