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