Trắc Nghiệm Bài 2 Công Thức Lượng Giác Mức Vận Dụng Giải Chi Tiết

Trắc nghiệm bài 2 công thức lượng giác mức vận dụng giải chi tiết được soạn dưới dạng file word và PDF gồm 6 trang. Các bạn xem và tải về ở dưới.

Câu 1. Cho ${\text{sin}}\alpha = \frac{1}{{\sqrt 3 }}$ với $0 < \alpha < \frac{\pi }{2}$. Giá trị của ${\text{cos}}\left( {\alpha + \frac{\pi }{3}} \right)$ bằng
A. $\frac{{2 – \sqrt 6 }}{{2\sqrt 6 }}$.
B. $\sqrt 6 – 3$.
C. $\frac{1}{{\sqrt 6 }} – \frac{1}{2}$.
D. $\sqrt 6 – \frac{1}{2}$.

Chọn A

Lời giải

Ta có: ${\text{si}}{{\text{n}}^2}\alpha + {\text{co}}{{\text{s}}^2}\alpha = 1 \Leftrightarrow {\text{co}}{{\text{s}}^2}\alpha = \frac{2}{3} \Leftrightarrow {\text{cos}}\alpha = \frac{{\sqrt 6 }}{3}$ (vì $0 < \alpha < \frac{1}{2}$ nên ${\text{cos}}\alpha > 0$ ).

Ta có: ${\text{cos}}\left( {\alpha + \frac{\pi }{3}} \right) = \frac{1}{2}{\text{cos}}\alpha – \frac{{\sqrt 3 }}{2}{\text{sin}}\alpha = \frac{1}{2} \cdot \frac{{\sqrt 6 }}{3} – \frac{{\sqrt 3 }}{2} \cdot \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 6 }} – \frac{1}{2} = \frac{{2 – \sqrt 6 }}{{2\sqrt 6 }}$.

Câu 2. Cho hai góc $\alpha ,\beta $ thỏa mãn ${\text{sin}}\alpha = \frac{5}{{13}},\left( {\frac{\pi }{2} < \alpha < \pi } \right)$ và ${\text{cos}}\beta = \frac{3}{5},\left( {0 < \beta < \frac{\pi }{2}} \right)$. Tính giá trị đúng của ${\text{cos}}\left( {\alpha – \beta } \right)$.
A. $\frac{{16}}{{65}}$.
B. $ – \frac{{18}}{{65}}$.
C. $\frac{{18}}{{65}}$.
D. $ – \frac{{16}}{{65}}$.

Chọn D

Lời giải

${\text{sin}}\alpha = \frac{5}{{13}},\left( {\frac{\pi }{2} < \alpha < \pi } \right)$ nên ${\text{cos}}\alpha = – \sqrt {1 – {{\left( {\frac{5}{{13}}} \right)}^2}} = – \frac{{12}}{{13}}$.

${\text{cos}}\beta = \frac{3}{5},\left( {0 < \beta < \frac{\pi }{2}} \right)$ nên ${\text{sin}}\beta = \sqrt {1 – {{\left( {\frac{3}{5}} \right)}^2}} = \frac{4}{5}$.

${\text{cos}}\left( {\alpha – \beta } \right) = {\text{cos}}\alpha {\text{cos}}\beta + {\text{sin}}\alpha {\text{sin}}\beta = – \frac{{12}}{{13}} \cdot \frac{3}{5} + \frac{5}{{13}} \cdot \frac{4}{5} = – \frac{{16}}{{65}}$.

Câu 3. Cho ${\text{sin}}\alpha = \frac{3}{5},\alpha \in \left( {\frac{\pi }{2};\frac{{3\pi }}{2}} \right)$. Tính giá trị ${\text{cos}}\left( {\alpha – \frac{{21\pi }}{4}} \right)$ ?
A. $\frac{{\sqrt 2 }}{{10}}$.
B. $\frac{{ – 7\sqrt 2 }}{{10}}$.
C. $\frac{{ – \sqrt 2 }}{{10}}$.
D. $\frac{{7\sqrt 2 }}{{10}}$.

Chọn A

Lời giải

Ta có: ${\text{co}}{{\text{s}}^2}\alpha = 1 – {\text{si}}{{\text{n}}^2}\alpha = \frac{{16}}{{25}} \Leftrightarrow {\text{cos}}\alpha = \pm \frac{4}{5}$. Do $\alpha \in \left( {\frac{\pi }{2};\frac{{3\pi }}{2}} \right) \Rightarrow {\text{cos}}\alpha < 0$ nên ${\text{cos}}\alpha = \frac{{ – 4}}{5}$.

Vậy: ${\text{cos}}\left( {\alpha – \frac{{21\pi }}{4}} \right) = {\text{cos}}\alpha {\text{cos}}\frac{{21\pi }}{4} + {\text{sin}}\alpha {\text{sin}}\frac{{21\pi }}{4} = \frac{{ – 4}}{5}\left( {\frac{{ – \sqrt 2 }}{2}} \right) + \frac{3}{5}\left( {\frac{{ – \sqrt 2 }}{2}} \right) = \frac{{\sqrt 2 }}{{10}}$.

