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)
25 câu hỏi
Giá trị biểu thức bằng
1;
−1;
;
.
Hướng dẫn giải:
Đáp án đúng là: A
.
Cho góc a thỏa mãn \(0 < \alpha < \frac{\pi }{2}\) và \(\sin \alpha = \frac{2}{3}\). Tính \(P = \frac{{1 + \sin 2\alpha + cos2\alpha }}{{\sin \alpha + cos\alpha }}\).
\(P = - \frac{{2\sqrt 5 }}{3}\);
\(P = \frac{3}{2}\);
\(P = \frac{{2\sqrt 5 }}{3}\);
\(P = - \frac{3}{2}\).
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 sin2a + cos2a = 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 \(P = \frac{{2\sqrt 5 }}{3}\).
Đơn giản biểu thức ta được kết quả là
A = sinx – cosx;
A = cosx + sinx;
A = −sinx – cosx;
A = cosx – sinx.
Hướng dẫn giải:
Đáp án đúng là: D
.
Rút gọn biểu thức với (sin4a + sin2a ≠ 0) ta được
P = 2cota;
P = 2cosa;
P = 2tana
P = 2sina.
Hướng dẫn giải:
Đáp án đúng là: D.
Rút gọn biểu thức \(A = \frac{{1 + cos\alpha + cos2\alpha + cos3\alpha }}{{2co{s^2}\alpha + cos\alpha - 1}}\) bằng
P = −2cosa;
P = cosa;
P = 2cosa;
P = 2sina.
Hướng dẫn giải:
Đáp án đúng là: C
\(A = \frac{{1 + cos\alpha + cos2\alpha + cos3\alpha }}{{2co{s^2}\alpha + cos\alpha - 1}}\)
\( = \frac{{\left( {1 + cos2\alpha } \right) + \left( {cos\alpha + cos3\alpha } \right)}}{{2co{s^2}\alpha + cos\alpha - 1}}\)
\( = \frac{{2co{s^2}\alpha + 2cos2\alpha cos\alpha }}{{\left( {2co{s^2}\alpha - 1} \right) + cos\alpha }}\)
\( = \frac{{2cos\alpha \left( {cos2\alpha + cos\alpha } \right)}}{{cos2\alpha + cos\alpha }}\)
\( = 2cos\alpha \).
Với điều kiện xác định, hãy rút gọn biểu thức
\(A = \frac{{{{\left( {tanx + \cot x} \right)}^2} - {{\left( {tanx - \cot x} \right)}^2}}}{{\cot x - tanx}}\).
\(A = \frac{2}{{\cot 2x}}\);
A = 4;
\(A = \frac{4}{{\cot 2x}}\);
\(A = \frac{8}{{\cot 2x}}\).
Hướng dẫn giải:
Đáp án đúng là: A
\(A = \frac{{{{\left( {tanx + \cot x} \right)}^2} - {{\left( {tanx - \cot x} \right)}^2}}}{{\cot x - tanx}}\)
\( = \frac{{ta{n^2}x + 2\tan x\cot x + {{\cot }^2}x - \left( {ta{n^2}x - 2\tan x\cot x + {{\cot }^2}x} \right)}}{{\frac{{{\rm{cosx}}}}{{\sin x}} - \frac{{\sin x}}{{{\rm{cosx}}}}}}\)
\( = \frac{{ta{n^2}x + 2\tan x\cot x + {{\cot }^2}x - ta{n^2}x + 2\tan x\cot x - {{\cot }^2}x}}{{\frac{{{\rm{cosx}}}}{{\sin x}} - \frac{{\sin x}}{{{\rm{cosx}}}}}}\)
\( = \frac{4}{{\frac{{{\rm{co}}{{\rm{s}}^2}{\rm{x}} - {{\sin }^2}x}}{{\sin x{\rm{cosx}}}}}}\)
\( = \frac{{4\sin x{\rm{cosx}}}}{{{\rm{cos2x}}}}\)
\( = \frac{{2\sin 2x}}{{{\rm{cos2x}}}} = 2\tan 2x = \frac{2}{{\cot 2x}}\).
