Bài tập Áp dụng công thức lượng giác vào các bài toán rút gọn, chứng minh đẳng thức lượng giác lớp 11 (có lời giải)

Hướng dẫn giải:

Đáp án đúng là: A

$\frac{\sin \frac{π}{15}cos\frac{π}{10}+\sin \frac{π}{10}cos\frac{π}{15}}{cos\frac{2π}{15}cos\frac{π}{5}-\sin \frac{2π}{15}\sin \frac{π}{5}}$$=\frac{\sin (\frac{π}{15}+\frac{π}{10})}{cos(\frac{2π}{15}+\frac{π}{5})}=\frac{\sin (\frac{π}{6})}{cos(\frac{π}{3})}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$.

Hướng dẫn giải:

Đáp án đúng là: C

$P = \frac{{1 + \sin 2\alpha + cos2\alpha }}{{\sin \alpha + cos\alpha }}$$= \frac{{\left( {1 + cos2\alpha } \right) + \sin 2\alpha }}{{\sin \alpha + cos\alpha }}$$= \frac{{2co{s^2}\alpha + 2\sin \alpha cos\alpha }}{{\sin \alpha + cos\alpha }}$

$= \frac{{2cos\alpha \left( {\sin \alpha + cos\alpha } \right)}}{{\sin \alpha + cos\alpha }}$$= 2cos\alpha$.

Có sin$^{2}$a + cos$^{2}$a = 1 mà $\sin \alpha = \frac{2}{3}$ nên $co{s^2}\alpha = \frac{5}{9}$.

Vì $0 < \alpha < \frac{\pi }{2}$ nên cosa > 0. Do đó $cos\alpha = \frac{{\sqrt 5 }}{3}$.

Vậy $P = \frac{{2\sqrt 5 }}{3}$.

Hướng dẫn giải:

Đáp án đúng là: D

$A=\frac{2co{s}^{2}x-1}{\sin x+\cos x}=\frac{cos2x}{\sin x+\cos x}=\frac{co{s}^{2}x-{\sin }^{2}x}{\sin x+\cos x}$

$=\frac{(cosx-\sin x)(cosx+\sin x)}{\sin x+\cos x}=cosx-\sin x$.

Hướng dẫn giải:

Đáp án đúng là: D.

$P=\frac{\cos a-cos5a}{\sin 4a+\sin 2a}=\frac{2\sin 3a\sin 2a}{2\sin 3a\cos a}=\frac{\sin 2a}{\cos a}=\frac{2\sin a\cos a}{\cos a}=2\sin a$

Hướng dẫn giải:

Đáp án đúng là: C

$A = \frac{{1 + cos\alpha + cos2\alpha + cos3\alpha }}{{2co{s^2}\alpha + cos\alpha - 1}}$

$= \frac{{\left( {1 + cos2\alpha } \right) + \left( {cos\alpha + cos3\alpha } \right)}}{{2co{s^2}\alpha + cos\alpha - 1}}$

$= \frac{{2co{s^2}\alpha + 2cos2\alpha cos\alpha }}{{\left( {2co{s^2}\alpha - 1} \right) + cos\alpha }}$

$= \frac{{2cos\alpha \left( {cos2\alpha + cos\alpha } \right)}}{{cos2\alpha + cos\alpha }}$

$= 2cos\alpha$.

Hướng dẫn giải:

Đáp án đúng là: A

$A = \frac{{{{\left( {tanx + \cot x} \right)}^2} - {{\left( {tanx - \cot x} \right)}^2}}}{{\cot x - tanx}}$

$= \frac{{ta{n^2}x + 2\tan x\cot x + {{\cot }^2}x - \left( {ta{n^2}x - 2\tan x\cot x + {{\cot }^2}x} \right)}}{{\frac{{{\rm{cosx}}}}{{\sin x}} - \frac{{\sin x}}{{{\rm{cosx}}}}}}$