Câu 4. Biểu thức $M = {\text{cos}}\left( { – {{53}^ \circ }} \right) \cdot {\text{sin}}\left( { – {{337}^ \circ }} \right) + {\text{sin}}{307^ \circ } \cdot {\text{sin}}{113^ \circ }$ có giá trị bằng:
A. $ – \frac{1}{2}$.
B. $\frac{1}{2}$.
C. $ – \frac{{\sqrt 3 }}{2}$.
D. $\frac{{\sqrt 3 }}{2}$.

Chọn A.

Lời giải

$M = {\text{cos}}\left( { – {{53}^ \circ }} \right) \cdot {\text{sin}}\left( { – {{337}^ \circ }} \right) + {\text{sin}}{307^ \circ } \cdot {\text{sin}}{113^ \circ }$

$ = {\text{cos}}\left( { – {{53}^ \circ }} \right) \cdot {\text{sin}}\left( {{{23}^ \circ } – {{360}^ \circ }} \right) + {\text{sin}}\left( { – {{53}^ \circ } + {{360}^ \circ }} \right) \cdot {\text{sin}}\left( {{{90}^ \circ } + {{23}^ \circ }} \right)$

$ = {\text{cos}}\left( { – {{53}^ \circ }} \right) \cdot {\text{sin}}{23^ \circ } + {\text{sin}}\left( { – {{53}^ \circ }} \right) \cdot {\text{cos}}{23^ \circ } = {\text{sin}}\left( {{{23}^ \circ } – {{53}^ \circ }} \right) = – {\text{sin}}{30^ \circ } = – \frac{1}{2}$.

Câu 5. Rút gọn biểu thức: ${\text{cos}}{54^ \circ } \cdot {\text{cos}}{4^ \circ } – {\text{cos}}{36^ \circ } \cdot {\text{cos}}{86^ \circ }$, ta được:
A. ${\text{cos}}{50^ \circ }$.
B. ${\text{cos}}{58^ \circ }$.
C. ${\text{sin}}{50^ \circ }$.
D. ${\text{sin}}{58^ \circ }$.

Chọn D.

Lời giải

Ta có: ${\text{cos}}{54^ \circ } \cdot {\text{cos}}{4^ \circ } – {\text{cos}}{36^ \circ } \cdot {\text{cos}}{86^ \circ } = {\text{cos}}{54^ \circ } \cdot {\text{cos}}{4^ \circ } – {\text{sin}}{54^ \circ } \cdot {\text{sin}}{4^ \circ } = {\text{cos}}{58^ \circ }$.

Câu 6. Cho hai góc nhọn $a$ và $b$ với ${\text{tan}}a = \frac{1}{7}$ và ${\text{tan}}b = \frac{3}{4}$. Tính $a + b$.
A. $\frac{\pi }{3}$
B. $\frac{\pi }{4}$.
C. $\frac{\pi }{6}$.
D. $\frac{{2\pi }}{3}$.

Chọn B.

Lời giải

${\text{tan}}\left( {a + b} \right) = \frac{{{\text{tan}}a + {\text{tan}}b}}{{1 – {\text{tan}}a \cdot {\text{tan}}b}} = 1$, suy ra $a + b = \frac{\pi }{4}$

Câu 7. Cho $x,y$ là các góc nhọn, ${\text{cot}}x = \frac{3}{4},{\text{cot}}y = \frac{1}{7}$. Tổng $x + y$ bằng:
A. $\frac{\pi }{4}$.
B. $\frac{{3\pi }}{4}$.
C. $\frac{\pi }{3}$.
D. $\pi $.

Chọn C.

Lời giải

Ta có :

${\text{tan}}\left( {x + y} \right) = \frac{{{\text{tan}}x + {\text{tan}}y}}{{1 – {\text{tan}}x \cdot {\text{tan}}y}} = \frac{{\frac{4}{3} + 7}}{{1 – \frac{4}{3} \cdot 7}} = – 1$

Suy ra, $x + y = \frac{{3\pi }}{4}$.

Câu 8. Biểu thức $A = {\text{co}}{{\text{s}}^2}x + {\text{co}}{{\text{s}}^2}\left( {\frac{\pi }{3} + x} \right) + {\text{co}}{{\text{s}}^2}\left( {\frac{\pi }{3} – x} \right)$ không phụ thuộc $x$ và bằng:
A. $\frac{3}{4}$.
B. $\frac{4}{3}$.
C. $\frac{3}{2}$.
D. $\frac{2}{3}$.

