Bài tập Áp dụng công thức biến đổi tổng thành tích lớp 11 (có lời giải)

Hướng dẫn giải:

Đáp án đúng là: B

Xét đáp án A

Có sin10° + sin11° + sin15° + sin16° = (sin10° + sin16°) + (sin11° + sin15°)

= 2sin13°cos3° + 2sin13°cos2°

= 2sin13°(cos3° + cos2°)

= 2sin13°×2cos2°30'cos0°30'

= 4sin13°cos2°30'cos0°30'.

Vậy đáp án A đúng

Xét đáp án B

sina + sin2a + sin3a + sin4a = (sina + sin4a) + (sin2a + sin3a)

= $2\sin \frac{5a}{2}cos\frac{3a}{2}+2\sin \frac{5a}{2}cos\frac{a}{2}$

$=2\sin \frac{5a}{2}\left(cos\frac{3a}{2}+cos\frac{a}{2}\right)$

$=2\sin \frac{5a}{2}\cdot 2\cos acos\frac{a}{2}$

$=4\sin \frac{5a}{2}\cos acos\frac{a}{2}$.

Vậy đáp án B sai

Xét đáp án C

cosa + cos2a + cos3a + cos4a = (cosa + cos4a) + (cos2a + cos3a)

Vậy đáp án C đúng

Xét đáp án D

Vậy đáp án D đúng.

Hướng dẫn giải:

Đáp án đúng là: C

P = cos(120° + x) + cos(120° − x) – cosx

= $2cos\frac{\left(120°+x\right)+\left(120°-x\right)}{2}cos\frac{\left(120°+x\right)-\left(120°-x\right)}{2}-\cos x$

$=2cos120°\cos x-\cos x$

$=2\cdot \left(\frac{-1}{2}\right)\cos x-\cos x$

$=-2\cos x$

Hướng dẫn giải:

Đáp án đúng là: C

$\frac{1}{\sin 18°}-\frac{1}{\sin 54°}=\frac{\sin 54°-\sin 18°}{\sin 18°\sin 54°}=\frac{2cos\frac{54°+18°}{2}\sin \frac{54°-18°}{2}}{\sin 18°\sin 54°}$

$=\frac{2cos36°\sin 18°}{\sin 18°\sin 54°}=\frac{2\sin 54°}{\sin 54°}=2$

Đáp án đúng là: D

\(\cos a + \cos b = 2\cos \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2}\)

Đáp án đúng là: A

A = cos3a – cosa = \( - 2\sin \frac{{3a + a}}{2} \cdot \sin \frac{{3a - a}}{2} = 2\sin 2a \cdot \sin a\).

Đáp án đúng là: D

\(\sin a - \sin b = 2\cos \frac{{a + b}}{2} \cdot \sin \frac{{a - b}}{2}\).

Đáp án đúng là: B

\(\sin \alpha - 1 = \sin \alpha - \sin \frac{\pi }{2} = 2\cos \left( {\frac{{\alpha + \frac{\pi }{2}}}{2}} \right) \cdot \sin \left( {\frac{{\alpha - \frac{\pi }{2}}}{2}} \right) = 2\cos \left( {\frac{\alpha }{2} + \frac{\pi }{4}} \right) \cdot \sin \left( {\frac{\alpha }{2} - \frac{\pi }{4}} \right)\)

Đáp án đúng là: D

\(P = \frac{{\cos a - \cos 5a}}{{\sin 4a + \sin 2a}} = \frac{{ - 2\sin 3a \cdot \sin \left( { - 2a} \right)}}{{2\sin 3a \cdot \cos a}} = \frac{{\sin 2a}}{{\cos a}} = \frac{{2\sin a \cdot \cos a}}{{\cos a}} = 2\sin a\).

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

a) Sai. \(\cos 3x + \cos x = 2\cos \frac{{3x + x}}{2} \cdot \cos \frac{{3x - x}}{2} = 2\cos 2x \cdot \cos x\).

b) Sai. \(\sin 3x + \sin 2x = 2\sin \frac{{3x + 2x}}{2} \cdot \cos \frac{{3x - 2x}}{2} = 2\sin \frac{{5x}}{2} \cdot \cos \frac{x}{2}\).

c) Đúng. \(\cos 4x - \cos x = - 2\sin \frac{{4x + x}}{2} \cdot \sin \frac{{4x - x}}{2} = - 2\sin \frac{{5x}}{2} \cdot \sin \frac{{3x}}{2}\).

d) Đúng. \(\sin 5x - {\mathop{\rm s}\nolimits} {\rm{inx}} = 2\cos \frac{{5x + x}}{2} \cdot \sin \frac{{5x - x}}{2} = 2\cos 3x \cdot \sin 2x\).