Biểu thức thu gọn của biểu thức \(B = \left( {\frac{1}{{cos2x}} + 1} \right) \cdot \tan x\) là
cot2x;
tan2x;
cos2x;
sinx.
Hướng dẫn giải:
Đáp án đúng là: B
\(B = \left( {\frac{1}{{cos2x}} + 1} \right) \cdot \tan x\)\[ = \frac{{1 + cos2x}}{{cos2x}} \cdot \frac{{\sin x}}{{\cos x}}\]\[ = \frac{{2{{\cos }^2}x}}{{cos2x}} \cdot \frac{{\sin x}}{{\cos x}}\]\[ = \frac{{2\cos x\sin x}}{{cos2x}}\]\[ = \frac{{\sin 2x}}{{cos2x}} = \tan 2x\]
Trong các hệ thức sau, hệ thức nào sai?
\(\sqrt 3 - 2\cos x = 4\sin \left( {\frac{x}{2} + 15^\circ } \right)\sin \left( {\frac{x}{2} - 15^\circ } \right)\);
\({\tan ^2}x - 3 = \frac{{4\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}\);
sin27x – cos25x = cos12xcos2x;
1 + sinx + cosx = \(2\sqrt 2 {\rm{cos}}\frac{x}{2}cos\left( {\frac{x}{2} - \frac{\pi }{4}} \right)\).
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.
Trong các khẳng định sau, khẳng định nào sai?
\(\frac{{\tan 30^\circ + \tan 40^\circ + \tan 50^\circ + \tan 60^\circ }}{{cos20^\circ }} = \frac{4}{{\sqrt 3 }}\);
\(cos\frac{\pi }{5} - cos\frac{{2\pi }}{5} = \frac{1}{2}\);
\(cos\frac{\pi }{7} - cos\frac{{2\pi }}{7} + cos\frac{{3\pi }}{7} = \frac{1}{2}\);
\(cos\frac{{2\pi }}{5} + cos\frac{{4\pi }}{5} + cos\frac{{6\pi }}{5} + cos\frac{{8\pi }}{5} = - 1\).
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.
Trong các hệ thức sau, hệ thức nào sai?
\(\sin \left( {x + \frac{\pi }{6}} \right)cos\left( {x - \frac{\pi }{6}} \right) = \frac{{2\sin 2x + \sqrt 3 }}{4}\);
\(\sin \frac{\pi }{5}\sin \frac{{2\pi }}{5} = \frac{1}{2}\left( {cos\frac{\pi }{5} + cos\frac{{2\pi }}{5}} \right)\);
\(\sin \left( {x + \frac{\pi }{6}} \right)\sin \left( {x - \frac{\pi }{6}} \right)cos2x = \frac{1}{4}cos2x - \frac{1}{8}cos4x - \frac{1}{8}\);
8cosxsin2xsin3x = 2(cos2x– cos4x – cos6x + 1).
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 + 2sin23x
= 2sin3x(sinx + sin3x)
= 2sin3x×2sin2xcosx
= 4sin3xsin2xcosx
Do đó 2(cos2x – cos4x – cos6x + 1) = 8sin3xsin2xcosx.
Vậy đáp án D đúng.
Chọn hệ thức sai trong các hệ thức dưới đây?
sin2a∙tana + cos2a∙cota + 2sina∙cosa = tana + cota.
3(sin4a + cos4a) – 2(sin6a + cos6a) = 1.
\(\frac{{\sin a}}{{\cos a + \sin a}} - \frac{{\cos a}}{{\cos a - \sin a}} = \frac{{1 - {{\cot }^2}a}}{{1 + {{\cot }^2}a}}\).
\(\frac{{1 + 2\sin a \cdot \cos a}}{{{{\sin }^2}a - {{\cos }^2}a}} = \frac{{\tan a + 1}}{{\tan a - 1}}\).