Lời giải

Chọn C

Ta có :

$A = {\text{co}}{{\text{s}}^2}x + {\text{co}}{{\text{s}}^2}\left( {\frac{\pi }{3} + x} \right) + {\text{co}}{{\text{s}}^2}{\left( {\frac{\pi }{3} – x} \right)^2}$
$ = {\text{co}}{{\text{s}}^2}x + {\left( {\frac{{\sqrt 3 }}{2}{\text{cos}}x – \frac{1}{2}{\text{sin}}x} \right)^2} + \left( {\frac{{\sqrt 3 }}{2}{\text{cos}}x + \frac{1}{2}{\text{sin}}x} \right)$
$ = \frac{3}{2}.$

Câu 9. Biết ${\text{sin}}\beta = \frac{4}{5},0 < \beta < \frac{\pi }{2}$ và $\alpha \ne k\pi $. Giá trị của biểu thức: $A = \frac{{\sqrt 3 {\text{sin}}\left( {\alpha + \beta } \right) – \frac{{4{\text{cos}}\left( {\alpha + \beta } \right)}}{{\sqrt 3 }}}}{{{\text{sin}}\alpha }}$ không phụ thuộc vào $\alpha $ và bằng
A. $\frac{{\sqrt 5 }}{3}$.
B. $\frac{5}{{\sqrt 3 }}$.
C. $\frac{{\sqrt 3 }}{5}$.
D. $\frac{3}{{\sqrt 5 }}$.

Chọn B.

Lời giải

Ta có $\left\{ {\begin{array}{*{20}{l}}
{0 < \beta < \frac{\pi }{2}} \\
{{\text{sin}}\beta = \frac{4}{5}}
\end{array} \Rightarrow {\text{cos}}\beta = \frac{3}{5}} \right.$,
thay vào biểu thức $A = \frac{{\sqrt 3 {\text{sin}}\left( {\alpha + \beta } \right) – \frac{{4{\text{cos}}\left( {\alpha + \beta } \right)}}{{\sqrt 3 }}}}{{{\text{sin}}\alpha }} = \frac{5}{{\sqrt 3 }}$.

Câu 10. Nếu ${\text{tan}}\frac{\beta }{2} = 4{\text{tan}}\frac{\alpha }{2}$ thì ${\text{tan}}\frac{{\beta – \alpha }}{2}$ bằng:
A. $\frac{{3{\text{sin}}\alpha }}{{5 – 3{\text{cos}}\alpha }}$.
B. $\frac{{3{\text{sin}}\alpha }}{{5 + 3{\text{cos}}\alpha }}$.
C. $\frac{{3{\text{cos}}\alpha }}{{5 – 3{\text{cos}}\alpha }}$.
D. $\frac{{3{\text{cos}}\alpha }}{{5 + 3{\text{cos}}\alpha }}$.

Chọn A.

Lời giải

Ta có:

${\text{tan}}\frac{{\beta – \alpha }}{2} = \frac{{{\text{tan}}\frac{\beta }{2} – {\text{tan}}\frac{\alpha }{2}}}{{1 + {\text{tan}}\frac{\beta }{2} \cdot {\text{tan}}\frac{\alpha }{2}}} = \frac{{3{\text{tan}}\frac{\alpha }{2}}}{{1 + 4{\text{ta}}{{\text{n}}^2}\frac{\alpha }{2}}} = \frac{{3{\text{sin}}\frac{\alpha }{2} \cdot {\text{cos}}\frac{\alpha }{2}}}{{1 + 3{\text{si}}{{\text{n}}^2}\frac{\alpha }{2}}} = \frac{{3{\text{sin}}\alpha }}{{5 – 3{\text{cos}}\alpha }}$

Câu 11. Cho ${\text{cos}}a = \frac{3}{4};{\text{sin}}a > 0;{\text{sin}}b = \frac{3}{5};{\text{cos}}b < 0$. Giá trị của ${\text{cos}}\left( {a + b} \right)$. bằng:
A. $\frac{3}{5}\left( {1 + \frac{{\sqrt 7 }}{4}} \right)$
B. $ – \frac{3}{5}\left( {1 + \frac{{\sqrt 7 }}{4}} \right)$
C. $\frac{3}{5}\left( {1 – \frac{{\sqrt 7 }}{4}} \right)$.
D. $ – \frac{3}{5}\left( {1 – \frac{{\sqrt 7 }}{4}} \right)$.

