22 câu trắc nghiệm Toán 11 Chân trời sáng tạo Bài 3. Các công thức lượng giác (Đúng sai - Trả lời ngắn) có đáp án

Đẳng thức đúng là: \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\].

Mệnh đề sai là: \[\sin 2a = \sin a\cos a\].

Ta có:\[\cos \left( {a + \frac{\pi }{3}} \right) = \cos a\cos \frac{\pi }{3} - \sin a\sin \frac{\pi }{3} = \frac{1}{2}\cos a - \frac{{\sqrt 3 }}{2}\sin a\].

Ta có:\[\sin \alpha  + \cos \beta  = \frac{5}{4} \Leftrightarrow {\left( {\sin \alpha  + \cos \beta } \right)^2} = {\left( {\frac{5}{4}} \right)^2} = \frac{{25}}{{16}}\]

\[ \Rightarrow {\sin ^2}\alpha  + 2\sin \alpha \cos \beta  + {\cos ^2}\beta  = \frac{{25}}{{16}}\]

\[ \Rightarrow 1 + \sin 2\alpha  = \frac{{25}}{{16}}\]

\[ \Rightarrow \sin 2\alpha  = \frac{{25}}{{16}} - 1 = \frac{9}{{16}}\].

Ta có\[\sin \alpha  = \frac{1}{{\sqrt 3 }}\], \[{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\cos ^2}\alpha  = 1 - \frac{1}{3} = \frac{2}{3}\].

Vì \(0 < \alpha  < \frac{\pi }{2}\)nên \[\cos \alpha  > 0 \Rightarrow \cos \alpha  = \sqrt {\frac{2}{3}} \]

\[ \Rightarrow \sin \left( {\alpha  + \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} + \cos \alpha \sin \frac{\pi }{3} = \frac{1}{{\sqrt 3 }}.\frac{1}{2} + \sqrt {\frac{2}{3}} .\frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{6} + \frac{{\sqrt 2 }}{2}\].

\[Q = \frac{{1 + \sin 4a - \cos 4a}}{{1 + \sin 4a + \cos 4a}} = \frac{{\sin 4a + \left( {1 - \cos 4a} \right)}}{{\sin 4a + \left( {1 + \cos 4a} \right)}}\]

\[ = \frac{{2\sin 2a\cos 2a + 2{{\sin }^2}2a}}{{2\sin 2a\cos 2a + 2{{\cos }^2}2a}} = \frac{{2\sin 2a\left( {\cos 2a + \sin 2a} \right)}}{{2\cos 2a\left( {\sin 2a + \cos 2a} \right)}}\]

\[ = \frac{{2\sin 2a}}{{2\cos 2a}} = \tan 2a\].

Do\[0 < a,b < \frac{\pi }{2} \Rightarrow 0 < a + b < \pi \]

Ta có\[\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = 1 \Leftrightarrow a + b = \frac{\pi }{4}\].

\[\frac{{\sin 2\alpha  + \sin 5\alpha  - \sin 3\alpha }}{{2{{\cos }^2}2\alpha  + \cos \alpha  - 1}} =  - 2\]

\[ \Leftrightarrow \frac{{2\sin \alpha \cos \alpha  + 2\cos 4\alpha \sin \alpha }}{{2.\frac{{1 + \cos 4\alpha }}{2} + \cos \alpha  - 1}} =  - 2\]

\[ \Leftrightarrow \frac{{2\sin \alpha \cos \alpha  + 2\cos 4\alpha \sin \alpha }}{{\cos 4\alpha  + \cos \alpha }} =  - 2\]

\[ \Leftrightarrow \frac{{2\sin \alpha \left( {\cos \alpha  + \cos 4\alpha } \right)}}{{\cos 4\alpha  + \cos \alpha }} =  - 2\]

\[ \Leftrightarrow 2\sin \alpha  =  - 2\]

\[ \Leftrightarrow \sin \alpha  =  - 1\].