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

a) Đúng. sin2x = 2sinx∙cosx.

b) Sai. \(\cos a - \cos b = - 2\sin \frac{{a + b}}{2} \cdot \sin \frac{{a - b}}{2}\).

c) Sai. sina + sin2a + sin3a + sin4a + sin5a + sin6a

= (sina + sin6a) + (sin2a + sin5a) + (sin3a + sin4a)

\( = 2\sin \frac{{7a}}{2} \cdot \cos \frac{{5a}}{2} + 2\sin \frac{{7a}}{2} \cdot \cos \frac{{3a}}{2} + 2\sin \frac{{7a}}{2} \cdot \cos \frac{a}{2}\)

\( = 2\left( {\cos \frac{{5a}}{2} + \cos \frac{{3a}}{2} + \cos \frac{a}{2}} \right) \cdot \sin \frac{{7a}}{2} = 2\left( {2\cos \frac{{3a}}{2} \cdot \cos a + \cos \frac{{3a}}{2}} \right) \cdot \sin \frac{{7a}}{2}\)

\( = 2\left( {2\cos a + 1} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2} = 4\left( {\cos a + \frac{1}{2}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}\)

\( = 4\left( {\cos a + \cos \frac{\pi }{3}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}\)

\( = 8\cos \left( {\frac{a}{2} + \frac{\pi }{6}} \right) \cdot \cos \left( {\frac{a}{2} - \frac{\pi }{6}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}\).

d) Đúng.

Vế trái:

\(\frac{{\sin A + \sin B}}{{\cos A + \cos B}} = \frac{{2\sin \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2}}}{{2\cos \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2}}} = \frac{{2\sin \frac{{A + B}}{2}}}{{2\cos \frac{{A + B}}{2}}} = \tan \frac{{A + B}}{2} = \frac{{\sin \frac{{A + B}}{2}}}{{\cos \frac{{A + B}}{2}}}\).

Vế phải:

\(\frac{1}{2}\left( {\tan A + \tan B} \right) = \frac{1}{2} \cdot \left( {\frac{{\sin A}}{{\cos A}} + \frac{{\sin B}}{{\cos B}}} \right) = \frac{1}{2}\left( {\frac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}} \right)\)

\( = \frac{1}{2}\frac{{\sin \left( {A + B} \right)}}{{\cos A\cos B}} = \frac{{\sin \frac{{A + B}}{2}\cos \frac{{A + B}}{2}}}{{\cos A\cos B}}\).

Khi đó \(\frac{{\sin \frac{{A + B}}{2}}}{{\cos \frac{{A + B}}{2}}} = \frac{{\sin \frac{{A + B}}{2}\cos \frac{{A + B}}{2}}}{{\cos A\cos B}}\)\( \Leftrightarrow \cos A\cos B = {\cos ^2}\frac{{A + B}}{2}\)

\( \Leftrightarrow \frac{1}{2}\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right] = \frac{{1 + \cos \left( {A + B} \right)}}{2}\)\( \Leftrightarrow \cos \left( {A - B} \right) = 1 \Leftrightarrow A - B = 0 \Leftrightarrow A = B\) hay \(\widehat A = \widehat B\) ⇔ tam giác ABC cân tại C.

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

a) Sai. \(\frac{{\sin 2x + \sin 4x + \sin 6x}}{{2\left( {1 - \cos x} \right)}}\)

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

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

b) Đúng. \(\frac{{1 + \cos x + \cos 2x + \cos 3x}}{{2{{\cos }^2}x + \cos x - 1}}\)

\( = \frac{{\left( {1 + \cos 2x} \right) + \left( {\cos x + \cos 3x} \right)}}{{\left( {2{{\cos }^2}x - 1} \right) + \cos x}} = \frac{{2{{\cos }^2}x + 2\cos 2x \cdot \cos x}}{{\cos x + \cos 2x}}\)

\( = \frac{{2\cos x \cdot \left( {\cos x + \cos 2x} \right)}}{{\cos x + \cos 2x}} = 2\cos x\).

c) Sai. \(\frac{{\cos x - \cos 2x + \cos 3x}}{{{\mathop{\rm s}\nolimits} {\rm{inx}} - \sin 2x + \sin 3x}}\)

\( = \frac{{\left( {\cos x + \cos 3x} \right) - \cos 2x}}{{\left( {{\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 3x} \right) - \sin 2x}} = \frac{{2\cos 2x\cos x - \cos 2x}}{{2\sin 2x\cos x - \sin 2x}} = \frac{{\cos 2x\left( {2\cos x - 1} \right)}}{{\sin 2x\left( {2\sin x - 1} \right)}} = \cot 2x\)

d) Đúng. \(D = \sin a + \sin b + \sin c - \sin \left( {a + b + c} \right)\)

\( = \left( {\sin a + \sin b} \right) - \left[ {\sin \left( {a + b + c} \right) - \sin c} \right]\)

