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

Hướng dẫn giải:

Đáp án đúng là: B

$\sin a\cdot \sin b=\frac{1}{2}[cos(a-b)-cos(a+b)]$.

Hướng dẫn giải:

Đáp án đúng là: A

$\cos 3x\cdot \cos 5x=\frac{1}{2}(cos8x+cos2x)$

$=\frac{1}{2}[cos(-2x)+cos8x]$ $=\frac{1}{2}(cos2x+cos8x)$.

Hướng dẫn giải:

Đáp án đúng là: B

$A = cos\frac{{2\pi }}{7} + cos\frac{{4\pi }}{7} + cos\frac{{6\pi }}{7}$$= \frac{{\sin \frac{\pi }{7}\left( {cos\frac{{2\pi }}{7} + cos\frac{{4\pi }}{7} + cos\frac{{6\pi }}{7}} \right)}}{{\sin \frac{\pi }{7}}}$

$$= \frac{{\sin \frac{\pi }{7}cos\frac{{2\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{4\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{6\pi }}{7}}}{{\sin \frac{\pi }{7}}}$$

Ta có $\sin \frac{\pi }{7}cos\frac{{2\pi }}{7}$$= \frac{1}{2}\left[ {sin\left( {\frac{\pi }{7} - \frac{{2\pi }}{7}} \right) + \sin \left( {\frac{\pi }{7} + \frac{{2\pi }}{7}} \right)} \right]$

$= \frac{1}{2}\left[ {sin\left( { - \frac{\pi }{7}} \right) + \sin \frac{{3\pi }}{7}} \right]$$= \frac{1}{2}\left( { - sin\frac{\pi }{7} + \sin \frac{{3\pi }}{7}} \right)$.

$\sin \frac{\pi }{7}cos\frac{{4\pi }}{7}$$= \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{7} - \frac{{4\pi }}{7}} \right) + \sin \left( {\frac{\pi }{7} + \frac{{4\pi }}{7}} \right)} \right]$

$= \frac{1}{2}\left[ {\sin \left( { - \frac{{3\pi }}{7}} \right) + \sin \frac{{5\pi }}{7}} \right]$$= \frac{1}{2}\left( { - \sin \frac{{3\pi }}{7} + \sin \frac{{5\pi }}{7}} \right)$.

$\sin \frac{\pi }{7}cos\frac{{6\pi }}{7}$$= \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{7} - \frac{{6\pi }}{7}} \right) + \sin \left( {\frac{\pi }{7} + \frac{{6\pi }}{7}} \right)} \right]$

$= \frac{1}{2}\left[ {\sin \left( { - \frac{{5\pi }}{7}} \right) + \sin \pi } \right]$$= \frac{1}{2}\left( { - \sin \frac{{5\pi }}{7} + \sin \pi } \right)$.

Do đó $\sin \frac{\pi }{7}cos\frac{{2\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{4\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{6\pi }}{7}$

$= \frac{1}{2}\left( { - sin\frac{\pi }{7} + \sin \frac{{3\pi }}{7}} \right)$$+ \frac{1}{2}\left( { - \sin \frac{{3\pi }}{7} + \sin \frac{{5\pi }}{7}} \right)$$+ \frac{1}{2}\left( { - \sin \frac{{5\pi }}{7} + \sin \pi } \right)$

$= \frac{1}{2}\left( { - sin\frac{\pi }{7} + \sin \pi } \right)$$= - \frac{1}{2}sin\frac{\pi }{7}$.

Do đó $A = \frac{{ - \frac{1}{2}sin\frac{\pi }{7}}}{{sin\frac{\pi }{7}}} = - \frac{1}{2}$.

Hướng dẫn giải:

Đáp án đúng là: A

$\tan \frac{\pi }{{24}} + \tan \frac{{7\pi }}{{24}}$$= \frac{{\sin \frac{\pi }{{24}}}}{{cos\frac{\pi }{{24}}}} + \frac{{\sin \frac{{7\pi }}{{24}}}}{{cos\frac{{7\pi }}{{24}}}}$$= \frac{{\sin \frac{\pi }{{24}}cos\frac{{7\pi }}{{24}} + cos\frac{\pi }{{24}}\sin \frac{{7\pi }}{{24}}}}{{cos\frac{\pi }{{24}}cos\frac{{7\pi }}{{24}}}}$

