20 câu trắc nghiệm Toán 11 Chân trời sáng tạo Các công thức lượng giác có đáp án

Đẳng thức sai là: ${\rm{sin}}\left( {{\rm{a + b}}} \right){\rm{ = sin}}\left( {\rm{a}} \right){\rm{cos}}\left( {\rm{b}} \right){\rm{ + cos}}\left( {\rm{a}} \right){\rm{sin}}\left( {\rm{b}} \right)$

Đáp án cần chọn là: A

$${\rm{sin}}\left( {{\rm{B + C}}} \right){\rm{ = sin}}\left( {{\rm{180}} - {\rm{A}}} \right){\rm{ = sin}}\left( {\rm{A}} \right)$$

$${\rm{cos}}\left( {{\rm{B + C}}} \right){\rm{ = cos}}\left( {{\rm{180}} - {\rm{A}}} \right){\rm{ = }} - {\rm{cos}}\left( {\rm{A}} \right)$$

$${\rm{tan}}\left( {{\rm{B + C}}} \right){\rm{ = tan}}\left( {{\rm{180}} - {\rm{A}}} \right){\rm{ = }} - {\rm{tan}}\left( {\rm{A}} \right)$$

$${\rm{cot}}\left( {{\rm{B + C}}} \right){\rm{ = cot}}\left( {{\rm{180}} - {\rm{A}}} \right){\rm{ = }} - {\rm{cot}}\left( {\rm{A}} \right)$$

Vậy đẳng thức sai là:$\sin \left( {B + C} \right){\rm{ = }} - \sin \left( A \right)$

Đáp án cần chọn là: A

Mệnh đề sai là: ${\rm{sin}}\left( {{\rm{2a}}} \right){\rm{ = sin}}\left( {\rm{a}} \right){\rm{cos}}\left( {\rm{a}} \right)$

Đáp án cần chọn là: B

Khẳng định đúng là ${\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{a}} \right){\rm{ + si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{a}} \right){\rm{ = 1}}$

Đáp án cần chọn là: C

Khẳng định sai: ${\rm{sin}}\left( {\rm{a}} \right) - {\rm{sin}}\left( {\rm{b}} \right){\rm{ = }} - {\rm{2sin}}\left( {\frac{{{\rm{a + b}}}}{{\rm{2}}}} \right){\rm{cos}}\left( {\frac{{{\rm{a}} - {\rm{b}}}}{{\rm{2}}}} \right)$

Đáp án cần chọn là: D

Ta có:${\rm{cos}}\left( {{\rm{a + }}\frac{{\rm{\pi }}}{{\rm{3}}}} \right){\rm{ = cos}}\left( {\rm{a}} \right){\rm{cos}}\left( {\frac{{\rm{\pi }}}{{\rm{3}}}} \right) - {\rm{sin}}\left( {\rm{a}} \right){\rm{sin}}\left( {\frac{{\rm{\pi }}}{{\rm{3}}}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{cos}}\left( {\rm{a}} \right) - \frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{sin}}\left( {\rm{a}} \right)$

Đáp án cần chọn là: D

Ta có:${\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ = 1}} \Rightarrow {\left( {\frac{{\rm{1}}}{{\rm{3}}}} \right)^{\rm{2}}}{\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ = 1}} \Rightarrow {\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ = 1}} - \frac{{\rm{1}}}{{\rm{9}}}{\rm{ = }}\frac{{\rm{8}}}{{\rm{9}}}$

Mà$\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < x < \pi }} \Rightarrow {\rm{cos}}\left( {\rm{x}} \right){\rm{ < 0}}$

$$\Rightarrow {\rm{cos}}\left( {\rm{x}} \right){\rm{ = }} - \sqrt {\frac{{\rm{8}}}{{\rm{9}}}} {\rm{ = }} - \frac{{2\sqrt 2 }}{3}$$

