22 câu trắc nghiệm Toán 11 Kết nối tri thức Bài 2: Công thức lượng giác có đáp án

Đáp án đúng là: B

Ta có $\tan \left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }} = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = 1$ suy ra $a + b = \frac{\pi }{4}.$

Đáp án đúng là: C
Từ giả thiết, ta có $\alpha + \beta + \gamma = \frac{\pi }{2} \Rightarrow \beta = \frac{\pi }{2} - \left( {\alpha + \gamma } \right).$
Suy ra $\cot \alpha + \cot \gamma = 2\cot \beta = 2.\cot \left[ {\frac{\pi }{2} - \left( {\alpha + \gamma } \right)} \right] = 2.\tan \left( {\alpha + \gamma } \right) = 2.\frac{{\tan \alpha + \tan \gamma }}{{1 - \tan \alpha .\tan \gamma }}$
Mặt khác $\frac{{\tan \alpha + \tan \gamma }}{{1 - \tan \alpha .\tan \gamma }} = \frac{{\frac{1}{{\cot \alpha }} + \frac{1}{{\cot \gamma }}}}{{1 - \frac{1}{{\cot \alpha }}.\frac{1}{{\cot \gamma }}}} = \frac{{\cot \alpha + \cot \gamma }}{{\cot \alpha .\cot \gamma - 1}}$ nên suy ra

$$\cot \alpha + \cot \gamma = 2.\frac{{\cot \alpha + \cot \gamma }}{{\cot \alpha .\cot \gamma - 1}} \Leftrightarrow \cot \alpha .\cot \gamma - 1 = 2 \Leftrightarrow \cot \alpha .\cot \gamma = 3.$$

Vì $a,b$ là các góc nhọn nên $\cos a > 0,\cos b > 0$.

Ta có $\cos a = \sqrt {1 - {{\sin }^2}a} = \frac{{15}}{{17}} \Rightarrow \tan a = \frac{{\sin a}}{{\cos a}} = \frac{8}{{15}}$;

$\cos b = \sqrt {\frac{1}{{1 + {{\tan }^2}b}}} = \frac{{12}}{{13}} \Rightarrow \sin b = \cos b\tan b = \frac{5}{{13}}{\rm{. }}$

Khi đó, $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b = \frac{8}{{17}} \cdot \frac{{12}}{{13}} - \frac{{15}}{{17}} \cdot \frac{5}{{13}} = \frac{{21}}{{221}}$.

$\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b = \frac{{15}}{{17}} \cdot \frac{{12}}{{13}} - \frac{8}{{17}} \cdot \frac{5}{{13}} = \frac{{140}}{{221}}$

$\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{\frac{8}{{15}} + \frac{5}{{12}}}}{{1 - \frac{8}{{15}} \cdot \frac{5}{{12}}}} = \frac{{171}}{{140}}.$

Đáp án: a) Đúng, b) Đúng, c) Sai, d) Sai.

Ta có $\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{2}$; $\cos 2\alpha = 2{\cos ^2}\alpha - 1 = \frac{2}{{{{\tan }^2}\alpha + 1}} - 1 = \frac{2}{5} - 1 = - \frac{3}{5}$;

$\sin 2\alpha = 2\sin \alpha \cos \alpha = 2\tan \alpha {\cos ^2}\alpha = \frac{{2\tan \alpha }}{{{{\tan }^2}\alpha + 1}} = \frac{4}{5};\tan 2\alpha = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} = - \frac{4}{3}$.

Đáp án: a) Sai, b) Đúng, c) Đúng, d) Đúng.

Vì $\frac{\pi }{2} < \alpha < \frac{{3\pi }}{2}$ nên $\cos \alpha < 0$. Ta có hệ: $\left\{ {\begin{array}{llllllllllllllllllll}{{{\sin }^2}\alpha + {{\cos }^2}\alpha = 1}\\{2\sin \alpha \cos \alpha = - \frac{4}{5}}\end{array}} \right.$

$\Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{\frac{4}{{25 {{\cos }^2}\alpha }} + {{\cos }^2}\alpha = 1}\\{ \sin \alpha = - \frac{2}{{5 \cos \alpha }}}\end{array} \Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{25{{\cos }^4}\alpha - 25{{\cos }^2}\alpha + 4 = 0}\\{\sin \alpha = - \frac{2}{{5\cos \alpha }}}\end{array}} \right.} \right.$

$\Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\cos { ^2}\alpha = \frac{4}{5}\\\cos { ^2}\alpha = \frac{1}{5}\end{array} \right.\\\sin \alpha = - \frac{2}{{5 \cos \alpha }}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\cos \alpha = \frac{{ - 2}}{{\sqrt 5 }}\\\cos \alpha = \frac{{ - 1}}{{\sqrt 5 }}\end{array} \right.\\\sin \alpha = - \frac{2}{{5 \cos \alpha }}\end{array} \right. \Rightarrow \left[ {\begin{array}{llllllllllllllllllll}{\cos \alpha = \frac{{ - 2}}{{\sqrt 5 }},\sin \alpha = \frac{1}{{\sqrt 5 }}}\\{\cos \alpha = \frac{{ - 1}}{{\sqrt 5 }},\sin \alpha = \frac{2}{{\sqrt 5 }}}\end{array}} \right.$

Đáp án: a) Đúng, b) Đúng, c) Đúng, d) Sai.

