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

A

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

B

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

D

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

C

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

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

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

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

D

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 $\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {\frac{2}{3}}$

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

C

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

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

$= \frac{{2\sin 2a}}{{2\cos 2a}} = \tan 2a$.

C

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

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

A

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

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

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

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

$$\Leftrightarrow 2\sin \alpha = - 2$$

$\Leftrightarrow \sin \alpha = - 1$.

A

Ta có M = cos2x.cosx + sin2x.sinx = cos(2x – x) = cosx.

C

Ta có $A = \tan \alpha + \cot \alpha$$= \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }}$$= \frac{1}{{\cos \alpha .\sin \alpha }}$$= \frac{2}{{\sin 2\alpha }} = \frac{2}{{\frac{3}{4}}} = \frac{8}{3}$.

B

$\sin \frac{\alpha }{4}\cos \frac{\alpha }{4}\cos \frac{\alpha }{2}\cos a$$= \frac{1}{2}\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\cos a$$= \frac{1}{4}\sin \alpha \cos a$$= \frac{1}{8}\sin 2\alpha$$= \frac{1}{8}.\frac{1}{3} = \frac{1}{{24}}$.

D

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

Khi đó $\tan 2\alpha = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{{2.\left( { - 2} \right)}}{{1 - {{\left( { - 2} \right)}^2}}} = \frac{4}{3}$.

a) tan(a + π) = tana = 3.

b) $\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{3 + \left( { - 2} \right)}}{{1 - 3.\left( { - 2} \right)}} = \frac{1}{7}$.

c) $\cot \left( {a - b} \right) = \frac{1}{{\tan \left( {a - b} \right)}} = \frac{1}{{\frac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}}} = - 1$.

d) Ta có $0 < a < \frac{\pi }{2}$; tana = 3 Þ cosa > 0.

Có $1 + {\tan ^2}a = \frac{1}{{{{\cos }^2}a}} \Rightarrow \cos a = \frac{{\sqrt {10} }}{{10}}$; $\sin a = \tan a.\cos a = \frac{{3\sqrt {10} }}{{10}}$.

Ta có $\frac{\pi }{2} < b < \pi ;\tan b = - 2 \Rightarrow \cos b < 0$.

$1 + {\tan ^2}b = \frac{1}{{{{\cos }^2}b}} \Rightarrow \cos b = - \frac{{\sqrt 5 }}{5}$; $\sin b = \tan b.\cos b = \frac{{2\sqrt 5 }}{5}$.

Do đó $\sin \left( {a - b} \right) = \sin a.\cos b - \cos a.\sin b = \frac{{3\sqrt {10} }}{{10}}.\left( {\frac{{ - \sqrt 5 }}{5}} \right) - \frac{{\sqrt {10} }}{{10}}.\frac{{2\sqrt 5 }}{5} = - \frac{{\sqrt 2 }}{2}$.

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

a) ${\rm{ V\`i }}\frac{\pi }{2} < \alpha < \pi {\rm{ n\^e n }}\cos \alpha < 0.{\rm{ }}$

Ta có: ${\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - \frac{1}{9} = \frac{8}{9} \Rightarrow \cos \alpha = - \frac{{2\sqrt 2 }}{3}$.

b) $\sin 2\alpha = 2\sin \alpha \cos \alpha = 2 \cdot \frac{1}{3} \cdot \left( { - \frac{{2\sqrt 2 }}{3}} \right) = - \frac{{4\sqrt 2 }}{9}$.

c) $\cos 2\alpha = 1 - 2{\sin ^2}\alpha = 1 - 2{\left( {\frac{1}{3}} \right)^2} = \frac{7}{9}$.

d) $\tan 2\alpha = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} = - \frac{{4\sqrt 2 }}{7}$; $\cot 2\alpha = \frac{1}{{\tan 2\alpha }} = - \frac{{7\sqrt 2 }}{8}$.

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

a) $\cot x = - \sqrt 3 ,\frac{{3\pi }}{2} < x < 2\pi$.

