Bài 43 trang 214 SGK Đại số 10 Nâng cao

LG a

\(\cos {75^0}\cos {15^0} = \sin {75^0}\sin {15^0} = {1 \over 4}\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \cos {75^0}\cos {15^0} \cr & = \frac{1}{2}\left[ {\cos \left( {{{75}^0} + {{15}^0}} \right) + \cos \left( {{{75}^0} - {{15}^0}} \right)} \right]\cr &= {1 \over 2}(\cos {90^0} + \cos {60^0})\cr &= \frac{1}{2}\left( {0 + \frac{1}{2}} \right) = {1 \over 4} \cr 
& \sin {75^0}\sin {15^0} \cr &= - \frac{1}{2}\left[ {\cos \left( {{{75}^0} + {{15}^0}} \right) - \cos \left( {{{75}^0} - {{15}^0}} \right)} \right]\cr &= {1 \over 2}(cos{60^0} - \cos {90^0}) \cr & = \frac{1}{2}\left( { \frac{1}{2}}-0 \right)= {1 \over 4} \cr} \)

Vậy \(\cos {75^0}\cos {15^0} = \sin {75^0}\sin {15^0} = {1 \over 4}\)

LG b

\(\cos {75^0}\sin {15^0} = {{2 - \sqrt 3 } \over 4}\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \cos {75^0}\sin {15^0} \cr & = \frac{1}{2}\left[ {\sin \left( {{{15}^0} + {{75}^0}} \right) + \sin \left( {{{15}^0} - {{75}^0}} \right)} \right] \cr &= \frac{1}{2}\left( {\sin {{90}^0} + \sin \left( { - {{60}^0}} \right)} \right)\cr &= {1 \over 2}(\sin {90^0} - \sin {60^0}) \cr 
& = {1 \over 2}(1 - {{\sqrt 3 } \over 2}) = {{2 - \sqrt 3 } \over 4} \cr} \) 

LG c

\(\sin {75^0}\cos {15^0} = {{2 + \sqrt 3 } \over 4}\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \sin {75^0}\cos {15^0} \cr & = \frac{1}{2}\left[ {\sin \left( {{{75}^0} + {{15}^0}} \right) + \sin \left( {{{75}^0} - {{15}^0}} \right)} \right]\cr &= {1 \over 2}(\sin {90^0} + \sin {60^0}) \cr 
& = {1 \over 2}(1 + {{\sqrt 3 } \over 2}) = {{2 + \sqrt 3 } \over 4} \cr} \)

LG d

 \(\cos \alpha \sin (\beta  - \gamma ) + \cos \beta \sin (\gamma  - \alpha ) \)

\(+ \cos \gamma \sin (\alpha  - \beta ) = 0\,\,\,\,\,\forall \alpha ,\beta ,\gamma \)

Lời giải chi tiết:

 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:

\(\cos \alpha \sin (\beta  - \gamma ) + \cos \beta \sin (\gamma  - \alpha ) \)

\(+ \cos \gamma \sin (\alpha  - \beta ) = 0\,\,\,\,\,\forall \alpha ,\beta ,\gamma \)

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