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

Hướng dẫn giải:

Đáp án đúng là: A

$\frac{\sin \frac{\pi }{15}cos\frac{\pi }{10}+\sin \frac{\pi }{10}cos\frac{\pi }{15}}{cos\frac{2\pi }{15}cos\frac{\pi }{5}-\sin \frac{2\pi }{15}\sin \frac{\pi }{5}}$$=\frac{\sin \left(\frac{\pi }{15}+\frac{\pi }{10}\right)}{cos\left(\frac{2\pi }{15}+\frac{\pi }{5}\right)}=\frac{\sin \left(\frac{\pi }{6}\right)}{cos\left(\frac{\pi }{3}\right)}=\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 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 \(P = \frac{{2\sqrt 5 }}{3}\).

Hướng dẫn giải:

Đáp án đúng là: D

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

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

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: 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: 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: 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.