$= \frac{{\sin \left( {\frac{\pi }{{24}} + \frac{{7\pi }}{{24}}} \right)}}{{\frac{1}{2}\left[ {cos\left( {\frac{\pi }{{24}} - \frac{{7\pi }}{{24}}} \right) + cos\left( {\frac{\pi }{{24}} + \frac{{7\pi }}{{24}}} \right)} \right]}}$$= \frac{{\sin \frac{\pi }{3}}}{{\frac{1}{2}\left[ {cos\left( { - \frac{\pi }{4}} \right) + cos\frac{\pi }{3}} \right]}}$

$= \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{1}{2}\left[ {\frac{{\sqrt 2 }}{2} + \frac{1}{2}} \right]}}$$= \frac{{2\sqrt 3 }}{{\sqrt 2 + 1}}$$= 2\sqrt 3 \left( {\sqrt 2 - 1} \right)$$= 2\left( {\sqrt 6 - \sqrt 3 } \right)$.

Hướng dẫn giải:

Đáp án đúng là: D

$P = cos(a + b)cos(a - b)$$= \frac{1}{2}\left\{ {cos\left[ {(a + b) - (a - b)} \right] + cos\left[ {(a + b) + (a - b)} \right]} \right\}$

$= \frac{1}{2}\left[ {cos(2b) + cos(2a)} \right]$$= \frac{1}{2}\left[ {2{{\cos }^2}a - 1 + 2{{\cos }^2}b - 1} \right]$$= {\cos ^2}a + {\cos ^2}b - 1$

$= {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{4}} \right)^2} - 1$$= - \frac{{119}}{{144}}$.

Hướng dẫn giải:

Đáp án đúng là: C

cos10°×cos30°×cos50°×cos70°

= $cos10^\circ \cdot cos30^\circ \cdot \frac{1}{2}\left[ {cos\left( {50^\circ - 70^\circ } \right) + cos\left( {50^\circ + 70^\circ } \right)} \right]$

$= cos10^\circ \cdot \frac{{\sqrt 3 }}{2} \cdot \frac{1}{2}\left( {cos20^\circ + cos120^\circ } \right)$

$= cos10^\circ \cdot \frac{{\sqrt 3 }}{2} \cdot \frac{1}{2}\left( {cos20^\circ - \frac{1}{2}} \right)$

$= \frac{{\sqrt 3 }}{4}cos10^\circ \left( {cos20^\circ - \frac{1}{2}} \right)$

$= \frac{{\sqrt 3 }}{4}cos10^\circ cos20^\circ - \frac{{\sqrt 3 }}{8}cos10^\circ$

$= \frac{{\sqrt 3 }}{4} \cdot \frac{1}{2}\left[ {cos\left( {10^\circ - 20^\circ } \right) + cos\left( {10^\circ + 20^\circ } \right)} \right] - \frac{{\sqrt 3 }}{8}cos10^\circ$$= \frac{{\sqrt 3 }}{4} \cdot \frac{1}{2}\left( {cos10^\circ + cos30^\circ } \right) - \frac{{\sqrt 3 }}{8}cos10^\circ$

$= \frac{3}{{16}}$.

Hướng dẫn giải:

Đáp án đúng là: A

$A = \frac{1}{{2\sin 10^\circ }} - 2\sin 70^\circ$$= \frac{{1 - 4\sin 70^\circ \sin 10^\circ }}{{2\sin 10^\circ }}$

$= \frac{{1 - 4 \cdot \frac{1}{2}\left[ {cos\left( {70^\circ - 10^\circ } \right) - cos\left( {70^\circ + 10^\circ } \right)} \right]}}{{2\sin 10^\circ }}$

$= \frac{{1 - 4 \cdot \frac{1}{2}\left( {cos60^\circ - cos80^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{1 - 4 \cdot \frac{1}{2}\left( {\frac{1}{2} - cos80^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{1 - \left( {1 - 2cos80^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{2cos80^\circ }}{{2\sin 10^\circ }}$

$= \frac{{2cos\left( {90^\circ - 10^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{2\sin 10^\circ }}{{2\sin 10^\circ }} = 1$.

Hướng dẫn giải:

Đáp án đúng là: D

$cos2x \cdot \sin 5x \cdot cos3x = cos2x \cdot \frac{1}{2} \cdot \left[ {\sin \left( {5x - 3x} \right) + \sin \left( {5x + 3x} \right)} \right]$

$= cos2x \cdot \frac{1}{2} \cdot \left( {\sin 2x + \sin 8x} \right)$

$= \frac{1}{2}cos2x\sin 2x + \frac{1}{2}cos2x\sin 8x$

$= \frac{1}{4}\sin 4x + \frac{1}{4}\left[ {\sin \left( {8x - 2x} \right) + \sin \left( {8x + 2x} \right)} \right]$

$= \frac{1}{4}\sin 4x + \frac{1}{4}\left( {\sin 6x + \sin 10x} \right)$

$= \frac{{\sin 10x + \sin 6x + \sin 4x}}{4}$.