$= \frac{{ta{n^2}x + 2\tan x\cot x + {{\cot }^2}x - ta{n^2}x + 2\tan x\cot x - {{\cot }^2}x}}{{\frac{{{\rm{cosx}}}}{{\sin x}} - \frac{{\sin x}}{{{\rm{cosx}}}}}}$

$= \frac{4}{{\frac{{{\rm{co}}{{\rm{s}}^2}{\rm{x}} - {{\sin }^2}x}}{{\sin x{\rm{cosx}}}}}}$

$= \frac{{4\sin x{\rm{cosx}}}}{{{\rm{cos2x}}}}$

$= \frac{{2\sin 2x}}{{{\rm{cos2x}}}} = 2\tan 2x = \frac{2}{{\cot 2x}}$.

Hướng dẫn giải:

Đáp án đúng là: B

$B = \left( {\frac{1}{{cos2x}} + 1} \right) \cdot \tan x$$= \frac{{1 + cos2x}}{{cos2x}} \cdot \frac{{\sin x}}{{\cos x}}$$= \frac{{2{{\cos }^2}x}}{{cos2x}} \cdot \frac{{\sin x}}{{\cos x}}$$= \frac{{2\cos x\sin x}}{{cos2x}}$$= \frac{{\sin 2x}}{{cos2x}} = \tan 2x$

Hướng dẫn giải:

Đáp án đúng là: C

Xét đáp án A.

Ta có $4\sin \left( {\frac{x}{2} + 15^\circ } \right)\sin \left( {\frac{x}{2} - 15^\circ } \right)$

$= 4 \cdot \frac{1}{2} \cdot \left\{ {co{\mathop{\rm s}\nolimits} \left[ {\left( {\frac{x}{2} + 15^\circ } \right) - \left( {\frac{x}{2} - 15^\circ } \right)} \right] - cos\left[ {\left( {\frac{x}{2} + 15^\circ } \right) + \left( {\frac{x}{2} - 15^\circ } \right)} \right]} \right\}$

$= 4 \cdot \frac{1}{2} \cdot \left( {cos30^\circ - c{\rm{os}}x} \right)$

$= \sqrt 3 - 2c{\rm{os}}x$.

Vậy đáp án A đúng

Xét đáp án B

$\frac{{4\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}$

$= \frac{{4 \cdot \frac{1}{2} \cdot \left\{ {co{\mathop{\rm s}\nolimits} \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x - \frac{\pi }{3}} \right)} \right] - co{\mathop{\rm s}\nolimits} \left[ {\left( {x + \frac{\pi }{3}} \right) + \left( {x - \frac{\pi }{3}} \right)} \right]} \right\}}}{{{\rm{co}}{{\rm{s}}^2}x}}$

$= \frac{{4 \cdot \frac{1}{2} \cdot \left( {co{\mathop{\rm s}\nolimits} \frac{{2\pi }}{3} - co{\mathop{\rm s}\nolimits} 2x} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}$

$= \frac{{4 \cdot \frac{1}{2} \cdot \left( { - \frac{1}{2} - co{\mathop{\rm s}\nolimits} 2x} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}$

$= \frac{{ - 1 - 2co{\mathop{\rm s}\nolimits} 2x}}{{{\rm{co}}{{\rm{s}}^2}x}}$

$= \frac{{ - \left( {si{n^2}x + co{s^2}x} \right) - 2\left( {co{s^2}x - si{n^2}x} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}$

$= \frac{{si{n^2}x - 3co{s^2}x}}{{{\rm{co}}{{\rm{s}}^2}x}}$

$= {\tan ^2}x - 3$.

Vậy đáp án B đúng.

Xét đáp án C

cos12xcos2x = $\frac{1}{2}\left( {cos10x + cos14x} \right)$

$= \frac{1}{2}\left( {2{{\cos }^2}5x - 1 + 2co{s^2}7x - 1} \right)$

$= co{s^2}5x + co{s^2}7x - 1$.

Vậy đáp án C sai.