Đáp án đúng là: C
\(\frac{{\sin a}}{{\cos a + \sin a}} - \frac{{\cos a}}{{\cos a - \sin a}} = \frac{{ - {{\sin }^2}a - {{\cos }^2}a}}{{{{\cos }^2}a - {{\sin }^2}a}} = \frac{{1 + {{\cot }^2}a}}{{1 - {{\cot }^2}a}}\)
Biểu thức \(D = \frac{{{{\cot }^2}x - {{\cos }^2}x}}{{{{\cot }^2}x}} + \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x}}{{\cot x}}\) bằng?
1.
−1.
\(\frac{1}{2}\).
\(\frac{{ - 1}}{2}\).
Đáp án đúng là: A
\(D = \frac{{{{\cot }^2}x - {{\cos }^2}x}}{{{{\cot }^2}x}} + \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x}}{{\cot x}} = 1 - {\sin ^2}x + {\sin ^2}x = 1\)
Biểu thức \(\frac{{1 + \sin 4a - \cos 4a}}{{1 + \sin 4a + \cos 4a}}\) có kết quả rút gọn bằng bao nhiêu?
sin2a.
cos2a.
tan2a.
cot 2a.
Đáp án đúng lả: C
\(\frac{{1 + \sin 4a - \cos 4a}}{{1 + \sin 4a + \cos 4a}} = \frac{{2{{\sin }^2}2a + 2\sin 2a \cdot \cos 2a}}{{2{{\cos }^2}2a + 2\sin 2a \cdot \cos 2a}}\)
\( = \frac{{2\sin 2a\left( {\sin 2a + \cos 2a} \right)}}{{2\cos 2a\left( {\sin 2a + \cos 2a} \right)}} = \tan 2a\).
Kết quả biến đổi nào dưới đây là sai?
\(1 + \cos x + \cos 2x = 4\cos x \cdot {\cos ^2}\frac{x}{2}\).
sinxcos3x + sin4xcos2x = sin5x∙cosx.
cos2x + cos22x + cos23x – 1 = 2cos3x∙cos2x∙cosx.
sin2x – sin22x – sin23x = 2sin3x∙sin2x∙sinx.
Đáp án đúng là: D
\({\sin ^2}x - {\sin ^2}2x - {\sin ^2}3x = \frac{{1 - \cos 2x - 1 + \cos 4x - 1 + \cos 6x}}{2}\)
\( = \frac{{\left( {\cos 6x + \cos 4x} \right) - \left( {\cos 2x + 1} \right)}}{2} = \frac{{2\cos 5x \cdot \cos x - 2{{\cos }^2}x}}{2} = \cos x\left( {\cos 5x - \cos x} \right)\)
\( = \cos x\left( { - 2\sin 3x \cdot \sin 2x} \right) = - 2\sin 3x \cdot \sin 2x \cdot \cos x\).
Gọi M = cos415° − sin415°có giá trị bằng?
1.
\(\frac{{\sqrt 3 }}{2}\).
\(\frac{1}{4}\).
0.
Đáp án đúng là: B
M = = cos415° − sin415° = (cos215°)2 – (sin215°)2
= (cos215° − sin215°)( cos215° + sin215°)
= cos215° − sin215° = cos(2∙15°) = cos30° = \(\frac{{\sqrt 3 }}{2}\).
Xét tính đúng sai của các mệnh đề sau:
\(A = \frac{{{{\cos }^3}a + {{\sin }^3}a}}{{\cos a + \sin a}} + \sin a \cdot \cos a = 1\).
\(B = 2\left( {{{\sin }^6}a + {{\cos }^6}a} \right) - 3\left( {{{\sin }^4}a + {{\cos }^4}a} \right) = 1\).
C = 8(cos8a – sin8a) – cos6a – 7cos2a = 8.
D = cos2x – 2cosa∙cosx∙cos(a + x) + cos2(a + x) = cos2a.