Hướng dẫn giải:

Đáp án đúng là: B

$2\sin (\frac{π}{4}+α)\sin (\frac{π}{4}-α)$

$=2\cdot \frac{1}{2}\{cos[(\frac{π}{4}+α)-(\frac{π}{4}-α)]-cos[(\frac{π}{4}+α)+(\frac{π}{4}-α)]\}$

$=2\cdot \frac{1}{2}(cos2α-cos\frac{π}{2})$

$=cos2α$

Hướng dẫn giải:

Đáp án đúng là: C

$A=cos\frac{π}{12}\cdot cos\frac{5π}{12}=\frac{1}{2}[cos(\frac{π}{12}-\frac{5π}{12})+cos(\frac{π}{12}+\frac{5π}{12})]$

$=\frac{1}{2}[cos(-\frac{π}{3})+cos\frac{π}{2}]$

$=\frac{1}{2}(cos\frac{π}{3}+cos\frac{π}{2})$

$=\frac{1}{4}$

Đáp án đúng là: B

M = cos(a + b)cos(a – b) – sin(a + b)sin(a −b)

$= \frac{1}{2}\left[ {\cos \left( {a + b + a - b} \right) + \cos \left( {a + b - a + b} \right)} \right] - \frac{1}{2}\left[ {\cos \left( {a + b - a + b} \right) - \cos \left( {a + b + a - b} \right)} \right]$

$= \frac{1}{2}\left( {\cos 2a + \cos 2b} \right) - \frac{1}{2}\left( {\cos 2b - \cos 2a} \right) = \frac{1}{2}\left( {\cos 2a + \cos 2b - \cos 2b + \cos 2a} \right)$

= $\frac{1}{2} \cdot 2 \cdot \cos 2a = \cos 2a = 1 - 2{\sin ^2}a$.

Đáp án đúng là: C

$\cos \frac{{3\alpha }}{2}\cos \frac{\alpha }{2} = \frac{1}{2}\left[ {\cos \left( {\frac{{3\alpha }}{2} + \frac{\alpha }{2}} \right) + \cos \left( {\frac{{3\alpha }}{2} - \frac{\alpha }{2}} \right)} \right] = \frac{1}{2}\left( {\cos 2\alpha + \cos \alpha } \right)$

$= \frac{1}{2}\left( {2{{\cos }^2}\alpha - 1 + \cos \alpha } \right) = \frac{1}{2}\left( {2 \cdot \frac{9}{{16}} - 1 + \frac{3}{4}} \right) = \frac{7}{{16}}$.

Đáp án đúng là: A

$\cos x\cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right) = \cos x \cdot \left[ {\cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right)} \right]$

$= \cos x \cdot \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{3} - x + \frac{\pi }{3} + x} \right) + \cos \left( {\frac{\pi }{3} - x - \frac{\pi }{3} - x} \right)} \right] = \cos x \cdot \frac{1}{2}\left( {\cos \frac{{2\pi }}{3} + \cos 2x} \right)$

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

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

Đáp án đúng là: D

$\sin a\cos b = \frac{1}{2}\left[ {\sin \left( {a - b} \right) + \sin \left( {a + b} \right)} \right]$.

Đáp án đúng là: D

$\cos 2\alpha \cdot \sin 5\alpha = \frac{1}{2}\left( {\sin 7\alpha + \sin 3\alpha } \right)$

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

a) Sai. sin$^{2}$α + cos$^{2}$α = 1 nên ${\sin ^2}\alpha = 1 - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} = \frac{2}{3}$.

