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Câu hỏi: Hãy tính sin α và tan α, nếu:
a) \(\cos \alpha =\displaystyle { 5 \over {13}}\);
b) \(\cos \alpha = \displaystyle {{15} \over {17}}\);
c) \(\cos \alpha = 0,6.\)
a) \(\cos \alpha =\displaystyle { 5 \over {13}}\);
b) \(\cos \alpha = \displaystyle {{15} \over {17}}\);
c) \(\cos \alpha = 0,6.\)
Phương pháp giải
Áp dụng kiến thức:
1) \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
2) \(tan\alpha = \displaystyle {{\sin \alpha } \over {\cos \alpha }}\)
Lời giải chi tiết
a) \(cos \alpha =\displaystyle {5 \over {13}}\)
* Ta có:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
Suy ra:
\(\eqalign{
& {\sin ^2}\alpha = 1 - {\cos ^2}\alpha = 1 - {\left( {{5 \over {13}}} \right)^2} \cr
& = 1 - {{25} \over {169}} = {{144} \over {169}} \cr} \)
Vì \(\sin \alpha > 0\) nên \(\sin \alpha =\displaystyle \sqrt {{{144} \over {169}}} = {{12} \over {13}}\)
* \(tan\alpha = \displaystyle {{\sin \alpha } \over {\cos \alpha }}\)\( = \displaystyle {\displaystyle {{{12} \over {13}}} \over {\displaystyle {5 \over {13}}}} = {{12} \over {13}}.{{13} \over 5} = {{12} \over 5}\)
b) \(\cos \alpha =\displaystyle {{15} \over {17}}\)
* Ta có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
Suy ra:
\(\eqalign{
& {\sin ^2}\alpha = 1 - {\cos ^2}\alpha = 1 - {\left( {{{15} \over {17}}} \right)^2} \cr
& = 1 - {{225} \over {289}} = {{64} \over {289}} \cr} \)
Vì \(\sin \alpha > 0\) nên \(\sin \alpha = \displaystyle \sqrt {{{64} \over {289}}} = {8 \over {17}}\)
* \(tan \alpha=\displaystyle {{\sin \alpha } \over {\cos \alpha }} = \displaystyle {{{8 \over {17}}} \over {{{15} \over {17}}}} \)\(=\displaystyle {8 \over {17}}.{{17} \over {15}} = {8 \over {15}}\)
c) \(\cos \alpha = 0,6\)
* Ta có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1.\)
Suy ra: \({\sin ^2}\alpha = 1 - {\cos ^2}\alpha \)
\( = 1 - {(0,6)^2} = 1 - 0,36 = 0,64\)
Vì \(\sin \alpha > 0\) nên \(\sin \alpha = \sqrt {0,64} = 0,8\)
* \(tan\alpha = \displaystyle {{\sin \alpha } \over {\cos \alpha }} = {{0,8} \over {0,6}} = {8 \over 6} = {4 \over 3}\)
Áp dụng kiến thức:
1) \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
2) \(tan\alpha = \displaystyle {{\sin \alpha } \over {\cos \alpha }}\)
Lời giải chi tiết
a) \(cos \alpha =\displaystyle {5 \over {13}}\)
* Ta có:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
Suy ra:
\(\eqalign{
& {\sin ^2}\alpha = 1 - {\cos ^2}\alpha = 1 - {\left( {{5 \over {13}}} \right)^2} \cr
& = 1 - {{25} \over {169}} = {{144} \over {169}} \cr} \)
Vì \(\sin \alpha > 0\) nên \(\sin \alpha =\displaystyle \sqrt {{{144} \over {169}}} = {{12} \over {13}}\)
* \(tan\alpha = \displaystyle {{\sin \alpha } \over {\cos \alpha }}\)\( = \displaystyle {\displaystyle {{{12} \over {13}}} \over {\displaystyle {5 \over {13}}}} = {{12} \over {13}}.{{13} \over 5} = {{12} \over 5}\)
b) \(\cos \alpha =\displaystyle {{15} \over {17}}\)
* Ta có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
Suy ra:
\(\eqalign{
& {\sin ^2}\alpha = 1 - {\cos ^2}\alpha = 1 - {\left( {{{15} \over {17}}} \right)^2} \cr
& = 1 - {{225} \over {289}} = {{64} \over {289}} \cr} \)
Vì \(\sin \alpha > 0\) nên \(\sin \alpha = \displaystyle \sqrt {{{64} \over {289}}} = {8 \over {17}}\)
* \(tan \alpha=\displaystyle {{\sin \alpha } \over {\cos \alpha }} = \displaystyle {{{8 \over {17}}} \over {{{15} \over {17}}}} \)\(=\displaystyle {8 \over {17}}.{{17} \over {15}} = {8 \over {15}}\)
c) \(\cos \alpha = 0,6\)
* Ta có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1.\)
Suy ra: \({\sin ^2}\alpha = 1 - {\cos ^2}\alpha \)
\( = 1 - {(0,6)^2} = 1 - 0,36 = 0,64\)
Vì \(\sin \alpha > 0\) nên \(\sin \alpha = \sqrt {0,64} = 0,8\)
* \(tan\alpha = \displaystyle {{\sin \alpha } \over {\cos \alpha }} = {{0,8} \over {0,6}} = {8 \over 6} = {4 \over 3}\)