$\frac{1}{{{{\sin }^2}x}} = 1 + {\cot ^2}x = 1 + {( - \sqrt 3 )^2} = 10 \Rightarrow {\sin ^2}x = \frac{1}{{10}} \Rightarrow \sin x = \pm \frac{{\sqrt {10} }}{{10}}$.

${\rm{V\`i }}\frac{{3\pi }}{2} < x < 2\pi {\rm{ n\^e n }}\sin x = - \frac{{\sqrt {10} }}{{10}}$.

b) $\cos x = \cot x \cdot \sin x = - \sqrt 3 .\left( { - \frac{{\sqrt {10} }}{{10}}} \right) = \frac{{\sqrt {30} }}{{10}}.$

c) $\sin \left( {\frac{{4\pi }}{3} - x} \right) = \sin \frac{{4\pi }}{3}\cos x - \cos \frac{{4\pi }}{3}\sin x = - \frac{{\sqrt 3 }}{2} \cdot \left( {\frac{{\sqrt {30} }}{{10}}} \right) - \frac{{ - 1}}{2} \cdot \frac{{ - \sqrt {10} }}{{10}} = \frac{{ - \sqrt {10} }}{5}$.

c) $\tan x = \frac{1}{{\cot x}} = - \frac{{\sqrt 3 }}{3}$

$\tan \left( {x + \frac{\pi }{3}} \right) = \frac{{\tan x + \tan \frac{\pi }{3}}}{{1 - \tan x \cdot \tan \frac{\pi }{3}}} = \frac{{\frac{{ - \sqrt 3 }}{3} + \sqrt 3 }}{{1 - \frac{{ - \sqrt 3 }}{3} \cdot \sqrt 3 }} = \frac{{\sqrt 3 }}{3}$.

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

a) Có ${\sin ^2}a = 1 - {\cos ^2}a = 1 - {\left( {\frac{3}{4}} \right)^2} = \frac{7}{{16}}$ mà sina > 0 nên $\sin a = \frac{{\sqrt 7 }}{4}$.

Do đó $\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\sqrt 7 }}{4}:\frac{3}{4} = \frac{{\sqrt 7 }}{3}$.

b) Có ${\cos ^2}b = 1 - {\sin ^2}b = 1 - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}}$ mà cosb < 0 Þ $\cos b = - \frac{4}{5}$.

Do đó $\cot b = \frac{{\cos b}}{{\sin b}} = - \frac{4}{5}:\frac{3}{5} = - \frac{4}{3}$.

c) cos2a + cos2b = 2cos2a – 1 + 2cos2b – 1 = $2.{\left( {\frac{3}{4}} \right)^2} + 2.{\left( {\frac{{ - 4}}{5}} \right)^2} - 2 = \frac{{81}}{{200}} \notin \left( {\frac{1}{2};1} \right)$.

d) cos(a + b) = cosacosb – sinasinb = $\frac{3}{4}.\left( {\frac{{ - 4}}{5}} \right) - \frac{{\sqrt 7 }}{4}.\frac{3}{5} = \frac{{ - 12 - 3\sqrt 7 }}{{20}} \notin \left( { - \frac{1}{2}; - \frac{1}{3}} \right)$.

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

Ta có ${\cos ^2}x = \frac{{1 + \cos 2x}}{2} = \frac{{1 + \frac{1}{7}}}{2} = \frac{4}{7}$.

Vì $\pi < x < \frac{{3\pi }}{2}$ nên $\cos x < 0 \Rightarrow \cos x = - \frac{{2\sqrt 7 }}{7}$.

a) $\cos x + \cos 3x = 2\cos 2x\cos x = 2.\frac{1}{7}.\frac{{ - 2\sqrt 7 }}{7} = \frac{{ - 4\sqrt 7 }}{{49}}$.

b) $\cos x = - \frac{{2\sqrt 7 }}{7}$.

c) ${\sin ^2}x = 1 - {\cos ^2}x = 1 - \frac{4}{7} = \frac{3}{7}$ mà sinx < 0 nên $\sin x = - \frac{{\sqrt {21} }}{7}$.

d) Có $\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{ - \sqrt {21} }}{7}:\frac{{ - 2\sqrt 7 }}{7} = \frac{{\sqrt 3 }}{2}$.