Đáp án cần chọn là: D

$${\rm{P = }}\frac{{{\rm{3sin}}\left( {\rm{x}} \right) - {\rm{4cos}}\left( {\rm{x}} \right)}}{{{\rm{cos}}\left( {\rm{x}} \right){\rm{ + 2sin}}\left( {\rm{x}} \right)}}{\rm{ = }}\frac{{\frac{{{\rm{3sin}}\left( {\rm{x}} \right) - {\rm{4cos}}\left( {\rm{x}} \right)}}{{{\rm{cos}}\left( {\rm{x}} \right)}}}}{{\frac{{{\rm{cos}}\left( {\rm{x}} \right){\rm{ + 2sin}}\left( {\rm{x}} \right)}}{{{\rm{cos}}\left( {\rm{x}} \right)}}}}{\rm{ = }}\frac{{{\rm{3tan}}\left( {\rm{x}} \right) - {\rm{4}}}}{{{\rm{1 + 2tan}}\left( {\rm{x}} \right)}}$$

Thay${\rm{tan}}\left( {\rm{x}} \right){\rm{ = 5}}$vào P ta được${\rm{P = }}\frac{{{\rm{3}}{\rm{.5}} - {\rm{4}}}}{{{\rm{1 + 2}}{\rm{.5}}}}{\rm{ = 1}}$

Đáp án cần chọn là: A

Ta có:${\rm{sin}}\left( {\rm{\alpha }} \right){\rm{ + cos}}\left( {\rm{\beta }} \right){\rm{ = }}\frac{{\rm{5}}}{{\rm{4}}} \Leftrightarrow {\left[ {{\rm{sin}}\left( {\rm{\alpha }} \right){\rm{ + cos}}\left( {\rm{\beta }} \right)} \right]^{\rm{2}}}{\rm{ = }}{\left( {\frac{{\rm{5}}}{{\rm{4}}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{25}}}}{{{\rm{16}}}}$

$$\Rightarrow {\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ + 2sin}}\left( {\rm{\alpha }} \right){\rm{cos}}\left( {\rm{\alpha }} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{\beta }} \right){\rm{ = }}\frac{{{\rm{25}}}}{{{\rm{16}}}}$$

$$\Rightarrow {\rm{1 + sin}}\left( {{\rm{2\alpha }}} \right){\rm{ = }}\frac{{{\rm{25}}}}{{{\rm{16}}}}$$

$$\Rightarrow {\rm{sin}}\left( {{\rm{2\alpha }}} \right){\rm{ = }}\frac{{{\rm{25}}}}{{{\rm{16}}}} - {\rm{1 = }}\frac{{\rm{9}}}{{{\rm{16}}}}$$

Đáp án cần chọn là: C

Ta có$\sin \left( \alpha \right) = \frac{1}{{\sqrt 3 }}$, ${\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ = 1}} \Rightarrow {\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ = 1}} - \frac{{\rm{1}}}{{\rm{3}}}{\rm{ = }}\frac{{\rm{2}}}{{\rm{3}}}$

Vì $0 < \alpha < \frac{\pi }{2}$nên $\cos \left( \alpha \right) > 0 \Rightarrow \cos \left( \alpha \right) = \sqrt {\frac{2}{3}}$</>

$$\Rightarrow {\rm{sin}}\left( {{\rm{\alpha + }}\frac{{\rm{\pi }}}{{\rm{3}}}} \right){\rm{ = sin}}\left( {\rm{\alpha }} \right){\rm{cos}}\left( {\frac{{\rm{\pi }}}{{\rm{3}}}} \right){\rm{ + cos}}\left( {\rm{\alpha }} \right){\rm{sin}}\left( {\frac{{\rm{\pi }}}{{\rm{3}}}} \right){\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{3}} }}{\rm{.}}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\sqrt {\frac{{\rm{2}}}{{\rm{3}}}} {\rm{.}}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ = }}\frac{{\sqrt {\rm{3}} }}{{\rm{6}}}{\rm{ + }}\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}$$