Xét đáp án D

$2\sqrt 2 {\rm{cos}}\frac{x}{2}cos\left( {\frac{x}{2} - \frac{\pi }{4}} \right)$

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

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

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

$= 1 + \sqrt 2 {\rm{cos}}\left( {x - \frac{\pi }{4}} \right)$

$= 1 + \sqrt 2 \left( {\cos x\cos \frac{\pi }{4} + \sin x\sin \frac{\pi }{4}} \right)$

$= 1 + \sqrt 2 \left( {\frac{{\sqrt 2 }}{2}\cos x + \frac{{\sqrt 2 }}{2}\sin x} \right)$

$= 1 + \cos x + \sin x$.

Vậy đáp án D đúng.

Hướng dẫn giải:

Đáp án đúng là: A

Xét đáp án A

$\frac{{\tan 30^\circ + \tan 40^\circ + \tan 50^\circ + \tan 60^\circ }}{{cos20^\circ }}$

$= \frac{{\frac{{\sin \left( {30^\circ + 40^\circ } \right)}}{{cos30^\circ cos40^\circ }} + \frac{{\sin \left( {50^\circ + 60^\circ } \right)}}{{cos50^\circ cos60^\circ }}}}{{cos20^\circ }}$

$= \frac{{\frac{{\sin 70^\circ }}{{cos30^\circ cos40^\circ }} + \frac{{\sin 110^\circ }}{{cos50^\circ cos60^\circ }}}}{{cos20^\circ }}$

$= \frac{{\sin 70^\circ }}{{cos20^\circ cos30^\circ cos40^\circ }} + \frac{{\sin 110^\circ }}{{cos20^\circ cos50^\circ cos60^\circ }}$

$= \frac{{\sin 70^\circ }}{{\sin 70^\circ cos30^\circ cos40^\circ }} + \frac{{cos20^\circ }}{{cos20^\circ cos50^\circ cos60^\circ }}$

$= \frac{1}{{cos30^\circ cos40^\circ }} + \frac{1}{{cos50^\circ cos60^\circ }}$

$= \frac{2}{{\sqrt 3 cos40^\circ }} + \frac{2}{{cos50^\circ }}$

$= 2 \cdot \frac{{cos50^\circ + \sqrt 3 cos40^\circ }}{{\sqrt 3 cos40^\circ cos50^\circ }}$

$= 2 \cdot \frac{{\sin 40^\circ + \sqrt 3 cos40^\circ }}{{\sqrt 3 cos40^\circ cos50^\circ }}$

$= 4 \cdot \frac{{\frac{1}{2} \cdot \sin 40^\circ + \frac{{\sqrt 3 }}{2} \cdot cos40^\circ }}{{\frac{{\sqrt 3 }}{2}\left( {cos10^\circ + cos90^\circ } \right)}}$

$= 4 \cdot \frac{{\sin 30^\circ \cdot \sin 40^\circ + co{\mathop{\rm s}\nolimits} 30^\circ \cdot cos40^\circ }}{{\frac{{\sqrt 3 }}{2}\left( {cos10^\circ + cos90^\circ } \right)}}$

$= \frac{8}{{\sqrt 3 }} \cdot \frac{{cos10^\circ }}{{cos10^\circ }} = \frac{8}{{\sqrt 3 }}$.

Vậy đáp án A sai

Xét đáp án B

$cos\frac{\pi }{5} - cos\frac{{2\pi }}{5}$$= 2\sin \frac{{3\pi }}{{10}}\sin \frac{\pi }{{10}}$$= 2\sin \left( {\frac{\pi }{2} - \frac{\pi }{5}} \right)\sin \left( {\frac{\pi }{2} - \frac{{2\pi }}{5}} \right)$$= 2cos\frac{\pi }{5}{\rm{cos}}\frac{{2\pi }}{5}$