\( = 2\sin \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} - \left[ {2\cos \frac{{\left( {c + a + b + c} \right)}}{2} \cdot \sin \frac{{\left( {a + b + c - c} \right)}}{2}} \right]\)

\( = 2\sin \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} - 2\cos \frac{{a + b + 2c}}{2} \cdot \sin \frac{{a + b}}{2}\)

\( = 2\sin \frac{{a + b}}{2} \cdot \left( {\cos \frac{{a - b}}{2} - \cos \frac{{a + b + 2c}}{2}} \right)\)

\( = 2\sin \frac{{a + b}}{2}\left[ { - 2 \cdot \sin \frac{{\frac{{a - b + a + b + 2c}}{2}}}{2} \cdot \sin \frac{{\frac{{a - b - 2c - a - b}}{2}}}{2}} \right]\)

\( = 2\sin \frac{{a + b}}{2} \cdot \left( { - 2} \right) \cdot \sin \frac{{a + c}}{2} \cdot \sin \frac{{ - \left( {b + c} \right)}}{2}\)

\( = 4\sin \frac{{a + b}}{2} \cdot \sin \frac{{b + c}}{2} \cdot \sin \frac{{c + a}}{2}\).

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

a) Sai. sina − sin2a + sin3a = (sina + sin3a) – sin2a

\( = 2\sin \frac{{a + 3a}}{2} \cdot \cos \frac{{a - 3a}}{2} - \sin 2a = 2\sin 2a\cos a - \sin 2a\)

\( = \sin 2a \cdot \left( {2\cos a - 1} \right)\).

b) Sai. sinx + 2sin3x + sin5x = (sinx + sin5x) + 2sin3x

\( = 2\sin \frac{{x + 5x}}{2} \cdot \cos \frac{{x - 5x}}{2} + 2\sin 3x = 2\sin 3x \cdot \cos 2x + 2\sin 3x = 2\sin 3x \cdot \left( {\cos 2x + 1} \right)\)

\( = 2\sin 3x \cdot 2{\cos ^2}x = 4\sin 3x \cdot {\cos ^2}x\).

c) Sai. 2cosx – cos3x – cos5x = 2cosx – (cos3x + cos5x)

\( = 2\cos x - 2\cos \frac{{3x + 5x}}{2} \cdot \cos \frac{{3x - 5x}}{2} = 2\cos x - 2\cos 4x \cdot \cos x = 2\cos x\left( {1 - \cos 4x} \right)\)

\( = 2\cos x \cdot 2{\sin ^2}2x = 4\cos x \cdot {\sin ^2}2x\).

d) Sai. \[\frac{{\cos 5x - \cos x}}{{\sin 4x + \sin 2x}} = \frac{{ - 2\sin \frac{{5x + x}}{2} \cdot \sin \frac{{5x - x}}{2}}}{{2\sin \frac{{4x + 2x}}{2} \cdot \cos \frac{{4x - 2x}}{2}}}\]

\[ = \frac{{ - 2\sin 3x \cdot \sin 2x}}{{2\sin 3x \cdot \cos x}} = - \frac{{2\sin x\cos x}}{{\cos x}} = - 2\sin x\].

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

a) Đúng. cosa + cos2a + cos3a + cos4a = (cosa + cos3a) + (cos2a + cos4a)

= 2cos2acosa + 2cos3acosa = 2(cos2a + cos3a)cosa = \(4\cos \frac{{5a}}{2} \cdot \cos \frac{a}{2} \cdot \cos a\).

b) Đúng. 1 + sina + cos3a – (cosa + sin2a + cos2a)

= (1 − cos2a) + (cos3a – cosa) + (sina – sin2a)

= 2sin²a – 2sin2asina + (sina – sin2a)

= 2sina(sina – sin2a) + (sina – sin2a) = (sina – sin2a)(2sina +1)

= \(2\cos \frac{{3a}}{2}\sin \left( {\frac{{ - a}}{2}} \right) \cdot \left( {2\sin a + 1} \right)\).

c) Đúng. cosa + cosb + cosc + cos(a + b + c)

= (cosa + cosb) + [cosc + cos(a + b +c )]

\( = 2\cos \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} + 2\cos \frac{{c + a + b + c}}{2} \cdot \cos \frac{{c + a + b - c}}{2}\)

\( = 2\cos \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} + 2\cos \frac{{a + b + 2c}}{2} \cdot \cos \frac{{a + b}}{2}\)

\( = 2\cos \frac{{a + b}}{2} \cdot \left( {\cos \frac{{a - b}}{2} + \cos \frac{{a + b + 2c}}{2}} \right) = 2\cos \frac{{a + b}}{2} \cdot 2\cos \frac{{b + c}}{2} \cdot \cos \frac{{c + a}}{2}\)