Đáp án cần chọn là: D

$${\rm{P = si}}{{\rm{n}}^{\rm{6}}}\left( {\rm{x}} \right){\rm{ + co}}{{\rm{s}}^{\rm{6}}}\left( {\rm{x}} \right){\rm{ = }}{\left[ {{\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{x}} \right)} \right]^{\rm{3}}}{\rm{ + }}{\left[ {{\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right)} \right]^{\rm{3}}}$$

$${\rm{ = }}{\left[ {{\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right)} \right]^{\rm{3}}} - {\rm{3si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right)\left[ {{\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right)} \right]$$

$${\rm{ = 1}} - {\rm{3si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ = 1}} - \frac{{\rm{3}}}{{\rm{4}}}{\left[ {{\rm{2sin}}\left( {\rm{x}} \right){\rm{cos}}\left( {\rm{x}} \right)} \right]^{\rm{2}}}{\rm{ = 1}} - \frac{{\rm{3}}}{{\rm{4}}}{\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{2x}}} \right)$$

Đáp án cần chọn là: C

$${\rm{Q = }}\frac{{{\rm{1 + sin}}\left( {{\rm{4a}}} \right) - {\rm{cos}}\left( {{\rm{4a}}} \right)}}{{{\rm{1 + sin}}\left( {{\rm{4a}}} \right){\rm{ + cos}}\left( {{\rm{4a}}} \right)}}{\rm{ = }}\frac{{{\rm{sin}}\left( {{\rm{4a}}} \right){\rm{ + }}\left[ {{\rm{1}} - {\rm{cos}}\left( {{\rm{4a}}} \right)} \right]}}{{{\rm{sin}}\left( {{\rm{4a}}} \right){\rm{ + }}\left[ {{\rm{1 + cos}}\left( {{\rm{4a}}} \right)} \right]}}$$

$${\rm{ = }}\frac{{{\rm{2sin}}\left( {{\rm{2a}}} \right){\rm{cos}}\left( {{\rm{2a}}} \right){\rm{ + 2si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{2a}}} \right)}}{{{\rm{2sin}}\left( {{\rm{2a}}} \right){\rm{cos}}\left( {{\rm{2a}}} \right){\rm{ + 2co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{2a}}} \right)}}{\rm{ = }}\frac{{{\rm{2sin}}\left( {{\rm{2a}}} \right)\left[ {{\rm{cos}}\left( {{\rm{2a}}} \right){\rm{ + sin}}\left( {{\rm{2a}}} \right)} \right]}}{{{\rm{2cos}}\left( {{\rm{2a}}} \right)\left[ {{\rm{sin}}\left( {{\rm{2a}}} \right){\rm{ + cos}}\left( {{\rm{2a}}} \right)} \right]}}$$

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

Đáp án cần chọn là: C

Do${\rm{0 < a, b < }}\frac{{\rm{\pi }}}{{\rm{2}}} \Rightarrow {\rm{0 < a + b < \pi }}$

Ta có${\rm{tan}}\left( {{\rm{a + b}}} \right){\rm{ = }}\frac{{{\rm{tan}}\left( {\rm{a}} \right){\rm{ + tan}}\left( {\rm{b}} \right)}}{{{\rm{1}} - {\rm{tan}}\left( {\rm{a}} \right){\rm{tan}}\left( {\rm{b}} \right)}}{\rm{ = }}\frac{{\frac{{\rm{1}}}{{\rm{7}}}{\rm{ + }}\frac{{\rm{3}}}{{\rm{4}}}}}{{{\rm{1}} - \frac{{\rm{1}}}{{\rm{7}}}{\rm{.}}\frac{{\rm{3}}}{{\rm{4}}}}}{\rm{ = 1}} \Leftrightarrow {\rm{a + b = }}\frac{{\rm{\pi }}}{{\rm{4}}}$