$= \frac{{2cos\frac{\pi }{5}{\rm{cos}}\frac{{2\pi }}{5}\sin \frac{\pi }{5}}}{{\sin \frac{\pi }{5}}}$$= \frac{{\sin \frac{{2\pi }}{5}{\rm{cos}}\frac{{2\pi }}{5}}}{{\sin \frac{\pi }{5}}}$$= \frac{1}{2} \cdot \frac{{\sin \frac{{4\pi }}{5}}}{{\sin \frac{\pi }{5}}}$$= \frac{1}{2} \cdot \frac{{\sin \left( {\pi - \frac{\pi }{5}} \right)}}{{\sin \frac{\pi }{5}}}$$= \frac{1}{2} \cdot \frac{{\sin \frac{\pi }{5}}}{{\sin \frac{\pi }{5}}} = \frac{1}{2}$.

Vậy đáp án B đúng.

Xét đáp án C

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

$= \frac{{\sin \frac{\pi }{7}cos\frac{\pi }{7} - \sin \frac{\pi }{7}cos\frac{{2\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{3\pi }}{7}}}{{\sin \frac{\pi }{7}}}$

$= \frac{1}{2} \cdot \frac{{\sin \frac{{2\pi }}{7} + \sin \frac{\pi }{7} - sin\frac{{3\pi }}{7} - \sin \frac{{2\pi }}{7} + \sin \frac{{4\pi }}{7}}}{{\sin \frac{\pi }{7}}}$

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

$= \frac{1}{2} \cdot \frac{{\sin \frac{\pi }{7} - sin\frac{{4\pi }}{7} + \sin \frac{{4\pi }}{7}}}{{\sin \frac{\pi }{7}}} = \frac{1}{2}$.

Vậy đáp án C đúng

Xét đáp án D

$cos\frac{{2\pi }}{5} + cos\frac{{4\pi }}{5} + cos\frac{{6\pi }}{5} + cos\frac{{8\pi }}{5}$

$= \frac{{2\sin \frac{{2\pi }}{5}cos\frac{{2\pi }}{5} + 2\sin \frac{{2\pi }}{5}cos\frac{{4\pi }}{5} + 2\sin \frac{{2\pi }}{5}cos\frac{{6\pi }}{5} + 2\sin \frac{{2\pi }}{5}cos\frac{{8\pi }}{5}}}{{2\sin \frac{{2\pi }}{5}}}$

$= \frac{{\sin \frac{{4\pi }}{5} - \sin \frac{{2\pi }}{5} + \sin \frac{{6\pi }}{5} - \sin \frac{{4\pi }}{5} + \sin \frac{{8\pi }}{5} - \sin \frac{{6\pi }}{5} + \sin 2\pi }}{{2\sin \frac{{2\pi }}{5}}}$

$= \frac{{ - \sin \frac{{2\pi }}{5} + \sin \frac{{8\pi }}{5}}}{{2\sin \frac{{2\pi }}{5}}}$$= \frac{{2cos\pi \sin \frac{{3\pi }}{5}}}{{2\sin \frac{{2\pi }}{5}}}$$= \frac{{ - 2\sin \left( {\pi - \frac{{2\pi }}{5}} \right)}}{{2\sin \frac{{2\pi }}{5}}}$$= \frac{{ - 2\sin \frac{{2\pi }}{5}}}{{2\sin \frac{{2\pi }}{5}}} = - 1$.

Vậy đáp án D đúng.

Hướng dẫn giải:

Đáp án đúng là: C

Xét đáp án A

$\sin \left( {x + \frac{\pi }{6}} \right)cos\left( {x - \frac{\pi }{6}} \right)$

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

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

Vậy đáp án A đúng.

Xét đáp án B

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

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

Vậy đáp án B đúng.