\( = 4\cos \frac{{a + b}}{2} \cdot \cos \frac{{b + c}}{2} \cdot \cos \frac{{c + a}}{2}\).

d) Sai. sin x + sin(x + a) + sin(x + 2a)

= [sin(x + 2a) + sinx] + sin(x + a) = \(2\sin \frac{{x + 2a + x}}{2} \cdot \cos \frac{{x + 2a - x}}{2} + \sin (x + a)\)

\( = 2\sin \frac{{2x + 2a}}{2} \cdot \cos \frac{{x + 2a - x}}{2} + \sin (x + a) = 2\sin (x + a)\cos a + \sin (x + a)\)

= (2cosa + 1)sin(x + a).

Đáp án đúng là: 3

\(P = \frac{{\cos \alpha + 2\cos 3\alpha + \cos 5\alpha }}{{\sin \alpha + 2\sin 3\alpha + \sin 5\alpha }} = \frac{{\left( {\cos \alpha + \cos 5\alpha } \right) + 2\cos 3\alpha }}{{\left( {\sin \alpha + \sin 5\alpha } \right) + 2\sin 3\alpha }} = \frac{{2\cos 3\alpha \cdot \cos 2\alpha + 2\cos 3\alpha }}{{2\sin 3\alpha \cdot \cos 2\alpha + 2\sin 3\alpha }}\)

\( = \frac{{2\cos 3\alpha \cdot \left( {\cos 2\alpha + 1} \right)}}{{2\sin 3\alpha \cdot \left( {\cos 2\alpha + 1} \right)}} = \cot 3\alpha \).

Vậy a = 3.

Đáp án: 2

\(P = \frac{{\sin \alpha + \sin 2\alpha + \sin 3\alpha }}{{\cos \alpha + \cos 2\alpha + \cos 3\alpha }} = \frac{{\left( {\sin \alpha + \sin 3\alpha } \right) + \sin 2\alpha }}{{\left( {\cos 2\alpha + \cos 3\alpha } \right) + \cos 2\alpha }} = \frac{{2\sin 2\alpha \cdot \cos \alpha + \sin 2\alpha }}{{2\cos 2\alpha \cdot \cos \alpha + \cos 2\alpha }}\)

\( = \frac{{\sin 2\alpha \cdot \left( {2\cos \alpha + 1} \right)}}{{\cos 2\alpha \cdot \left( {2\cos \alpha + 1} \right)}} = \tan 2\alpha \).

Vậy k = 2.

Đáp án: 6

B = sina + sin2a + sin3a + sin4a = (sina + sin3a) + (sin2a + sin4a)

\( = 2\sin \frac{{a + 3a}}{2} \cdot \cos \frac{{a - 3a}}{2} + 2\sin \frac{{2a + 4a}}{2} \cdot \cos \frac{{2a - 4a}}{2} = 2\sin 2a\cos a + 2\sin 3a\cos a\)

\( = 2\cos a\left( {\sin 2a + \sin 3a} \right) = 2\cos a \cdot 2\sin \frac{{5a}}{2} \cdot \cos \frac{a}{2} = 4\cos a \cdot \sin \frac{{5a}}{2} \cdot \cos \frac{a}{2}\).

Vậy m + n = 6.

Đáp án: 6

\(C = {\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x - \sin \left( {x - \frac{\pi }{6}} \right) + \cos \left( {x - \frac{\pi }{6}} \right)\)

\( = \sqrt 2 \cos \left( {x - \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{3}} \right) + \cos \left( {x - \frac{\pi }{6}} \right)\)

\( = \sqrt 2 \cos \left( {x - \frac{\pi }{4}} \right) + 2\cos \left( {x + \frac{\pi }{{12}}} \right) \cdot \cos \frac{\pi }{4}\)

\( = \sqrt 2 \left[ {\cos \left( {x - \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{{12}}} \right)} \right] = 2\sqrt 2 \cos \left( {x - \frac{\pi }{{12}}} \right) \cdot \cos \frac{\pi }{6}\)

\( = \sqrt 6 \cos \left( {x - \frac{\pi }{{12}}} \right)\).

Vậy a = 6.

Đáp án: 2

\(M = \cos \left( {x + \frac{\pi }{4}} \right) - \cos \left( {x - \frac{\pi }{4}} \right) = - 2\sin \left( {\frac{{x + \frac{\pi }{4} + x - \frac{\pi }{4}}}{2}} \right) \cdot \sin \left( {\frac{{x + \frac{\pi }{4} - x + \frac{\pi }{4}}}{2}} \right)\)

\( = - 2\sin x \cdot \sin \frac{\pi }{4} = - \sqrt 2 {\mathop{\rm s}\nolimits} {\rm{inx}}\).

Vậy a = 2.