Ta có $\cos 2\alpha = \frac{5}{9},0^\circ < \alpha < 90^\circ$

${\sin ^2}\alpha = \frac{{1 - \cos 2\alpha }}{2} = \frac{{1 - {{\left( {\frac{5}{9}} \right)}^2}}}{2} = \frac{{28}}{{81}} \Rightarrow \sin \alpha = \pm \frac{{\sqrt {28} }}{9}$

Vì $0^\circ < \alpha < 90^\circ$ nên $\sin \alpha = \frac{{\sqrt {28} }}{9}$

${\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{{\sqrt {28} }}{9}} \right)^2} = \frac{{53}}{{81}} \Rightarrow \cos \alpha = \pm \frac{{\sqrt {53} }}{{81}}$

Vì $0^\circ < \alpha < 90^\circ$ nên $\cos \alpha = \frac{{\sqrt {53} }}{9}$

$\begin{array}{l}\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{2\sqrt {371} }}{{53}}\\\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{{\sqrt {371} }}{{14}}.\end{array}$

Đáp án: a) Đúng, b) Đúng, c) Sai, d) Đúng.

$\begin{array}{l}\frac{\pi }{2} < x < \pi \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \sin \frac{x}{2} > 0.\\\sin \frac{x}{2} = \sqrt {\frac{{1 - \cos x}}{2}} = \sqrt {\frac{{1 - \frac{1}{5}}}{2}} = \frac{{\sqrt {10} }}{5}\end{array}$.

$\begin{array}{l}\frac{\pi }{2} < x < \pi \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \cos \frac{x}{2} > 0.\\\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} = \sqrt {\frac{{1 + \frac{1}{5}}}{2}} = \frac{{\sqrt {15} }}{5}.\\\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\frac{{\sqrt {10} }}{5}}}{{\frac{{\sqrt {15} }}{5}}} = \frac{{\sqrt 6 }}{3}.\\\cot \frac{x}{2} = \frac{{\cos \frac{x}{2}}}{{\sin \frac{x}{2}}} = \frac{{\frac{{\sqrt {15} }}{5}}}{{\frac{{\sqrt {10} }}{5}}} = \frac{{\sqrt 6 }}{2}.\end{array}$

Đáp án: a) Sai, b) Sai, c) Đúng, d) Đúng.

Do $\frac{\pi }{2} < \alpha ,\beta < \pi \Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{\cos \alpha < 0}\\{\sin \beta > 0}\end{array}} \right.$.

Ta có $\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3};\sin \beta = \sqrt {1 - {{\cos }^2}\beta } = \sqrt {1 - \frac{4}{9}} = \frac{{\sqrt 5 }}{3}$.

Suy ra $\sin \left( {\alpha + \beta } \right) = \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) + \left( { - \frac{{2\sqrt 2 }}{3}} \right) \cdot \frac{{\sqrt 5 }}{3} = - \frac{{2 + 2\sqrt {10} }}{9}$.

Suy ra $\sin \left( {\alpha + \beta } \right) = - \frac{{2 + 2\sqrt {10} }}{9} = - \frac{{2\left( {1 + \sqrt {10} } \right)}}{9}$.

Vậy$a + b = 2 + 9 = 11$.

Đáp án: 11.

Ta có $\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7} \cdot \frac{3}{4}}} = 1$, suy ra $a + b = 45^\circ$.

Vậy $n = 45$.

Đáp án: 45.

Ta có $4\sin 3x\sin 2x\cos x = 4\left( {\sin 3x\cos x} \right)\sin 2x = 4 \cdot \frac{1}{2}\left( {\sin 4x + \sin 2x} \right)\sin 2x$

$= 2\sin 4x\sin 2x + 2{\sin ^2}2x = \left( {\cos 2x - \cos 6x} \right) + 2\left( {\frac{{1 - \cos 4x}}{2}} \right)$ $= \cos 2x - \cos 4x - \cos 6x + 1$.

Vậy $a + b + c + d = 0$.

Đáp án: 0.

Ta có $\tan \alpha = \tan \left( {\widehat {BAD} - \widehat {CAD}} \right) = \frac{{\tan \widehat {BAD} - \tan \widehat {CAD}}}{{1 + \tan \widehat {BAD}\tan \widehat {CAD}}} = \frac{{\frac{{15}}{{12}} - \frac{9}{{12}}}}{{1 + \frac{{15}}{{12}} \cdot \frac{9}{{12}}}} = \frac{8}{{31}}.$

Vì vậy $\alpha \approx 14{,}5^\circ$.

Đáp án: $14{,}5$.

$\frac{1}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha = 1 + 18 = 19$$\to {\sin ^2}\alpha = \frac{1}{{19}}$$\to \sin \alpha = \pm \frac{1}{{\sqrt {19} }}$.

Vì $\frac{\pi }{2} < \alpha < \pi$$\Rightarrow \sin \alpha > 0$$\Rightarrow \sin \alpha = \frac{1}{{\sqrt {19} }}$.

Suy ra $\tan \frac{\alpha }{2} + \cot \frac{\alpha }{2} = \frac{{{{\sin }^2}\frac{\alpha }{2} + {{\cos }^2}\frac{\alpha }{2}}}{{\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}} = \frac{2}{{\sin \alpha }} = 2\sqrt {19} \approx 8{,}72$.

Đáp án: $8{,}72$.