Xét đáp án C

$\sin \left( {x + \frac{\pi }{6}} \right)\sin \left( {x - \frac{\pi }{6}} \right)cos2x$$= \frac{1}{2}\left\{ {cos\left[ {\left( {x + \frac{\pi }{6}} \right) - \left( {x - \frac{\pi }{6}} \right)} \right] - cos\left[ {\left( {x + \frac{\pi }{6}} \right) + \left( {x - \frac{\pi }{6}} \right)} \right]} \right\}cos2x$

$= \frac{1}{2}\left( {cos\frac{\pi }{3} - cos2x} \right)cos2x$

$= \frac{1}{2}\left( {\frac{1}{2} - cos2x} \right)cos2x$

$= \frac{1}{4}cos2x - \frac{1}{2}co{s^2}2x$

$= \frac{1}{4}cos2x - \frac{1}{2}\left( {\frac{{1 + cos4x}}{2}} \right)$

$= \frac{1}{4}cos2x - \frac{1}{4}cos4x - \frac{1}{4}$

Vậy đáp án C sai.

Xét đáp án D

cos2x – cos4x – cos6x + 1

= 2sin3xsinx + 1 – cos6x

= 2sin3xsinx + 2sin$^{2}$3x

= 2sin3x(sinx + sin3x)

= 2sin3x×2sin2xcosx

= 4sin3xsin2xcosx

Do đó 2(cos2x – cos4x – cos6x + 1) = 8sin3xsin2xcosx.

Vậy đáp án D đúng.

Đáp án đúng là: C

$\frac{{\sin a}}{{\cos a + \sin a}} - \frac{{\cos a}}{{\cos a - \sin a}} = \frac{{ - {{\sin }^2}a - {{\cos }^2}a}}{{{{\cos }^2}a - {{\sin }^2}a}} = \frac{{1 + {{\cot }^2}a}}{{1 - {{\cot }^2}a}}$

Đáp án đúng là: A

$D = \frac{{{{\cot }^2}x - {{\cos }^2}x}}{{{{\cot }^2}x}} + \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x}}{{\cot x}} = 1 - {\sin ^2}x + {\sin ^2}x = 1$

Đáp án đúng lả: C

$\frac{{1 + \sin 4a - \cos 4a}}{{1 + \sin 4a + \cos 4a}} = \frac{{2{{\sin }^2}2a + 2\sin 2a \cdot \cos 2a}}{{2{{\cos }^2}2a + 2\sin 2a \cdot \cos 2a}}$

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

Đáp án đúng là: D

${\sin ^2}x - {\sin ^2}2x - {\sin ^2}3x = \frac{{1 - \cos 2x - 1 + \cos 4x - 1 + \cos 6x}}{2}$

$= \frac{{\left( {\cos 6x + \cos 4x} \right) - \left( {\cos 2x + 1} \right)}}{2} = \frac{{2\cos 5x \cdot \cos x - 2{{\cos }^2}x}}{2} = \cos x\left( {\cos 5x - \cos x} \right)$

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

Đáp án đúng là: B

M = = cos$^{4}$15° − sin$^{4}$15° = (cos$^{2}$15°)$^{2}$ – (sin$^{2}$15°)$^{2}$

= (cos$^{2}$15° − sin$^{2}$15°)( cos$^{2}$15° + sin$^{2}$15°)

= cos$^{2}$15° − sin$^{2}$15° = cos(2∙15°) = cos30° = $\frac{{\sqrt 3 }}{2}$.

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

a) Đúng. $A = \frac{{{{\cos }^3}a + {{\sin }^3}a}}{{\cos a + \sin a}} + \sin a \cdot \cos a$

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

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

b) Sai. $B = 2\left( {{{\sin }^6}a + {{\cos }^6}a} \right) - 3\left( {{{\sin }^4}a + {{\cos }^4}a} \right)$

= $2\left( {{{\sin }^2}a + {{\cos }^2}a} \right) \cdot \left( {{{\sin }^4}a - {{\sin }^2}a \cdot {{\cos }^2}a + {{\cos }^4}a} \right)$

$- 3 \cdot \left[ {{{\left( {{{\sin }^2}a + {{\cos }^2}a} \right)}^2} - 2{{\sin }^2}a \cdot {{\cos }^2}a} \right]$

$= 2 \cdot \left[ {1 - 3{{\sin }^2}a \cdot {{\cos }^2}a} \right] - 3\left( {1 - 2{{\sin }^2}a \cdot {{\cos }^2}a} \right)$