Do α ∈ $\left( {\frac{{ - \pi }}{2};0} \right)$ nên $\sin \alpha = \frac{{ - \sqrt 6 }}{3}$.

b) Đúng. $\sin \left( {\alpha + \frac{\pi }{4}} \right) = \sin \alpha \cdot \cos \frac{\pi }{4} + \cos \alpha \cdot \sin \frac{\pi }{4} = \frac{{ - \sqrt 6 }}{3} \cdot \frac{{\sqrt 2 }}{2} + \frac{1}{{\sqrt 3 }} \cdot \frac{{\sqrt 2 }}{2}$

= $\frac{{\sqrt 6 \cdot \left( {1 - \sqrt 2 } \right)}}{6}$.

c) Sai. $A = \sin 3\alpha \cdot \cos \alpha - \frac{1}{4}\sin 4\alpha = \left( {\sin 3\alpha \cdot \cos \alpha } \right) - \frac{1}{4}\sin 4\alpha$

= $\frac{1}{2}\left[ {\sin \left( {3\alpha + \alpha } \right) + \sin \left( {3\alpha - \alpha } \right)} \right] - \frac{1}{4}\sin 4\alpha = \frac{1}{2}\left( {\sin 4\alpha + \sin 2\alpha } \right) - \frac{1}{4}\sin 4\alpha$

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

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

= $2 \cdot {\left( {\frac{1}{{\sqrt 3 }}} \right)^3} \cdot \left( {\frac{{ - \sqrt 6 }}{3}} \right) = \frac{{ - 2\sqrt 2 }}{9}$.

d) Đúng. $B = \sin \left( {\alpha + \frac{\pi }{6}} \right) \cdot \cos \left( {\alpha - \frac{\pi }{6}} \right) - \frac{1}{2} \cdot \sin 2\alpha$

= $\frac{1}{2} \cdot \sin 2\alpha + \frac{1}{2} \cdot \sin \frac{\pi }{3} - \frac{1}{2} \cdot \sin 2\alpha = \frac{1}{2} \cdot \sin \frac{\pi }{3} = \frac{1}{2} \cdot \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{4}$.

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

a) Đúng. $\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{3}$.

b) Sai. Ta có $A = \frac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha = 10$.

c) Đúng. Do tanα = 3 nên cosα ≠ 0.

$B = \frac{{\sin \alpha + 5\cos \alpha }}{{3\sin \alpha + 6\cos \alpha }} = \frac{{\frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{5\cos \alpha }}{{\cos \alpha }}}}{{\frac{{3\sin \alpha }}{{\cos \alpha }} + \frac{{6\cos \alpha }}{{\cos \alpha }}}} = \frac{{\tan \alpha + 5}}{{3\tan \alpha + 6}} = \frac{{3 + 5}}{{3 \cdot 3 + 6}} = \frac{8}{{15}}$.

d) Đúng. C = $\frac{{\sin 3\alpha \cdot \cos 2\alpha + \sin \alpha \cdot \cos 6\alpha }}{{\sin 4\alpha }}$

= $\frac{{\frac{1}{2}\left[ {\sin \left( {3\alpha + 2\alpha } \right) + \sin \left( {3\alpha - 2\alpha } \right)} \right] + \frac{1}{2}\left[ {\sin \left( {\alpha + 6\alpha } \right) - \sin \left( {\alpha - 6\alpha } \right)} \right]}}{{\sin 4\alpha }}$

$= \frac{{\frac{1}{2}\left( {\sin 5\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha } \right) + \frac{1}{2}\left( {\sin 7\alpha - \sin 5\alpha } \right)}}{{\sin 4\alpha }} = \frac{{\frac{1}{2}\left( {\sin \alpha + \sin 7\alpha } \right)}}{{\sin 4\alpha }} = \frac{{{\mathop{\rm s}\nolimits} {\rm{in}}\alpha + \sin 7\alpha }}{{2\sin 4\alpha }}$

= $\frac{{\left( {\sin 7\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha } \right) \cdot \cos 3\alpha }}{{2\sin 4\alpha \cdot \cos 3\alpha }} = \frac{{\left( {\sin 7\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha } \right) \cdot \cos 3\alpha }}{{\sin 7\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha }} = \cos 3\alpha$.

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

a) Sai. Ta có: $\sin \left( {x - \frac{\pi }{6}} \right) = {\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos \frac{\pi }{6} - \cos x \cdot \sin \frac{\pi }{6} = \frac{{\sqrt 3 }}{2} \cdot {\mathop{\rm s}\nolimits} {\rm{inx}} - \frac{1}{2} \cdot \cos x$.

b) Đúng. $4\sin x \cdot \sin 2x \cdot \sin 3x = 4 \cdot \left( {{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \sin 2x} \right) \cdot \sin 3x$

= $4 \cdot \frac{1}{2} \cdot \left[ {\cos \left( {x - 2x} \right) - \cos \left( {x + 2x} \right)} \right] \cdot \sin 3x = 2 \cdot \left( {\cos x - \cos 3x} \right) \cdot \sin 3x$