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 – 6sin2acos2a – 3+ 6sin2acos2a = −1.
c) Sai. C = 8(cos8a – sin8a) – cos6a – 7cos2a
= 8(cos4a – sin4a)(sin4a + cos4a) – cos6a – 7cos2a
= 8(cos2a – sin2a)(sin2a + cos2a)[(sin2a + cos2a)2 – 2sin2acos2a] – cos6a – 7cos2a
= 8(cos2a – sin2a)(1 − 2sin2acos2a) – 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 = cos2x – 2cosa∙cosx∙cos(a + x) + cos2(a + x).
= cos2x – [cos(a + x) + cos(a – x)]∙cos(a + x) + cos2(a + x).
= cos2x – cos2(a + x) – cos(a – x)cos(a + x) + cos2(a + x)
= cos2x – cos(a – x)cos(a + x)
= cos2x – (cosa∙cosx − sina∙sinx)( cosa∙cosx + sina∙sinx)
= cos2x – (cos2a∙cos2x) + (sin2asin2x) = cos2x(1 – cos2a) + sin2asin2x
= cos2x∙sin2a + sin2asin2x = sin2a(cos2x + sin2x) = sin2a.
Xét tính đúng sai của các mệnh đề sau:
\({S_n} = {\sin ^3}\frac{a}{3} + 3{\sin ^3}\frac{a}{{{3^2}}} + ... + {3^{n - 1}}{\sin ^3}\frac{a}{{{3^n}}}\)\( = \frac{1}{4}\left( {{3^n}\sin \frac{a}{{{3^n}}} - \sin a} \right)\).
\(S = \frac{1}{{\sin \alpha }} + \frac{1}{{\sin 2\alpha }} + ... + \frac{1}{{\sin {2^{n - 1}}\alpha }}\)= \(\cot \frac{\alpha }{2} - \cot {2^{n - 1}}\alpha \).
\({S_n} = {\tan ^2}\frac{a}{2} \cdot \tan a + 2{\tan ^2}\frac{a}{{{2^2}}} \cdot \tan \frac{a}{2} + ... + {2^{n - 1}}{\tan ^2}\frac{a}{{{2^n}}} \cdot \tan \frac{a}{{{2^{n - 1}}}} = \tan a - {2^n}\tan \frac{a}{{{2^n}}}\) .
\({P_n} = \cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}} \cdot ... \cdot c{\rm{os}}\frac{x}{{{2^n}}}\)\( = \frac{{\sin x}}{{{2^n} \cdot \sin \frac{x}{{{2^n}}}}}\) .
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. \({P_n} = \cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}} \cdot ... \cdot c{\rm{os}}\frac{x}{{{2^n}}}\)
\({P_n} \cdot \sin \frac{x}{{{2^n}}} = \cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}}...\cos \frac{x}{{{2^{n - 1}}}} \cdot \left( {\cos \frac{x}{{{2^n}}} \cdot \sin \frac{x}{{{2^n}}}} \right)\)
\({P_n} \cdot \sin \frac{x}{{{2^n}}} = \frac{1}{{{2^{n - 1}}}} \cdot \cos \frac{x}{2} \cdot \sin \frac{x}{2} = \frac{1}{{{2^n}}} \cdot {\mathop{\rm s}\nolimits} {\rm{inx}}\)
\({P_n} = \frac{{\sin x}}{{{2^n} \cdot \sin \frac{x}{{{2^n}}}}}\).
Xét tính đúng sai của mệnh đề sai.
\(\cot x - \tan x - 2\tan 2x = 4\cot 4x\).
sin2(a + b) – sin2a – sin2b = 2sina∙sinb∙cos(a + b).
\(\sin \alpha \cdot \sin \left( {\frac{\pi }{3} - \alpha } \right) \cdot \sin \left( {\frac{\pi }{3} + \alpha } \right) = \frac{1}{4} \cdot \cos 3\alpha \).
cos(a + b)∙sin(a – b) + cos(b + c)∙sin(b – c) + cos(a + c)∙sin(c – a) = 1.
