Đề bài
Cho \(\cos \alpha = \frac{{11}}{{61}}\) và \( - \frac{\pi }{2} < \alpha < 0\), tính giá trị của các biểu thức sau:
a) \(\sin \left( {\frac{\pi }{6} - \alpha } \right)\);
b) \(\cot \left( {\alpha + \frac{\pi }{4}} \right)\);
c) \(\cos \left( {2\alpha + \frac{\pi }{3}} \right)\);
d) \(\tan \left( {\frac{{3\pi }}{4} - 2\alpha } \right)\).
Phương pháp giải - Xem chi tiết
Sử dụng kiến thức về công thức cộng để tính:
a) \(\sin \left( {\alpha - \beta } \right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \)
b) \(\tan \left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }},\cot \alpha = \frac{1}{{\tan \alpha }}\)
c) \(\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
d) \(\tan \left( {\alpha - \beta } \right) = \frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha .\tan \beta }}\)
Lời giải chi tiết
Vì \( - \frac{\pi }{2} < \alpha < 0 \Rightarrow \sin \alpha < 0\)
Do đó, \(\sin \alpha \) \( = - \sqrt {1 - {{\cos }^2}\alpha } \) \( = - \sqrt {1 - {{\left( {\frac{{11}}{{61}}} \right)}^2}} \) \( = \frac{{ - 60}}{{61}}\)
a) \(\sin \left( {\frac{\pi }{6} - \alpha } \right) \) \( = \sin \frac{\pi }{6}\cos \alpha - \cos \frac{\pi }{6}\sin \alpha \) \( = \frac{1}{2}.\frac{{11}}{{61}} - \frac{{\sqrt 3 }}{2}.\frac{{ - 60}}{{61}} \) \( = \frac{{11 + 60\sqrt 3 }}{{122}}\);
b) Ta có: \(\tan \alpha \) \( = \frac{{\sin \alpha }}{{\cos \alpha }} \) \( = \frac{{\frac{{ - 60}}{{61}}}}{{\frac{{11}}{{61}}}} \) \( = \frac{{ - 60}}{{11}}\)
\(\cot \left( {\alpha + \frac{\pi }{4}} \right) \) \( = \frac{1}{{\tan \left( {\alpha + \frac{\pi }{4}} \right)}} \) \( = \frac{{1 - \tan \alpha \tan \frac{\pi }{4}}}{{\tan \alpha + \tan \frac{\pi }{4}}} \) \( = \frac{{1 - \left( {\frac{{ - 60}}{{11}}} \right).1}}{{\left( {\frac{{ - 60}}{{11}}} \right) + 1}} \) \( = \frac{{ - 71}}{{49}}\);
c) Ta có: \(\cos 2\alpha \) \( = 2{\cos ^2}\alpha - 1 \) \( = 2.{\left( {\frac{{11}}{{61}}} \right)^2} - 1 \) \( = \frac{{ - 3479}}{{3721}}\), \(\sin 2\alpha \) \( = 2\sin \alpha \cos \alpha \) \( = 2.\frac{{11}}{{61}}.\frac{{ - 60}}{{61}} \) \( = \frac{{ - 1320}}{{3721}}\)
\(\cos \left( {2\alpha + \frac{\pi }{3}} \right) \) \( = \cos 2\alpha \cos \frac{\pi }{3} - \sin 2\alpha \sin \frac{\pi }{3} \) \( = \frac{{ - 3479}}{{3721}}.\frac{1}{2} - \frac{{ - 1320}}{{3721}}.\frac{{\sqrt 3 }}{2} \) \( = \frac{{ - 3479 + 1320\sqrt 3 }}{{7442}}\)
d) Ta có: \(\tan 2\alpha \) \( = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} \) \( = \frac{{\frac{{ - 1320}}{{3721}}}}{{\frac{{ - 3479}}{{3721}}}} \) \( = \frac{{1320}}{{3479}}\)
\(\tan \left( {\frac{{3\pi }}{4} - 2\alpha } \right) \) \( = \frac{{\tan \frac{{3\pi }}{4} - \tan 2\alpha }}{{1 + \tan \frac{{3\pi }}{4}.\tan 2\alpha }} \) \( = \frac{{ - 1 - \frac{{1320}}{{3479}}}}{{1 + \left( { - 1} \right).\frac{{1320}}{{3479}}}} \) \( = \frac{{ - 4799}}{{2159}}\)