$\tan \left( {x - \frac{\pi }{4}} \right) = \frac{{\tan x - \tan \frac{\pi }{4}}}{{1 + \tan x\tan \frac{\pi }{4}}} = \frac{{\frac{{\sqrt 3 }}{2} - 1}}{{1 + \frac{{\sqrt 3 }}{2}}} = - 7 + 4\sqrt 3$.

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

2sin5xcos2x = sin(5x – 2x) + sin(5x + 2x) = sin7x.sin3x.

Suy ra k = 3.

Trả lời: 3.

A = (cosα + cosβ)2 + (sinα + sinβ)2

= cos2α + 2cosαcosβ + cos2β + sin2α + 2sinαsinβ + sin2β

= 2 + 2(cosαcosβ + sinαsinβ) = 2 + 2cos(α – β) = $2 + 2\cos \frac{\pi }{3} = 2 + 2.\frac{1}{2} = 3$.

Trả lời: 3.

$A = \frac{{\left( {\cos 10x + \cos 7x} \right) - \left( {\cos 9x + \cos 8x} \right)}}{{\left( {\sin 10x + \sin 7x} \right) - \left( {\sin 9x + \sin 8x} \right)}}$

$= \frac{{2\cos \frac{{17x}}{2}\cos \frac{{3x}}{2} - 2\cos \frac{{17x}}{2}\cos \frac{x}{2}}}{{2\sin \frac{{17x}}{2}\cos \frac{{3x}}{2} - 2\sin \frac{{17x}}{2}\cos \frac{x}{2}}}$$= \frac{{2\cos \frac{{17x}}{2}\left( {\cos \frac{{3x}}{2} - \cos \frac{x}{2}} \right)}}{{2\sin \frac{{17x}}{2}\left( {\cos \frac{{3x}}{2} - \cos \frac{x}{2}} \right)}}$$= \frac{{cos\frac{{17x}}{2}}}{{\sin \frac{{17x}}{2}}}$$= \cot \frac{{17x}}{2}$.

Suy ra m = 17; n = 2. Suy ra m + n = 19.

Trả lời: 19.

$\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = \frac{1}{2}\left[ {\cos \left( {\frac{A}{2} - \frac{B}{2}} \right) - \cos \left( {\frac{A}{2} + \frac{B}{2}} \right)} \right]\sin \frac{C}{2}$

$= \frac{1}{2}\left[ {\cos 60^\circ - \cos \left( {\frac{{180^\circ - C}}{2}} \right)} \right]\sin \frac{C}{2}$$= \frac{1}{2}\left[ {\frac{1}{2} - \sin \frac{C}{2}} \right]\sin \frac{C}{2}$$= \frac{1}{2}\left[ {\frac{1}{2}\sin \frac{C}{2} - {{\sin }^2}\frac{C}{2}} \right] = \frac{1}{{32}}$

$\Rightarrow \frac{1}{2}\sin \frac{C}{2} - {\sin ^2}\frac{C}{2} = \frac{1}{{16}}$$\Leftrightarrow \sin \frac{C}{2} = \frac{1}{4}$.

Suy ra ${\cos ^2}\frac{C}{2} = 1 - {\sin ^2}\frac{C}{2} = 1 - \frac{1}{{16}} = \frac{{15}}{{16}}$.

Có $\cos 2C = 1 - 2{\sin ^2}C = 1 - 2.{\left( {2\sin \frac{C}{2}.\cos \frac{C}{2}} \right)^2}$$= 1 - 8.{\sin ^2}\frac{C}{2}.{\cos ^2}\frac{C}{2}$$= 1 - 8.\frac{1}{{16}}.\frac{{15}}{{16}} = \frac{{17}}{{32}} \approx 0{,}53$.

Trả lời: 0,53.

$\begin{array}{l}\cos x\cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right) = \frac{1}{2}\cos x\left( {\cos \frac{{2\pi }}{3} + \cos 2x} \right)\\ = \frac{1}{2}\cos x\cos 2x - \frac{1}{4}\cos x = \frac{1}{4}(\cos 3x + \cos x) - \frac{1}{4}\cos x = \frac{1}{4}\cos 3x\end{array}$

Trả lời: 0,25