Đáp án cần chọn là: C

Từ${\rm{cot}}\left( {\rm{\alpha }} \right){\rm{ = }}\frac{{\rm{2}}}{{\rm{3}}} \Rightarrow {\rm{tan}}\left( {\rm{\alpha }} \right){\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}$

$${\rm{sin}}\left( {{\rm{2\alpha }}} \right){\rm{ = }}\frac{{{\rm{2tan}}\left( {\rm{\alpha }} \right)}}{{{\rm{ta}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ + 1}}}}{\rm{ = }}\frac{{{\rm{2}}{\rm{.}}\frac{{\rm{3}}}{{\rm{2}}}}}{{{{\left( {\frac{{\rm{3}}}{{\rm{2}}}} \right)}^{\rm{2}}}{\rm{ + 1}}}}{\rm{ = }}\frac{{\rm{3}}}{{\frac{{{\rm{13}}}}{{\rm{4}}}}}{\rm{ = }}\frac{{{\rm{12}}}}{{{\rm{13}}}}$$

$${\rm{cos}}\left( {{\rm{2\alpha }}} \right){\rm{ = }}\frac{{{\rm{1}} - {\rm{ta}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right)}}{{{\rm{1 + ta}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right)}}{\rm{ = }}\frac{{{\rm{1}} - {{\left( {\frac{{\rm{3}}}{{\rm{2}}}} \right)}^{\rm{2}}}}}{{{\rm{1 + }}{{\left( {\frac{{\rm{3}}}{{\rm{2}}}} \right)}^{\rm{2}}}}}{\rm{ = }}\frac{{ - {\rm{5}}}}{{{\rm{13}}}}$$

Ta có${\rm{sin}}\left( {{\rm{2\alpha + }}\frac{{{\rm{7\pi }}}}{{\rm{4}}}} \right){\rm{ = sin}}\left( {{\rm{2\alpha }} - \frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + 2\pi }}} \right){\rm{ = sin}}\left( {{\rm{2\alpha }} - \frac{{\rm{\pi }}}{{\rm{4}}}} \right)$

$${\rm{ = sin}}\left( {{\rm{2\alpha }}} \right){\rm{cos}}\left( {\frac{{\rm{\pi }}}{{\rm{4}}}} \right) - {\rm{cos}}\left( {{\rm{2\alpha }}} \right){\rm{sin}}\left( {\frac{{\rm{\pi }}}{{\rm{4}}}} \right){\rm{ = }}\frac{{{\rm{12}}}}{{{\rm{13}}}}{\rm{.}}\frac{{\sqrt {\rm{2}} }}{{\rm{2}}} - \left( {\frac{{ - {\rm{5}}}}{{{\rm{13}}}}} \right){\rm{.}}\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}{\rm{ = }}\frac{{{\rm{17}}\sqrt {\rm{2}} }}{{{\rm{26}}}}$$

Đáp án cần chọn là: A

$${\rm{A = co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{\alpha + \beta }}} \right) - {\rm{2cos}}\left( {\rm{\alpha }} \right){\rm{cos}}\left( {\rm{\beta }} \right){\rm{cos}}\left( {{\rm{\alpha + \beta }}} \right)$$

$${\rm{ = co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{\alpha + \beta }}} \right) - {\rm{2}}{\rm{.}}\frac{1}{2}\left[ {{\rm{cos}}\left( {{\rm{\alpha + \beta }}} \right) + {\rm{cos}}\left( {{\rm{\alpha }} - {\rm{\beta }}} \right)} \right]{\rm{cos}}\left( {{\rm{\alpha + \beta }}} \right)$$

$${\rm{ = co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{\alpha + \beta }}} \right) - {\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{\alpha + \beta }}} \right) - {\rm{cos}}\left( {{\rm{\alpha }} - {\rm{\beta }}} \right){\rm{cos}}\left( {{\rm{\alpha + \beta }}} \right)$$