Chọn A

Lời giải

Ta có :

$\begin{array}{*{20}{r}}
{}&{\left\{ {\begin{array}{*{20}{c}}
{{\text{cos}}a = \frac{3}{4}} \\
{{\text{sin}}a > 0}
\end{array} \Rightarrow {\text{sin}}a = \sqrt {1 – {\text{co}}{{\text{s}}^2}a} = \frac{{\sqrt 7 }}{4}.} \right.} \\
{}&{\left\{ {\begin{array}{*{20}{c}}
{{\text{sin}}b = \frac{3}{5}} \\
{{\text{cos}}b < 0}
\end{array} \Rightarrow {\text{cos}}b = – \sqrt {1 – {\text{si}}{{\text{n}}^2}b} = – \frac{4}{5}.} \right.}
\end{array}$

${\text{cos}}\left( {a + b} \right) = {\text{cos}}a{\text{cos}}b – {\text{sin}}a{\text{sin}}b = \frac{3}{4} \cdot \left( { – \frac{4}{5}} \right) – \frac{{\sqrt 7 }}{4} \cdot \frac{3}{5} = – \frac{3}{5}\left( {1 + \frac{{\sqrt 7 }}{4}} \right)$

Câu 12. Biết ${\text{cos}}\left( {a – \frac{b}{2}} \right) = \frac{1}{2}$ và ${\text{sin}}\left( {a – \frac{b}{2}} \right) > 0;{\text{sin}}\left( {\frac{a}{2} – b} \right) = \frac{3}{5}$ và ${\text{cos}}\left( {\frac{a}{2} – b} \right) > 0$. Giá trị ${\text{cos}}\left( {a + b} \right)$ bằng:
A. $\frac{{24\sqrt 3 – 7}}{{50}}$.
B. $\frac{{7 – 24\sqrt 3 }}{{50}}$.
C. $\frac{{22\sqrt 3 – 7}}{{50}}$.
D. $\frac{{7 – 22\sqrt 3 }}{{50}}$.

Chọn A.

Lời giải

Ta có :

$\left\{ {\begin{array}{*{20}{l}}
{{\text{cos}}\left( {a – \frac{b}{2}} \right) = \frac{1}{2}} \\
{{\text{sin}}\left( {a – \frac{b}{2}} \right) > 0}
\end{array} \Rightarrow {\text{sin}}\left( {a – \frac{b}{2}} \right) = \sqrt {1 – {\text{co}}{{\text{s}}^2}\left( {a – \frac{b}{2}} \right)} = \frac{{\sqrt 3 }}{2}} \right.$.

$\left\{ {\begin{array}{*{20}{l}}
{{\text{sin}}\left( {\frac{a}{2} – b} \right) = \frac{3}{5}} \\
{{\text{cos}}\left( {\frac{a}{2} – b} \right)}
\end{array} \Rightarrow {\text{cos}}\left( {\frac{a}{2} – b} \right) = \sqrt {1 – {\text{si}}{{\text{n}}^2}\left( {\frac{a}{2} – b} \right)} = \frac{4}{5}} \right.$.

${\text{cos}}\frac{{a + b}}{2} = {\text{cos}}\left( {a – \frac{b}{2}} \right){\text{cos}}\left( {\frac{a}{2} – b} \right) + {\text{sin}}\left( {a – \frac{b}{2}} \right){\text{sin}}\left( {\frac{a}{2} – b} \right) = \frac{1}{2} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 + 4}}{{10}}$.

${\text{cos}}\left( {a + b} \right) = 2{\text{co}}{{\text{s}}^2}\frac{{a + b}}{2} – 1 = \frac{{24\sqrt 3 – 7}}{{50}}$.

Câu 13. Rút gọn biểu thức: ${\text{cos}}\left( {{{120}^ \circ } – x} \right) + {\text{cos}}\left( {{{120}^ \circ } + x} \right) – {\text{cos}}x$ ta được kết quả là
A. 0 .
B. $ – {\text{cos}}x$.
C. $ – 2{\text{cos}}x$.
D. ${\text{sin}}x – {\text{cos}}x$.

Chọn C.

Lời giải

${\text{cos}}\left( {{{120}^ \circ } – x} \right) + {\text{cos}}\left( {{{120}^ \circ } + x} \right) – {\text{cos}}x $
$= – \frac{1}{2}{\text{cos}}x + \frac{{\sqrt 3 }}{2}{\text{sin}}x – \frac{1}{2}{\text{cos}}x + \frac{{\sqrt 3 }}{2}{\text{sin}}x – {\text{cos}}x = – 2{\text{cos}}x$

Câu 14. Cho ${\text{sin}}a = \frac{3}{5};{\text{cos}}a < 0;\cos b = \frac{3}{4};\,\sin b > 0$. Giá trị ${\text{sin}}\left( {a – b} \right)$ bằng:
A. $ – \frac{1}{5}\left( {\sqrt 7 + \frac{9}{4}} \right)$
B. $ – \frac{1}{5}\left( {\sqrt 7 – \frac{9}{4}} \right)$
C. $\frac{1}{5}\left( {\sqrt 7 + \frac{9}{4}} \right)$.
D. $\frac{1}{5}\left( {\sqrt 7 – \frac{9}{4}} \right)$.

Chọn A.