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. sin2(a + b) – sin2a – sin2b
= sin2(a + b) – (sin2a + sin2b) = sin2(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)
\( = \frac{1}{2}\left( {\sin 2a - \sin 2b} \right) + \frac{1}{2}\left( {\sin 2b - \sin 2c} \right) + \frac{1}{2}\left( {\sin 2c - \sin 2a} \right)\)
\( = \frac{1}{2}\left( {\sin 2a - \sin 2b + \sin 2b - \sin 2c + \sin 2c - \sin 2a} \right) = \frac{1}{2} \cdot 0 = 0\).
Xét tính đúng sai của các mệnh đề sau:
\({\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) = 2 - {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) - {\sin ^2}\left( {x + \pi } \right)\).
\({\sin ^4}x\left( {1 + {{\sin }^2}x} \right) + {\cos ^4}x\left( {1 + {{\cos }^2}x} \right) + 5\left( {404 + {{\sin }^2}x \cdot {{\cos }^2}x} \right) = 2020\).
\({\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x \cdot \cos 2x \cdot \cos 4x = \frac{1}{8}\cos 8x\).
\(\cos \left( {x - \frac{\pi }{3}} \right) \cdot \cos \left( {x + \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{6}} \right) \cdot \cos \left( {x + \frac{{3\pi }}{4}} \right) = \frac{{\sqrt 2 }}{2} \cdot \left( {\frac{{1 - \sqrt 3 }}{2}} \right)\).
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. \({\sin ^4}x\left( {1 + {{\sin }^2}x} \right) + {\cos ^4}x\left( {1 + {{\cos }^2}x} \right) + 5\left( {404 + {{\sin }^2}x \cdot {{\cos }^2}x} \right)\)
= \({\sin ^4}x + {\sin ^6}x + {\cos ^4}x + {\cos ^6}x + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x\)
\( = \left( {{{\sin }^4}x + {{\cos }^4}x} \right) + \left( {{{\sin }^6}x + {{\cos }^6}x} \right) + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x\)
\( = \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x \cdot {{\cos }^2}x} \right] + \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x \cdot {{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\)
\( + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x\)
\( = 1 - 2{\sin ^2}x \cdot {\cos ^2}x + 1 - 3{\sin ^2}x \cdot {\cos ^2}x + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x = 2022\).
c) Sai. \({\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x \cdot \cos 2x \cdot \cos 4x\)
\( = \left( {{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x} \right) \cdot \cos 2x \cdot \cos 4x = \frac{1}{2}\sin 2x \cdot \cos 2x \cdot \cos 4x = \frac{1}{4}\sin 4x \cdot \cos 4x = \frac{1}{8}\sin 8x\)d) Đúng. \(\cos \left( {x - \frac{\pi }{3}} \right) \cdot \cos \left( {x + \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{6}} \right) \cdot \cos \left( {x + \frac{{3\pi }}{4}} \right)\)
\( = \frac{1}{2}\left[ {\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right)} \right] + \frac{1}{2}\left[ {\cos \frac{{7\pi }}{{12}} + \cos \left( {2x + \frac{{11\pi }}{{12}}} \right)} \right]\)
\( = \frac{1}{2}\left[ {2\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right) + \cos \left( {2x + \frac{{11\pi }}{{12}}} \right)} \right]\)
\( = \frac{1}{2}\left[ {2\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right) + \cos \left( {2x - \frac{\pi }{{12}} + \pi } \right)} \right]\)
\( = \frac{1}{2}\left[ {2\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right) - \cos \left( {2x - \frac{\pi }{{12}}} \right)} \right]\)
\( = \cos \frac{{7\pi }}{{12}} = \cos \left( {\frac{\pi }{3} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3} \cdot \cos \frac{\pi }{4} - \sin \frac{\pi }{3} \cdot \sin \frac{\pi }{4}\)
\( = \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2} \cdot \left( {\frac{{1 - \sqrt 3 }}{2}} \right)\).
Cho tam giác ABC
Khi đó:
\(\sin B \cdot \cos C + \sin C \cdot \cos B = \sin A\).
cos2A + cos2B + cos2C = 1 − 2cosA∙cosB∙cosC.
\[\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C\].