= 2 – 6sin$^{2}$acos$^{2}$a – 3+ 6sin$^{2}$acos$^{2}$a = −1.

c) Sai. C = 8(cos$^{8}$a – sin$^{8}$a) – cos6a – 7cos2a

= 8(cos$^{4}$a – sin$^{4}$a)(sin$^{4}$a + cos$^{4}$a) – cos6a – 7cos2a

= 8(cos$^{2}$a – sin$^{2}$a)(sin$^{2}$a + cos$^{2}$a)[(sin$^{2}$a + cos$^{2}$a)$^{2}$ – 2sin$^{2}$acos$^{2}$a] – cos6a – 7cos2a

= 8(cos$^{2}$a – sin$^{2}$a)(1 − 2sin$^{2}$acos$^{2}$a) – cos6a – 7cos2a

$= 8 \cdot \left( {\cos 2a} \right)\left( {1 - \frac{1}{2}{{\sin }^2}2a} \right) - \cos 6a - 7\cos 2a$

$= 8\cos 2a - 4\cos 2a \cdot {\sin ^2}2a - \cos 6a - 7\cos 2a = \cos 2a - 4\cos 2a\left( {1 - {{\cos }^2}2a} \right) - \cos 6a$

$= \cos 2a - 4\cos 2a + 4{\cos ^3}2a - \cos 6a = \left( {4{{\cos }^3}2a - 3\cos 2a} \right) - \cos 6a = \cos 6a - \cos 6a = 0$ d) Sai. D = cos$^{2}$x – 2cosa∙cosx∙cos(a + x) + cos$^{2}$(a + x).

= cos$^{2}$x – [cos(a + x) + cos(a – x)]∙cos(a + x) + cos$^{2}$(a + x).

= cos$^{2}$x – cos$^{2}$(a + x) – cos(a – x)cos(a + x) + cos$^{2}$(a + x)

= cos$^{2}$x – cos(a – x)cos(a + x)

= cos$^{2}$x – (cosa∙cosx − sina∙sinx)( cosa∙cosx + sina∙sinx)

= cos$^{2}$x – (cos$^{2}$a∙cos$^{2}$x) + (sin$^{2}$asin$^{2}$x) = cos$^{2}$x(1 – cos$^{2}$a) + sin$^{2}$asin$^{2}$x

= cos$^{2}$x∙sin$^{2}$a + sin$^{2}$asin$^{2}$x = sin$^{2}$a(cos$^{2}$x + sin$^{2}$x) = sin$^{2}$a.

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

a) Đúng. ${S_n} = {\sin ^3}\frac{a}{3} + 3{\sin ^3}\frac{a}{{{3^2}}} + ... + {3^{n - 1}}{\sin ^3}\frac{a}{{{3^n}}}$

$= \frac{1}{4}\left( {3\sin \frac{a}{3} - \sin a} \right) + \frac{1}{4}\left( {{3^2}\sin \frac{a}{{{3^2}}} - 3\sin \frac{a}{3}} \right) + ... + \frac{1}{4}\left( {{3^n}\sin \frac{a}{{{3^n}}} - {3^{n - 1}}\sin \frac{a}{{{3^{n - 1}}}}} \right)$

$= \frac{1}{4}\left( {{3^n}\sin \frac{a}{{{3^n}}} - \sin a} \right)$.

b) Đúng. $S = \frac{1}{{\sin \alpha }} + \frac{1}{{\sin 2\alpha }} + ... + \frac{1}{{\sin {2^{n - 1}}\alpha }}$

$= \cot \frac{\alpha }{2} - \cot \alpha + \cot \alpha - \cot 2\alpha + ... + \cot {2^{n - 2}}\alpha - \cot {2^{n - 1}}\alpha = \cot \frac{\alpha }{2} - \cot {2^{n - 1}}\alpha$.