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

$$= \frac{1}{2} - \frac{{{\rm{cos}}\left( {2{\rm{\beta }}} \right)}}{2} = {\sin ^2}\left( {\rm{\beta }} \right)$$

Đáp án cần chọn là: D

$$\frac{{{\rm{sin}}\left( {{\rm{2\alpha }}} \right){\rm{ + sin}}\left( {{\rm{5\alpha }}} \right) - {\rm{sin}}\left( {{\rm{3\alpha }}} \right)}}{{{\rm{2co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{2\alpha }}} \right){\rm{ + cos}}\left( {\rm{\alpha }} \right) - {\rm{1}}}}{\rm{ = }} - {\rm{2}}$$

$$\Leftrightarrow \frac{{2\sin \left( \alpha \right)\cos \left( \alpha \right) + 2\cos \left( {4\alpha } \right)\sin \left( \alpha \right)}}{{2.\frac{{1 + \cos \left( {4\alpha } \right)}}{2} + \cos \left( \alpha \right) - 1}}{\rm{ = }} - 2$$

$$\Leftrightarrow \frac{{{\rm{2sin}}\left( {\rm{\alpha }} \right){\rm{cos}}\left( {\rm{\alpha }} \right){\rm{ + 2cos}}\left( {{\rm{4\alpha }}} \right){\rm{sin}}\left( {\rm{\alpha }} \right)}}{{{\rm{cos}}\left( {{\rm{4\alpha }}} \right){\rm{ + cos}}\left( {\rm{\alpha }} \right)}}{\rm{ = }} - 2$$

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

$$\Leftrightarrow {\rm{2sin(\alpha ) = }} - 2$$

$$\Leftrightarrow {\rm{sin(\alpha ) = }} - 1$$

Đáp án cần chọn là: A

$${\rm{S = si}}{{\rm{n}}^{\rm{2}}}{{\rm{5}}^{\rm{0}}}{\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{1}}{{\rm{0}}^{\rm{0}}}{\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{1}}{{\rm{5}}^{\rm{0}}}{\rm{ + }}...{\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{8}}{{\rm{5}}^{\rm{0}}}$$

$$= (si{n^2}{5^0} + si{n^2}{85^0}) + (si{n^2}{10^0} + si{n^2}{80^0}) + ... + (si{n^2}{40^0} + si{n^2}{50^0}) + si{n^2}{45^0}$$

$${\rm{ = }}\left[ {co{s^2}\left( {{{180}^0} - {5^0}} \right) + si{n^2}{{85}^0}} \right] + \left[ {co{s^2}\left( {{{180}^0} - {{10}^0}} \right) + si{n^2}{{80}^0}} \right] + ..$$

$$+ \left[ {co{s^2}\left( {{{180}^0} - {{40}^0}} \right) + si{n^2}{{50}^0}} \right] + si{n^2}{45^0}$$

$${\rm{ = }}\left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{8}}{{\rm{5}}^{{\rm{0 }}}}{\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{8}}{{\rm{5}}^{\rm{0}}}} \right){\rm{ + (co}}{{\rm{s}}^{\rm{2}}}{\rm{8}}{{\rm{0}}^{{\rm{0 }}}}{\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{8}}{{\rm{0}}^{\rm{0}}}{\rm{) + }}...{\rm{ + }}\left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{5}}{{\rm{0}}^{{\rm{0 }}}}{\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{5}}{{\rm{0}}^{\rm{0}}}} \right){\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{4}}{{\rm{5}}^{\rm{0}}}$$

${\rm{ = 1 + 1 + }}...{\rm{ + 1 + }}{\left( {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} \right)^{{\rm{2 }}}}{\rm{ = 8 + }}{\left( {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} \right)^{\rm{2}}}$

