Giải bài 2 trang 19 sách bài tập toán 11 – Chân trời sáng tạo tập 1

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) \(\tan \left( {\frac{{3\pi }}{4} – 2\alpha } \right)\).

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: \(\tan 2\alpha  \) \( = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} \) \( = \frac{{\frac{{ – 1320}}{{3721}}}}{{\frac{{ – 3479}}{{3721}}}} \) \( = \frac{{1320}}{{3479}}\)

\(\tan \left( {\frac{{3\pi }}{4} – 2\alpha } \right) \) \( = \frac{{\tan \frac{{3\pi }}{4} – \tan 2\alpha }}{{1 + \tan \frac{{3\pi }}{4}.\tan 2\alpha }} \) \( = \frac{{ – 1 – \frac{{1320}}{{3479}}}}{{1 + \left( { – 1} \right).\frac{{1320}}{{3479}}}} \) \( = \frac{{ – 4799}}{{2159}}\)