Bài 56 trang 218 SGK Đại số 10 Nâng cao

Đề bài

Tính

a) 

\(sin\alpha ,{\rm{ }}cos2\alpha ,{\rm{ }}sin2\alpha ,\,\cos {\alpha \over 2},\sin {\alpha \over 2}\) biết

\(\cos \alpha = {4 \over 5} \) và \(- {\pi \over 2} < \alpha < 0 \)

b) \(\tan ({\pi  \over 4} - \alpha )\) biết 

\(\left\{ \matrix{
\cos \alpha = - {9 \over {11}} \hfill \cr
\pi < \alpha < {{3\pi } \over 2} \hfill \cr} \right.\)

c) \({\sin ^4}\alpha  - {\cos ^4}\alpha \) biết \(cos2\alpha  = {3 \over 5}\)

d) 

\(\cos (\alpha - \beta )\) biết \(\left\{ \matrix{
\sin \alpha - \sin \beta = {1 \over 3} \hfill \cr
\cos \alpha - \cos \beta = {1 \over 2} \hfill \cr} \right.\)

e) \(\sin {\pi  \over {16}}\sin {{3\pi } \over {16}}\sin {{5\pi } \over {16}}\sin {{7\pi } \over {16}}\)

Hướng dẫn giải

a) Ta có:

\(\eqalign{
& - {\pi \over 2} < \alpha < 0 \Rightarrow \sin \alpha < 0\cr& \Rightarrow \sin \alpha = - \sqrt {1 - {{\cos }^2}\alpha } = - {3 \over 5} \cr
& \sin 2\alpha = 2\sin \alpha \cos \alpha = - {{24} \over {25}} \cr
& \cos 2\alpha = 2{\cos ^2}\alpha - 1 = {7 \over {25}} \cr
& \cos {\alpha \over 2} = \sqrt {{{1 + \cos \alpha } \over 2}} = {{3\sqrt {10} } \over {10}};\cr&\sin {\alpha \over 2} =- \sqrt {{{1 - \cos \alpha } \over 2}} = - {{\sqrt {10} } \over {10}} \cr} \)

b) Vì \(\pi  < \alpha  < {{3\pi } \over 2} \Rightarrow \tan \alpha  > 0\)

Do đó:

\(\eqalign{
& \tan \alpha = \sqrt {{1 \over {{{\cos }^2}}} - 1} = {{2\sqrt {10} } \over 9} \cr
& \tan ({\pi \over 4} - \alpha ) = {{1 - \tan \alpha } \over {1 + \tan \alpha }} = {{121 - 36\sqrt {10} } \over {41}} \cr} \) 

c) Ta có:

\(\eqalign{
& {\sin ^4}\alpha - {\cos ^4}\alpha = ({\sin ^2}\alpha - {\cos ^2}\alpha )({\sin ^2}\alpha + {\cos ^2}\alpha ) \cr
& = {\sin ^2}\alpha - {\cos ^2}\alpha = - \cos 2\alpha = - {3 \over 5} \cr} \)

d) Ta có:

\(\eqalign{
& {(\sin \alpha - \sin \beta )^2} = {({1 \over 3})^2}\cr& \Rightarrow {\sin ^2}\alpha + {\sin ^2}\beta - 2\sin \alpha \sin \beta = {1 \over 9}\,\,\,\,\,\,(1) \cr
& {(cos\alpha - \cos \beta )^2} = {({1 \over 2})^2}\cr& \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta - 2\cos \alpha \cos \beta = {1 \over 4}\,\,\,(2) \cr} \)

Cộng từng vế của (1) và (2), ta được:

 \(1 + 1 - 2(cos\alpha \cos \beta  + \sin \alpha \sin \beta ) = {1 \over 9} + {1 \over 4} = {{13} \over {36}}\)

Từ đó: \(\cos (\alpha  - \beta ) = {{59} \over {72}}\)

e) Ta có:

\(\eqalign{
& \sin {\pi \over {16}}\sin {{3\pi } \over {16}}\sin {{5\pi } \over {16}}\sin {{7\pi } \over {16}}\cr& = \sin {\pi \over {16}}\sin {{3\pi } \over {16}}\sin ({\pi \over 2} - {{3\pi } \over 6})\sin ({\pi \over 2} - {\pi \over {16}}) \cr
& = \sin {\pi \over {16}}\sin {{3\pi } \over {16}}\cos {{3\pi } \over {16}}\cos {\pi \over {16}}\cr& = ({1 \over 2}\sin {\pi \over 8})({1 \over 2}\sin {{3\pi } \over 8}) \cr
& = {1 \over 4}\sin {\pi \over 8}\sin ({\pi \over 2} - {\pi \over 8}) \cr&= {1 \over 4}sin{\pi \over 8}\cos {\pi \over 8} = {1 \over 8}\sin {\pi \over 4} = {{\sqrt 2 } \over {16}} \cr} \)