= $2 \cdot \cos x \cdot \sin 3x - 2 \cdot \cos 3x \cdot \sin 3x$

= $2 \cdot \frac{1}{2}\left[ {\sin \left( {x + 3x} \right) - \sin \left( {x - 3x} \right)} \right] - 2 \cdot \frac{1}{2}\left[ {\sin \left( {3x + 3x} \right) - \sin \left( {3x - 3x} \right)} \right]$

= $\sin 4x + \sin 2x - \sin 6x$.

c) Sai. $2\sqrt 2 {\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos \left( {x - \frac{\pi }{4}} \right) = 2\sqrt 2 {\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \left( {\cos x \cdot \cos \frac{\pi }{4} + {\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \sin \frac{\pi }{4}} \right)$

= $2\sqrt 2 {\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \left( {\cos x \cdot \frac{{\sqrt 2 }}{2} + {\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \frac{{\sqrt 2 }}{2}} \right)$

= 2sinx∙cosx + 2sinx∙sinx = 2sinx(sinx + cosx).

d) Sai. 4sin3x∙sin2x∙sinx = −2sin2x∙(cos4x – cos2x)

= −2cos4x∙sin2x + 2sin2x∙cos2x = −(sin6x – sin2x) + sin4x

= −sin6x + sin4x + sin2x

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

a) Sai. 2cos2x∙cos3x = 2∙$\frac{1}{2}$[cos(2x – 3x) + cos(2x + 3x)] = cos(x) + cos5x.

b) Sai. 2sin2x∙cos$\frac{x}{2}$ = $\frac{1}{2} \cdot 2\left[ {\sin \left( {2x + \frac{x}{2}} \right) + \sin \left( {2x - \frac{x}{2}} \right)} \right] = \sin \frac{{5x}}{2} + \sin \frac{{3x}}{2}$.

c) Đúng. $- 2\sin \frac{{5x}}{2} \cdot \sin \frac{{3x}}{2} = - 2 \cdot \frac{1}{2}\left[ {\cos \left( {\frac{{5x}}{2} - \frac{{3x}}{2}} \right) - \cos \left( {\frac{{5x}}{2} + \frac{{3x}}{2}} \right)} \right]$

= −cosx + cos4x.

d) Đúng. 2cos3x∙sin2x = $\frac{1}{2} \cdot 2$$\left[ {\sin \left( {2x + 3x} \right) + \sin \left( {2x - 3x} \right)} \right] = \sin 5x - \sin x$.

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

a) Đúng. Do A, B, C là ba góc trong tam giác nên A + B + C = 180°.

Vậy ta có tan(2A + B + C) = tan(A + 180°) = tan A.

b) Sai. Ta có cosB∙cosC = sinB∙sinC

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

$\Leftrightarrow \frac{1}{2}\cos \left( {B - C} \right) + \frac{1}{2}\cos \left( {B + C} \right) = \frac{1}{2}\cos \left( {B - C} \right) - \frac{1}{2}\cos \left( {B + C} \right)$

$\Leftrightarrow \frac{1}{2}\cos \left( {B + C} \right) + \frac{1}{2}\cos \left( {B + C} \right) = 0 \Leftrightarrow \cos \left( {B + C} \right) = 0$

B + C = 90°.

Vậy tam giác ABC vuông tại A.

c) Đúng. $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90$°

$\Rightarrow \tan \left( {\frac{A}{2} + \frac{B}{2} - \frac{C}{2}} \right) = \cot \left[ {90^\circ - \left( {\frac{A}{2} + \frac{B}{2} - \frac{C}{2}} \right)} \right] = \cot \left[ {\left( {\frac{A}{2} + \frac{B}{2} + \frac{C}{2}} \right) - \left( {\frac{A}{2} + \frac{B}{2} - \frac{C}{2}} \right)} \right]$

= cotC.

d) Sai. $4\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{1}{2} \cdot 4\left[ {\cos \left( {\frac{A}{2} + \frac{B}{2}} \right) + \cos \left( {\frac{A}{2} - \frac{B}{2}} \right)} \right] \cdot \sin \frac{C}{2}$

Vì A + B + C = 180 nên $\frac{{A + B}}{2} = 90^\circ - \frac{C}{2}$ do đó $\cos \left( {\frac{{A + B}}{2}} \right) = \sin \frac{C}{2}$.