${\rm{ = }}\frac{{{\rm{17}}}}{{\rm{2}}}$

Đáp án cần chọn là: B

$$\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{sinA}}}}} \right)\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{sinB}}}}} \right)\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{sinC}}}}} \right){\rm{ = }}{\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\sqrt[{\rm{3}}]{{{\rm{sinA}}{\rm{.sinB}}{\rm{.sinC}}}}}}} \right)^{\rm{3}}}$$

$$\Leftrightarrow \frac{{\left( {sinA + 1} \right)\left( {sinB + 1} \right)\left( {sinC + 1} \right)}}{{sinA.sinB.sinC}} = {\left( {\frac{{\sqrt[3]{{sinA.sinB.sinC}} + 1}}{{\sqrt[3]{{sinA.sinB.sinC}}}}} \right)^3}$$

$$\Leftrightarrow \left( {sinA + 1} \right)\left( {sinB + 1} \right)\left( {sinC + 1} \right) = {\left( {\sqrt[3]{{sinA.sinB.sinC}} + 1} \right)^3}$$

$$\Leftrightarrow sinA.sinB.sinC + sinA.sinB + sinB.sinC + \sin A.sinC + \sin A + \sin B + \sin C + 1$$

$$= sinA.sinB.sinC + 3\sqrt[3]{{sinA.sinB.sinC}} + 3\sqrt[3]{{{{\left( {sinA.sinB.sinC} \right)}^2}}} + 1$$

$$\Leftrightarrow sinA.sinB + sinB.sinC + \sin A.sinC + \sin A + \sin B + \sin C$$

$$= 3\sqrt[3]{{sinA.sinB.sinC}} + 3\sqrt[3]{{{{\left( {sinA.sinB.sinC} \right)}^2}}}$$

Ta có A, B, C là các góc trong tam giác $\Rightarrow {\rm{0 < sinA, sinB, sinC}} \le 1$

Áp dụng bất đẳng sức cô si ta có:

$\left\{ {\begin{array}{cccccccccccccccccccc}{sinA.sinB + sinB.sinC + \sin A.sinC \ge 3\sqrt[3]{{si{n^2}A.si{n^2}B.si{n^2}C}}}\\{sinA + sinB + sinC \ge 3\sqrt[3]{{sinA.sinB.sinC}}}\end{array}} \right.$

$$\Rightarrow sinA.sinB + sinB.sinC + sinAsinC + sinA + sinB + sinC$$

$$\ge 3\sqrt[3]{{si{n^2}A.si{n^2}B.si{n^2}C}} + 3\sqrt[3]{{sinA.sinB.sinC}}$$

Dấu = xảy ra $\Leftrightarrow {\rm{sinA = sinB = sinC}}$

$$\widehat A{\rm{ = }}\widehat B{\rm{ = }}\widehat C{\rm{ = 6}}{{\rm{0}}^{\rm{0}}}$$

Đáp án cần chọn là: A

Ta có ${\rm{tan}}\left( {\rm{\alpha }} \right)$và ${\rm{tan}}\left( {\rm{\beta }} \right)$là nghiệm của phương trình ${{\rm{x}}^{\rm{2}}} - {\rm{px + q = 0, (q}} \ne 1)$

Theo định lí Vi-ét ta có:

$\left\{ {\begin{array}{cccccccccccccccccccc}{{\rm{tan(\alpha ) + tan(\beta ) = p}}}\\{{\rm{tan(\alpha )}}{\rm{.tan(\beta ) = q}}}\end{array}} \right. \Rightarrow tan(\alpha + \beta )\,\,{\rm{ = }}\frac{{{\rm{tan(\alpha ) + tan(\beta )}}}}{{{\rm{1}} - {\rm{tan(\alpha )}}{\rm{.tan(\beta )}}}}{\rm{ = }}\frac{p}{{1 - q}}$