\(\tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1\).
Đáp án: a) Sai. b) Đúng. c) Đúng. d) Đúng.
a) Sai. \(\sin B \cdot \cos C + \sin C \cdot \cos B\)
= sin(B + C) = sin(180° − A) = sinA.
b) Đúng. cos2A + cos2B + cos2C
\( = \frac{{1 + \cos 2A}}{2} + \frac{{1 + \cos 2B}}{2} + {\cos ^2}C = 1 + \frac{{\cos 2A + \cos 2B}}{2} + {\cos ^2}C\)
\( = 1 + \frac{1}{2} \cdot 2\left[ {\cos \left( {A + B} \right) \cdot \cos \left( {A - B} \right)} \right] + {\cos ^2}C = 1 + \cos \left( {\pi - C} \right) \cdot \cos \left( {A - B} \right) + {\cos ^2}C\)
\( = 1 - \cos C \cdot \cos \left( {A - B} \right) + {\cos ^2}C = 1 - \cos C \cdot \left[ {\cos \left( {A - B} \right) - \cos C} \right]\)
\( = 1 - \cos C\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right] = 1 - 2\cos A \cdot \cos B \cdot \cos C\).
c) Đúng. \[\tan A + \tan B + \tan C\]
\( = \tan \left( {A + B} \right) \cdot \left( {1 - \tan A \cdot \tan B} \right) + \tan C = - \tan C \cdot \left( {1 - \tan A \cdot \tan B} \right) + \tan C\)
\( = - \tan C + \tan A \cdot \tan B \cdot \tan C + \tan C = \tan A \cdot \tan B \cdot \tan C\).
d) Đúng. \(\tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1\)
Ta có \(\tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2} \cdot \tan \frac{B}{2}}} \Leftrightarrow \frac{1}{{\tan \frac{C}{2}}} = \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2} \cdot \tan \frac{B}{2}}}\)
\( \Leftrightarrow \tan \frac{C}{2} \cdot \left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right) = 1 - \tan \frac{A}{2} \cdot \tan \frac{B}{2}\)
\( \Leftrightarrow \tan \frac{C}{2} \cdot \tan \frac{A}{2} + \tan \frac{C}{2} \cdot \tan \frac{B}{2} + \tan \frac{A}{2} \cdot \tan \frac{B}{2} = 1\).
Giả sử A \( = \cos \frac{\pi }{{15}} \cdot \cos \frac{{2\pi }}{{15}} \cdot \cos \frac{{3\pi }}{{15}} \cdot \cos \frac{{4\pi }}{{15}} \cdot \cos \frac{{5\pi }}{{15}} \cdot \cos \frac{{6\pi }}{{15}} \cdot \cos \frac{{7\pi }}{{15}} = \frac{a}{b}\) . Khi đó giá trị của biểu thức a + b bằng bao nhiêu?