Lời giải

Ta có :

$\begin{array}{*{20}{r}}
{}&{\left\{ {\begin{array}{*{20}{c}}
{{\text{sin}}a = \frac{3}{5}} \\
{{\text{cos}}a < 0}
\end{array} \Rightarrow {\text{cos}}a = – \sqrt {1 – {\text{si}}{{\text{n}}^2}a} = – \frac{4}{5}.} \right.} \\
{}&{\left\{ {\begin{array}{*{20}{c}}
{{\text{cos}}b = \frac{3}{4}} \\
{{\text{sin}}b > 0}
\end{array} \Rightarrow {\text{sin}}b = \sqrt {1 – {\text{co}}{{\text{s}}^2}b} = \frac{{\sqrt 7 }}{4}.} \right.}
\end{array}$

${\text{sin}}\left( {a – b} \right) = {\text{sin}}a{\text{cos}}b – {\text{cos}}a{\text{sin}}b = \frac{3}{5} \cdot \frac{3}{4} – \left( { – \frac{4}{5}} \right) \cdot \frac{{\sqrt 7 }}{4} = \frac{1}{5}\left( {\sqrt 7 + \frac{9}{4}} \right).$

Câu 15. Biết $\alpha + \beta + \gamma = \frac{\pi }{2}$ và ${\text{cot}}\alpha ,{\text{cot}}\beta ,{\text{cot}}\gamma $ theo thứ tự lập thành một cấp số cộng. Tích số ${\text{cot}}\alpha \cdot {\text{cot}}\gamma $ bằng:
A. 2 .
B. -2 .
C. 3 .
D. -3 .

Chọn C.

Lời giải

Ta có :

$\alpha + \beta + \gamma = \frac{\pi }{2}$,

${\text{cot}}\beta = {\text{tan}}\left( {\alpha + \gamma } \right) = \frac{{{\text{tan}}\alpha + {\text{tan}}\gamma }}{{1 – {\text{tan}}\alpha {\text{tan}}\gamma }} = \frac{{{\text{cot}}\alpha + {\text{cot}}\gamma }}{{{\text{cot}}\alpha {\text{cot}}\gamma – 1}} = \frac{{2{\text{cot}}\beta }}{{{\text{cot}}\alpha {\text{cot}}\gamma – 1}}$

$ \Rightarrow {\text{cot}}\alpha {\text{cot}}\gamma = 3.$

Câu 16. Cho ${\text{sin}}2\alpha = \frac{3}{4}$. Tính giá trị biểu thức $A = {\text{tan}}\alpha + {\text{cot}}\alpha $
A. $A = \frac{4}{3}$.
B. $A = \frac{2}{3}$.
C. $A = \frac{8}{3}$.
D. $A = \frac{{16}}{3}$.

Chọn C

Lời giải

$A = {\text{tan}}\alpha + {\text{cot}}\alpha = \frac{{{\text{sin}}\alpha }}{{{\text{cos}}\alpha }} + \frac{{{\text{cos}}\alpha }}{{{\text{sin}}\alpha }} = \frac{{{\text{si}}{{\text{n}}^2}\alpha + {\text{co}}{{\text{s}}^2}\alpha }}{{{\text{sin}}\alpha {\text{cos}}\alpha }} = \frac{1}{{\frac{1}{2}{\text{sin}}2\alpha }} = \frac{1}{{\frac{1}{2} \cdot \frac{3}{4}}} = \frac{8}{3}$.

Câu 17. Cho $a,b$ là hai góc nhọn. Biết ${\text{cos}}a = \frac{1}{3},{\text{cos}}b = \frac{1}{4}$. Giá trị của biểu thức ${\text{cos}}\left( {a + b} \right){\text{cos}}\left( {a – b} \right)$ bằng
A. $ – \frac{{119}}{{144}}$.
B. $ – \frac{{115}}{{144}}$.
C. $ – \frac{{113}}{{144}}$.
D. $ – \frac{{117}}{{144}}$.

Chọn A

Lời giải

Từ ${\text{cos}}a = \frac{1}{3} \Rightarrow {\text{cos}}2a = 2{\text{co}}{{\text{s}}^2}a – 1 = – \frac{7}{9}$

${\text{cos}}b = \frac{1}{4} \Rightarrow {\text{cos}}2b = 2{\text{co}}{{\text{s}}^2}b – 1 = – \frac{7}{8}$

Ta có ${\text{cos}}\left( {a + b} \right){\text{cos}}\left( {a – b} \right) = \frac{1}{2}\left( {{\text{cos}}2a + {\text{cos}}2b} \right) = \frac{1}{2}\left( { – \frac{7}{9} – \frac{7}{8}} \right) = – \frac{{119}}{{144}}$.

Câu 18. Cho số thực $\alpha $ thỏa mãn ${\text{sin}}\alpha = \frac{1}{4}$. Tính $\left( {{\text{sin}}4\alpha + 2{\text{sin}}2\alpha } \right){\text{cos}}\alpha $
A. $\frac{{25}}{{128}}$.
B. $\frac{1}{{16}}$.
C. $\frac{{255}}{{128}}$.
D. $\frac{{225}}{{128}}$.