$$\Rightarrow {\cos ^2}\left( {\alpha + \beta } \right){\rm{ = }}\frac{1}{{1 + ta{n^2}(\alpha + \beta )}}{\rm{ = }}\frac{1}{{1 + \frac{{{p^2}}}{{{{\left( {1 - q} \right)}^2}}}}}{\rm{ = }}\frac{{{{\left( {1 - q} \right)}^2}}}{{{{\left( {1 - q} \right)}^2} + {p^2}}}$$

$$q \ne 1 \Rightarrow \frac{{sin(\alpha )sin(\beta )}}{{cos(\alpha )cos(\beta )}} \ne 1 \Rightarrow sin(\alpha )sin(\beta ) \ne cos(\alpha )cos(\beta )$$

$$\Rightarrow {\rm{cos(\alpha + \beta ) = cos(\alpha )cos(\beta )}} - {\rm{sin(\alpha )sin(\beta )}} \ne 0$$

$\Rightarrow Q\,\,{\rm{ = }}co{s^2}(\alpha + \beta )\left[ {1 + p.\frac{{sin(\alpha + \beta )}}{{cos(\alpha + \beta )}} + q.\frac{{si{n^2}(\alpha + \beta )}}{{co{s^2}(\alpha + \beta )}}} \right]$

$${\rm{ = co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{\alpha + \beta }}} \right)\left[ {{\rm{1 + p}}{\rm{.tan}}\left( {{\rm{\alpha + \beta }}} \right){\rm{ + q}}{\rm{.ta}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{\alpha + \beta }}} \right)} \right]$$

${\rm{ = }}\frac{{{{\left( {1 - q} \right)}^2}}}{{{{\left( {1 - q} \right)}^2} + {p^2}}}\left[ {1 + \frac{{{p^2}}}{{1 - q}} + \frac{{{p^2}q}}{{{{\left( {1 - q} \right)}^2}}}} \right]{\rm{ = }}\frac{{{{\left( {1 - q} \right)}^2}}}{{{{\left( {1 - q} \right)}^2} + {p^2}}}.\frac{{{{\left( {1 - q} \right)}^2} + {p^2}\left( {1 - q} \right) + {p^2}q}}{{{{\left( {1 - q} \right)}^2}}}$

${\rm{ = 1}}$

Đáp án cần chọn là: D

$${\rm{sin}}\left( {\rm{A}} \right){\rm{ + sin}}\left( {\rm{B}} \right){\rm{ = cos}}\left( {\rm{A}} \right){\rm{ + cos}}\left( {\rm{B}} \right)$$

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

Nếu ${\rm{cos}}\left( {\frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}} \right){\rm{ = 0}} \Rightarrow \frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}{\rm{ = 9}}{{\rm{0}}^{\rm{0}}} \Rightarrow {\rm{A}} - {\rm{B = 18}}{{\rm{0}}^{\rm{0}}}{\rm{ = A + B + C}} \Leftrightarrow {\rm{2B + C = 0}}$

Nếu${\rm{cos}}\left( {\frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}} \right) \ne 0$ khi đó

$\left( {\rm{1}} \right) \Leftrightarrow {\rm{sin}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right){\rm{ = cos}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right) \Leftrightarrow {\rm{sin}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right){\rm{ = sin}}\left( {\frac{{\rm{C}}}{{\rm{2}}}} \right)$do $\frac{{{\rm{A + B}}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{C}}}{{\rm{2}}}{\rm{ = }}{90^0}$

$$\Rightarrow \frac{{{\rm{A + B}}}}{{\rm{2}}}{\rm{ = }}\frac{{\rm{C}}}{{\rm{2}}} \Leftrightarrow {\rm{A + B = C}} \Leftrightarrow {\rm{18}}{{\rm{0}}^{\rm{0}}} - {\rm{C = C}} \Rightarrow {\rm{C = 9}}{{\rm{0}}^{\rm{0}}}$$

Đáp án cần chọn là: B