Đáp án: 129
\(A = \cos \frac{\pi }{{15}} \cdot \cos \frac{{2\pi }}{{15}} \cdot \cos \frac{{3\pi }}{{15}} \cdot \cos \frac{{4\pi }}{{15}} \cdot \cos \frac{{5\pi }}{{15}} \cdot \cos \frac{{6\pi }}{{15}} \cdot \cos \frac{{7\pi }}{{15}}\)
\( = \frac{{\sin \frac{\pi }{{15}} \cdot \cos \frac{\pi }{{15}} \cdot \cos \frac{{2\pi }}{{15}} \cdot \cos \frac{{3\pi }}{{15}} \cdot \cos \frac{{4\pi }}{{15}} \cdot \cos \frac{{5\pi }}{{15}} \cdot \cos \frac{{6\pi }}{{15}} \cdot \cos \frac{{7\pi }}{{15}} \cdot \sin \frac{{3\pi }}{{15}}}}{{\sin \frac{\pi }{{15}} \cdot \sin \frac{{3\pi }}{{15}}}}\)
\(\)\( = \frac{{\sin \frac{{2\pi }}{{15}} \cdot \cos \frac{{2\pi }}{{15}} \cdot \cos \frac{{4\pi }}{{15}} \cdot \frac{1}{2} \cdot \sin \frac{{6\pi }}{{15}} \cdot \cos \frac{{6\pi }}{{15}} \cdot \cos \frac{{7\pi }}{{15}}}}{{4\sin \frac{\pi }{{15}} \cdot \sin \frac{{3\pi }}{{15}}}}\)
\( = \frac{{\sin \frac{{4\pi }}{{15}} \cdot \cos \frac{{4\pi }}{{15}} \cdot \sin \frac{{12\pi }}{{15}} \cdot \cos \frac{{7\pi }}{{15}}}}{{32\sin \frac{\pi }{{15}} \cdot \sin \frac{{3\pi }}{{15}}}} = = \frac{{ - \sin \frac{{8\pi }}{{15}} \cdot \cos \frac{{8\pi }}{{15}} \cdot \sin \frac{{12\pi }}{{15}}}}{{64\sin \frac{\pi }{{15}} \cdot \sin \frac{{3\pi }}{{15}}}}\)
\( = \frac{{ - \sin \frac{{16\pi }}{{15}} \cdot \sin \frac{{12\pi }}{{15}}}}{{128\sin \frac{\pi }{{15}} \cdot \sin \frac{{3\pi }}{{15}}}} = \frac{1}{{128}}\).
Vậy a + b = 129.
Cho biểu thức sau:
\(B = \left[ {\tan \left( {\pi - x} \right) \cdot \tan \left( {\frac{{3\pi }}{2} + x} \right) \cdot \frac{1}{{{{\cos }^2}\left( {x - \frac{{3\pi }}{2}} \right)}} + \cos \left( {x - \frac{{3\pi }}{2}} \right) \cdot \frac{1}{{\sin \left( {\pi - x} \right)}}} \right] \cdot {\sin ^2}\left( {2x - x} \right)\)
Biết biểu thức B có dạng cos2(ax). Khi đó giá trị của a bằng bao nhiêu?
Đáp án: 1
\(B = \left[ {\tan \left( {\pi - x} \right) \cdot \tan \left( {\frac{{3\pi }}{2} + x} \right) \cdot \frac{1}{{{{\cos }^2}\left( {x - \frac{{3\pi }}{2}} \right)}} + \cos \left( {x - \frac{{3\pi }}{2}} \right) \cdot \frac{1}{{\sin \left( {\pi - x} \right)}}} \right] \cdot {\sin ^2}\left( {2x - x} \right)\)
\(B = \left[ { - \tan x \cdot \tan \left( {\frac{\pi }{2} + \pi + x} \right) \cdot \frac{1}{{{{\cos }^2}\left( {\frac{\pi }{2} + \pi - x} \right)}} + \cos \left( {\frac{\pi }{2} + \pi - x} \right) \cdot \frac{1}{{\sin x}}} \right] \cdot {\sin ^2}x\)
\(B = \left[ { - \tan x \cdot \left( { - \cot x} \right) \cdot \frac{1}{{{{\sin }^2}x}} - \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{{\sin x}}} \right] \cdot {\sin ^2}x = \left( {\frac{1}{{{{\sin }^2}x}} - 1} \right) \cdot {\sin ^2}x\)
\( = {\cot ^2}x \cdot {\sin ^2}x = {\cos ^2}x\).
Vậy a = 1.
Cho biểu thức sau:
\[\tan x \cdot \tan \left( {x + \frac{\pi }{3}} \right) + \tan \left( {x + \frac{\pi }{3}} \right) \cdot \tan \left( {x + \frac{{2\pi }}{3}} \right) + \tan \left( {x + \frac{{2\pi }}{3}} \right) \cdot \tan x\]. Kết quả rút gọn biểu thức có dạng a (a ∈ ℤ). Khi đó giá trị của a bằng bao nhiêu?