Lời giải

Ta có $\left( {{\text{sin}}4\alpha + 2{\text{sin}}2\alpha } \right){\text{cos}}\alpha = 2{\text{sin}}2\alpha \left( {{\text{cos}}2\alpha + 1} \right){\text{cos}}\alpha = 4{\text{sin}}\alpha {\text{cos}}\alpha \left( {1 – 2{\text{si}}{{\text{n}}^2}\alpha + 1} \right){\text{cos}}\alpha $

$ = 4{\text{sin}}\alpha \left( {1 – {\text{si}}{{\text{n}}^2}\alpha } \right)\left( {2 – 2{\text{si}}{{\text{n}}^2}\alpha } \right) = 8{\left( {1 – {\text{si}}{{\text{n}}^2}\alpha } \right)^2}{\text{sin}}\alpha = 8{\left( {1 – \frac{1}{{16}}} \right)^2} \cdot \frac{1}{4} = \frac{{225}}{{128}}$.

Câu 19. Cho ${\text{cot}}a = 15$, giá trị sin $2a$ có thể nhận giá trị nào dưới đây:
A. $\frac{{11}}{{113}}$.
B. $\frac{{13}}{{113}}$.
C. $\frac{{15}}{{113}}$.
D. $\frac{{17}}{{113}}$.

Chọn C.

Lời giải

${\text{cot}}a = 15 \Rightarrow \frac{1}{{{\text{si}}{{\text{n}}^2}a}} = 226 \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{{\text{si}}{{\text{n}}^2}a = \frac{1}{{226}}} \\
{{\text{co}}{{\text{s}}^2}a = \frac{{225}}{{226}}}
\end{array} \Rightarrow {\text{sin}}2a = \pm \frac{{15}}{{113}}} \right.$.

Câu 20. Giá trị đúng của ${\text{cos}}\frac{{2\pi }}{7} + {\text{cos}}\frac{{4\pi }}{7} + {\text{cos}}\frac{{6\pi }}{7}$ bằng:
A. $\frac{1}{2}$.
B. $ – \frac{1}{2}$.
C. $\frac{1}{4}$.
D. $ – \frac{1}{4}$.

Chọn B.

Lời giải

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

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

Câu 21. Giá trị đúng của ${\text{tan}}\frac{\pi }{{24}} + {\text{tan}}\frac{{7\pi }}{{24}}$ bằng:
A. $2\left( {\sqrt 6 – \sqrt 3 } \right)$
B. $2\left( {\sqrt 6 + \sqrt 3 } \right)$.
C. $2\left( {\sqrt 3 – \sqrt 2 } \right)$.
D. $2\left( {\sqrt 3 + \sqrt 2 } \right)$.

Chọn A.

Lời giải

${\text{tan}}\frac{\pi }{{24}} + {\text{tan}}\frac{{7\pi }}{{24}} = \frac{{{\text{sin}}\frac{\pi }{3}}}{{{\text{cos}}\frac{\pi }{{24}} \cdot {\text{cos}}\frac{{7\pi }}{{24}}}} = \frac{{\sqrt 3 }}{{{\text{cos}}\frac{\pi }{3} + {\text{cos}}\frac{\pi }{4}}} = 2\left( {\sqrt 6 – \sqrt 3 } \right){\text{.\;}}$

Câu 22. Biểu thức $A = \frac{1}{{2{\text{sin}}{{10}^0}}} – 2{\text{sin}}{70^ \circ }$ có giá trị đúng bằng:
A. 1 .
B. -1 .
C. 2 .
D. -2 .

Chọn A.

Lời giải

$A = \frac{1}{{2{\text{sin}}{{10}^ \circ }}} – 2{\text{sin}}{70^ \circ } = \frac{{1 – 4{\text{sin}}{{10}^ \circ } \cdot {\text{sin}}{{70}^ \circ }}}{{2{\text{sin}}{{10}^ \circ }}} = \frac{{2{\text{sin}}{{80}^ \circ }}}{{2{\text{sin}}{{10}^ \circ }}} = \frac{{2{\text{sin}}{{10}^ \circ }}}{{2{\text{sin}}{{10}^ \circ }}} = 1.$

Câu 23. Tích số ${\text{cos}}{10^ \circ } \cdot {\text{cos}}{30^ \circ } \cdot {\text{cos}}{50^ \circ } \cdot {\text{cos}}{70^ \circ }$ bằng:
A. $\frac{1}{{16}}$.
B. $\frac{1}{8}$.
C. $\frac{3}{{16}}$.
D. $\frac{1}{4}$.

Chọn C.

