Bài 2 trang 19 SBT Toán 11 - Chân trời sáng tạo tập 1: Cho $\cos \alpha = \frac{{11}}{{61}}$ và $- \frac{\pi }{2} \(\sin \left( {\frac{\pi }{6} - \alpha } \right)$; \(\cot \left(...

Cho $\cos \alpha = \frac{{11}}{{61}}$ và \( - \frac{\pi }{2}

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

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) $\tan \left( {\alpha - \beta } \right) = \frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha .\tan \beta }}$

Vì \( - \frac{\pi }{2}

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: $\tan 2\alpha$ $= \frac{{\sin 2\alpha }}{{\cos 2\alpha }}$ $= \frac{{\frac{{ - 1320}}{{3721}}}}{{\frac{{ - 3479}}{{3721}}}}$ $= \frac{{1320}}{{3479}}$

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