Đáp án: −3
Từ \(\tan \left( {a - b} \right) = \frac{{\tan a - \tan b}}{{1 + \tan a \cdot \tan b}} \Rightarrow \tan a \cdot \tan b = \frac{{\tan a - \tan b}}{{\tan \left( {a - b} \right)}} - 1\)
Áp dụng ta có:
\[\tan x \cdot \tan \left( {x + \frac{\pi }{3}} \right) = \frac{{\tan x - \tan \left( {x + \frac{\pi }{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1\]
\[\tan \left( {x + \frac{\pi }{3}} \right) \cdot \tan \left( {x + \frac{{2\pi }}{3}} \right) = \frac{{\tan \left( {x + \frac{\pi }{3}} \right) - \tan \left( {x + \frac{{2\pi }}{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1\]
\[\tan \left( {x + \frac{{2\pi }}{3}} \right) \cdot {\rm{tanx}} = \frac{{\tan \left( {x + \frac{{2\pi }}{3}} \right) - \tan x}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1\]
Khi đó \[\tan x \cdot \tan \left( {x + \frac{\pi }{3}} \right) + \tan \left( {x + \frac{\pi }{3}} \right) \cdot \tan \left( {x + \frac{{2\pi }}{3}} \right) + \tan \left( {x + \frac{{2\pi }}{3}} \right) \cdot {\rm{tanx}}\]
\[ = \frac{{\tan x - \tan \left( {x + \frac{\pi }{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1 + \frac{{\tan \left( {x + \frac{\pi }{3}} \right) - \tan \left( {x + \frac{{2\pi }}{3}} \right)}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1 + \frac{{\tan \left( {x + \frac{{2\pi }}{3}} \right) - \tan x}}{{\tan \left( {\frac{{ - \pi }}{3}} \right)}} - 1 = - 3\]
Giả sử \(M = {\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 2x + sin3x + \cos x + \cos 2x + \cos 3x\). Biết biểu thức M có dạng \(a\sqrt b \cos \left( {\frac{x}{2} + \frac{\pi }{6}} \right) \cdot \cos \left( {\frac{x}{2} - \frac{\pi }{6}} \right) \cdot \cos \left( {2x - \frac{\pi }{4}} \right)\). Khi đó giá trị của biểu thức a + b bằng bao nhiêu?
Đáp án: 6
\(M = {\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 2x + sin3x + \cos x + \cos 2x + \cos 3x\)
\( = \left( {\sin 3x + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right) + \left( {\cos 3x + \cos x} \right) + \sin 2x + \cos 2x\)
\( = 2\sin 2x \cdot \cos x + 2\cos 2x \cdot \cos x + \sin 2x + \cos 2x\)
= \(\left( {2\sin 2x\cos x + \sin 2x} \right) + \left( {2\cos 2x\cos x + \cos 2x} \right)\)
\( = \sin 2x\left( {2\cos x + 1} \right) + \cos 2x\left( {2\cos x + 1} \right) = \left( {{\mathop{\rm s}\nolimits} {\rm{in2x}} + \cos 2x} \right) \cdot \left( {2\cos x + 1} \right)\)
\( = 2\left( {\cos x + \frac{1}{2}} \right) \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right) = 2\left( {\cos x + \cos \frac{\pi }{3}} \right) \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right)\)
\( = 2\left[ {2\cos \left( {\frac{{x + \frac{\pi }{3}}}{2}} \right) \cdot \cos \left( {\frac{{x - \frac{\pi }{3}}}{2}} \right)} \right] \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right)\)
\( = 4\cos \left( {\frac{x}{2} - \frac{\pi }{6}} \right) \cdot \cos \left( {\frac{x}{2} + \frac{\pi }{6}} \right) \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right)\).
Vậy a + b = 6.
Cho biểu thức Q = cos54°cos4° – cos36°cos86° = cosa°. Khi đó giá trị của a bằng bao nhiêu?
Đáp án: 58
Q = cos54°cos4° – cos36°cos86° = cos54°cos4° – sin54°sin4° = cos(54° + 4°) = cos58°.
Vậy a = 58.