Lời giải

${\text{cos}}{10^ \circ } \cdot {\text{cos}}{30^ \circ } \cdot {\text{cos}}{50^ \circ } \cdot {\text{cos}}{70^ \circ } = {\text{cos}}{10^ \circ } \cdot {\text{cos}}{30^ \circ } \cdot \frac{1}{2}\left( {{\text{cos}}{{120}^ \circ } + {\text{cos}}{{20}^ \circ }} \right)$

$ = \frac{{\sqrt 3 }}{4}\left( { – \frac{{{\text{cos}}{{10}^ \circ }}}{2} + \frac{{{\text{cos}}{{30}^ \circ } + {\text{cos}}{{10}^ \circ }}}{2}} \right) = \frac{{\sqrt 3 }}{4} \cdot \frac{1}{4} = \frac{{\sqrt 3 }}{{16}}.$

Câu 24. Tích số ${\text{cos}}\frac{\pi }{7} \cdot {\text{cos}}\frac{{4\pi }}{7} \cdot {\text{cos}}\frac{{5\pi }}{7}$ bằng:
A. $\frac{1}{8}$.
B. $ – \frac{1}{8}$.
C. $\frac{1}{4}$.
D. $ – \frac{1}{4}$.

Lời giải

Chọn A.

${\text{cos}}\frac{\pi }{7} \cdot {\text{cos}}\frac{{4\pi }}{7} \cdot {\text{cos}}\frac{{5\pi }}{7} = \frac{{{\text{sin}}\frac{{2\pi }}{7} \cdot {\text{cos}}\frac{{4\pi }}{7} \cdot {\text{cos}}\frac{{5\pi }}{7}}}{{2{\text{sin}}\frac{\pi }{7}}}$

$ = – \frac{{{\text{sin}}\frac{{2\pi }}{7} \cdot {\text{cos}}\frac{{2\pi }}{7} \cdot {\text{cos}}\frac{{4\pi }}{7}}}{{2{\text{sin}}\frac{\pi }{7}}}$

$ = – \frac{{{\text{sin}}\frac{{4\pi }}{7} \cdot {\text{cos}}\frac{{4\pi }}{7}}}{{4{\text{sin}}\frac{\pi }{7}}}$

$ = – \frac{{{\text{sin}}\frac{{8\pi }}{7}}}{{8{\text{sin}}\frac{\pi }{7}}} = \frac{1}{8}$

Câu 25. Giá trị đúng của biểu thức $A = \frac{{{\text{tan}}{{30}^ \circ } + {\text{tan}}{{40}^ \circ } + {\text{tan}}{{50}^ \circ } + {\text{tan}}{{60}^ \circ }}}{{{\text{cos}}{{20}^ \circ }}}$ bằng:
A. $\frac{2}{{\sqrt 3 }}$.
B. $\frac{4}{{\sqrt 3 }}$.
C. $\frac{6}{{\sqrt 3 }}$.
D. $\frac{8}{{\sqrt 3 }}$.

Chọn D.

Lời giải

$A = \frac{{{\text{tan}}{{30}^ \circ } + {\text{tan}}{{40}^ \circ } + {\text{tan}}{{50}^ \circ } + {\text{tan}}{{60}^ \circ }}}{{{\text{cos}}{{20}^ \circ }}}$
$ = \frac{{\frac{{{\text{sin}}{{70}^ \circ }}}{{{\text{cos}}{{30}^ \circ } \cdot {\text{cos}}{{40}^ \circ }}} + \frac{{{\text{sin}}{{110}^ \circ }}}{{{\text{cos}}{{50}^ \circ } \cdot {\text{cos}}{{60}^ \circ }}}}}{{{\text{cos}}{{20}^ \circ }}}$

$ = \frac{1}{{{\text{cos}}{{30}^ \circ } \cdot {\text{cos}}{{40}^ \circ }}} + \frac{2}{{{\text{cos}}{{50}^ \circ } \cdot {\text{cos}}{{60}^ \circ }}} = \frac{2}{{\sqrt 3 {\text{cos}}{{40}^ \circ }}} + \frac{2}{{{\text{cos}}{{50}^ \circ }}} = 2\left( {\frac{{{\text{cos}}{{50}^ \circ } + \sqrt 3 {\text{cos}}{{40}^ \circ }}}{{\sqrt 3 {\text{cos}}{{40}^ \circ } \cdot {\text{cos}}{{50}^ \circ }}}} \right)$

$ = 2\left( {\frac{{{\text{sin}}{{40}^ \circ } + \sqrt 3 {\text{cos}}{{40}^ \circ }}}{{\sqrt 3 {\text{cos}}{{40}^ \circ } \cdot {\text{cos}}{{50}^ \circ }}}} \right) $
$= 4\frac{{{\text{sin}}{{100}^ \circ }}}{{\frac{{\sqrt 3 }}{2}\left( {{\text{cos}}{{10}^ \circ } + {\text{cos}}{{90}^ \circ }} \right)}}$
$ = \frac{{8{\text{cos}}{{10}^ \circ }}}{{\sqrt 3 {\text{cos}}{{10}^ \circ }}} = \frac{8}{{\sqrt 3 }}.$

Câu 26. Cho hai góc nhọn $a$ và $b$. Biết ${\text{cos}}a = \frac{1}{3},{\text{cos}}b = \frac{1}{4}$. Giá trị ${\text{cos}}\left( {a + b} \right) \cdot {\text{cos}}\left( {a – b} \right)$ bằng:
A. $ – \frac{{113}}{{144}}$.
B. $ – \frac{{115}}{{144}}$.
C. $ – \frac{{117}}{{144}}$.
D. $ – \frac{{119}}{{144}}$.