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

= $2{\sin ^2}\frac{C}{2} + 2\cos \left( {\frac{A}{2} - \frac{B}{2}} \right) \cdot \cos \left( {\frac{A}{2} + \frac{B}{2}} \right)$

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

= 2${\sin ^2}\frac{C}{2} + \cos A + \cos B$ = 1 − $\cos C$ + cosA + cosB.

Đáp án: 0

$P = 4\sin 3x \cdot \sin 2x \cdot \cos x = 4 \cdot \left( {\sin 3x \cdot \cos x} \right) \cdot \sin 2x$

= $4 \cdot \frac{1}{2}\left[ {\sin \left( {3x + x} \right) + \sin \left( {3x - x} \right)} \right] \cdot \sin 2x = 2\left( {\sin 4x + \sin 2x} \right) \cdot \sin 2x$

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

= $\cos 2x - \cos 6x + 1 - \cos 4x$.

Vậy a + b + c + d = 1 – 1 – 1 + 1 = 0.

Đáp án: 1

$B = \sin 5x - 2\sin x \cdot \left( {\cos 4x + \cos 2x} \right) = \sin 5x - 2\sin x \cdot \cos 4x - 2\sin x \cdot \cos 2x$

= $= \sin 5x - 2 \cdot \frac{1}{2}\left[ {\sin \left( {x + 4x} \right) + \sin \left( {x - 4x} \right)} \right] - 2 \cdot \frac{1}{2}\left[ {\sin \left( {x + 2x} \right) + \sin \left( {x - 2x} \right)} \right]$

= $\sin 5x - \sin 5x + \sin 3x - \sin 3x + {\mathop{\rm s}\nolimits} {\rm{inx = sinx}}$.

Vậy k = 1.

Đáp án: 6

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

= $2\sin \alpha \cdot \cos 4\alpha + 2\sin \alpha \cdot \cos 2\alpha + \sin \alpha$

= $2 \cdot \frac{1}{2}\left[ {\sin \left( {\alpha + 4\alpha } \right) + \sin \left( {\alpha - 4\alpha } \right)} \right] + 2 \cdot \frac{1}{2}\left[ {\sin \left( {\alpha + 2\alpha } \right) + \sin \left( {\alpha - 2\alpha } \right)} \right] + \sin \alpha$

= $\sin 5\alpha - \sin 3\alpha + \sin 3\alpha - \sin \alpha + \sin \alpha = \sin 5\alpha$

Vậy m + k = 1 + 5 = 6.

Đáp án: −1,5

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

= $\left[ {\cos \frac{{2\pi }}{3} + \cos \left( { - 2x} \right)} \right] - 2{\cos ^2}x = \frac{{ - 1}}{2} + \cos 2x - 2{\cos ^2}x$

$= \frac{{ - 1}}{2} + 2{\cos ^2}x - 1 - 2{\cos ^2}x = \frac{{ - 3}}{2} = - 1{,}5$.

Đáp án: 3

$E = \tan x \cdot \tan \left( {\frac{\pi }{3} - x} \right) \cdot \tan \left( {\frac{\pi }{3} + x} \right) = \tan x\frac{{\sin \left( {\frac{\pi }{3} - x} \right) \cdot \sin \left( {\frac{\pi }{3} + x} \right)}}{{\cos \left( {\frac{\pi }{3} - x} \right) \cdot \cos \left( {\frac{\pi }{3} + x} \right)}}$

$E = \tan x\frac{{\frac{{ - 1}}{2}\left( {\cos \frac{{2\pi }}{3} - \cos 2x} \right)}}{{\frac{1}{2}\left( {\cos \frac{{2\pi }}{3} + \cos 2x} \right)}} = \tan x\frac{{ - \left( {\frac{{ - 1}}{2} - \cos 2x} \right)}}{{\frac{{ - 1}}{2} + \cos 2x}}$

$E = \tan x\frac{{2\cos 2x + 1}}{{2\cos 2x - 1}} = \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{{\cos x}} \cdot \frac{{2\left( {1 - 2{{\sin }^2}x} \right) + 1}}{{2\left( {2{{\cos }^2}x - 1} \right) - 1}} = \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}\left( { - 4{{\sin }^2}x + 3} \right)}}{{\cos x\left( {4{{\cos }^2}x - 3} \right)}}$

$E = \frac{{3\sin x - 4\sin {}^3x}}{{4{{\cos }^3}x - 3\cos x}} = \frac{{\sin 3x}}{{\cos 3x}} = \tan 3x$.