c) Đúng. ${S_n} = {\tan ^2}\frac{a}{2} \cdot \tan a + 2{\tan ^2}\frac{a}{{{2^2}}} \cdot \tan \frac{a}{2} + ... + {2^{n - 1}}{\tan ^2}\frac{a}{{{2^n}}} \cdot \tan \frac{a}{{{2^{n - 1}}}}$

$$= \tan a - 2\tan \frac{a}{2} + 2 \cdot \left( {\tan \frac{a}{2} - 2\tan \frac{a}{{{2^2}}}} \right) + ... + {2^{n - 1}}\tan \frac{a}{{{2^{n - 1}}}} - {2^n}\tan \frac{a}{{{2^n}}}$$

${S_n} = \tan a - {2^n}\tan \frac{a}{{{2^n}}}$.

d) Đúng. ${P_n} = \cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}} \cdot ... \cdot c{\rm{os}}\frac{x}{{{2^n}}}$

${P_n} \cdot \sin \frac{x}{{{2^n}}} = \cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}}...\cos \frac{x}{{{2^{n - 1}}}} \cdot \left( {\cos \frac{x}{{{2^n}}} \cdot \sin \frac{x}{{{2^n}}}} \right)$

${P_n} \cdot \sin \frac{x}{{{2^n}}} = \frac{1}{{{2^{n - 1}}}} \cdot \cos \frac{x}{2} \cdot \sin \frac{x}{2} = \frac{1}{{{2^n}}} \cdot {\mathop{\rm s}\nolimits} {\rm{inx}}$

${P_n} = \frac{{\sin x}}{{{2^n} \cdot \sin \frac{x}{{{2^n}}}}}$.

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

a) Đúng. $\cot x - \tan x - 2\tan 2x$

$= \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x}} - 2\tan 2x = \frac{{\cos 2x}}{{\frac{1}{2}\sin 2x}} - 2\tan 2x$

$= 2 \cdot \frac{{\cos 2x}}{{\sin 2x}} - 2\tan 2x = 2\cot 2x - 2\tan 2x = 2\left( {\cot 2x - \tan 2x} \right)$

$= 2 \cdot \left( {2\cot 4x} \right) = 4\cot 4x$.

b) Đúng. sin$^{2}$(a + b) – sin$^{2}$a – sin$^{2}$b

= sin$^{2}$(a + b) – (sin$^{2}$a + sin$^{2}$b) = sin$^{2}$(a + b) − $\frac{{1 - \cos 2a}}{2} - \frac{{1 - \cos 2b}}{2}$

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

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

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

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

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

$= \cos \left( {a + b} \right) \cdot \left( {2\sin a \cdot \sin b} \right) = 2\sin a \cdot \sin b \cdot \cos \left( {a + b} \right)$.

c) Sai. $\sin \alpha \cdot \sin \left( {\frac{\pi }{3} - \alpha } \right) \cdot \sin \left( {\frac{\pi }{3} + \alpha } \right)$

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

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

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

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

d) Sai. cos(a + b)∙sin(a – b) + cos(b + c)∙sin(b – c) + cos(a + c)∙sin(c – a)

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

$= \frac{1}{2}\left( {\sin 2a - \sin 2b + \sin 2b - \sin 2c + \sin 2c - \sin 2a} \right) = \frac{1}{2} \cdot 0 = 0$.

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

a) Đúng. ${\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) = 2 - {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) - {\sin ^2}\left( {x + \pi } \right)$

$\Leftrightarrow {\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) + {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) + {\sin ^2}\left( {x + \pi } \right) = 2$

Xét biểu thức: ${\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) + {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) + {\sin ^2}\left( {x + \pi } \right)$

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

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

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

b) Sai. ${\sin ^4}x\left( {1 + {{\sin }^2}x} \right) + {\cos ^4}x\left( {1 + {{\cos }^2}x} \right) + 5\left( {404 + {{\sin }^2}x \cdot {{\cos }^2}x} \right)$

= ${\sin ^4}x + {\sin ^6}x + {\cos ^4}x + {\cos ^6}x + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x$