Chọn D.

Lời giải

Ta có :

${\text{cos}}\left( {a + b} \right) \cdot {\text{cos}}\left( {a – b} \right) $
$= \frac{1}{2}\left( {{\text{cos}}2a + {\text{cos}}2b} \right) $
$= {\text{co}}{{\text{s}}^2}a + {\text{co}}{{\text{s}}^2}b – 1 $
$= {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{4}} \right)^2} – 1 = – \frac{{119}}{{144}}$

Câu 27. Rút gọn biểu thức $A = \frac{{{\text{sin}}x + {\text{sin}}2x + {\text{sin}}3x}}{{{\text{cos}}x + {\text{cos}}2x + {\text{cos}}3x}}$
A. $A = {\text{tan}}6x$.
B. $A = {\text{tan}}3x$.
C. $A = {\text{tan}}2x$.
D. $A = {\text{tan}}x + {\text{tan}}2x + {\text{tan}}3x$.

Chọn C.

Lời giải

Ta có :

$A = \frac{{{\text{sin}}x + {\text{sin}}2x + {\text{sin}}3x}}{{{\text{cos}}x + {\text{cos}}2x + {\text{cos}}3x}} = \frac{{2{\text{sin}}2x \cdot {\text{cos}}x + {\text{sin}}2x}}{{2{\text{cos}}2x \cdot {\text{cos}}x + {\text{cos}}2x}} $
$= \frac{{{\text{sin}}2x\left( {2{\text{cos}}x + 1} \right)}}{{{\text{cos}}2x\left( {2{\text{cos}}x + 1} \right)}} = {\text{tan}}2x$

Câu 28. Biến đổi biểu thức ${\text{sin}}a + 1$ thành tích.
A. ${\text{sin}}a + 1 = 2{\text{sin}}\left( {\frac{a}{2} + \frac{\pi }{4}} \right){\text{cos}}\left( {\frac{a}{2} – \frac{\pi }{4}} \right)$.
B. ${\text{sin}}a + 1 = 2{\text{cos}}\left( {\frac{a}{2} + \frac{\pi }{4}} \right){\text{sin}}\left( {\frac{a}{2} – \frac{\pi }{4}} \right)$.
C. ${\text{sin}}a + 1 = 2{\text{sin}}\left( {a + \frac{\pi }{2}} \right){\text{cos}}\left( {a – \frac{\pi }{2}} \right)$.
D. ${\text{sin}}a + 1 = 2{\text{cos}}\left( {a + \frac{\pi }{2}} \right){\text{sin}}\left( {a – \frac{\pi }{2}} \right)$.

Chọn D.

Lời giải

Ta có ${\text{sin}}a + 1 = 2{\text{sin}}\frac{a}{2}{\text{cos}}\frac{a}{2} + {\text{si}}{{\text{n}}^2}\frac{a}{2} + {\text{co}}{{\text{s}}^2}\frac{a}{2} = {\left( {{\text{sin}}\frac{a}{2} + {\text{cos}}\frac{a}{2}} \right)^2} = 2{\text{si}}{{\text{n}}^2}\left( {\frac{a}{2} + \frac{\pi }{4}} \right)$

$ = 2{\text{sin}}\left( {\frac{a}{2} + \frac{\pi }{4}} \right){\text{cos}}\left( {\frac{\pi }{4} – \frac{a}{2}} \right) = 2{\text{sin}}\left( {\frac{a}{2} + \frac{\pi }{4}} \right){\text{cos}}\left( {\frac{a}{2} – \frac{\pi }{4}} \right)$

Câu 29. Cho góc $\alpha $ thỏa mãn $\frac{\pi }{2} < \alpha < \pi $ và ${\text{sin}}\frac{\alpha }{2} = \frac{2}{{\sqrt 5 }}$.Tính giá trị của biểu thức $A = {\text{tan}}\left( {\frac{\alpha }{2} – \frac{\pi }{4}} \right)$.
A. $A = \frac{1}{3}$.
B. $A = – \frac{1}{3}$.
C. $A = 3$.
D. $A = – 3$.

Chọn A

Lời giải

Vì góc $\alpha $ thỏa mãn $\frac{\pi }{2} < \alpha < \pi $ nên $\frac{\pi }{4} < \frac{\alpha }{2} < \frac{\pi }{2}$ suy ra ${\text{cos}}\frac{\alpha }{2} > 0$.

Do ${\text{sin}}\frac{\alpha }{2} = \frac{2}{{\sqrt 5 }}$ nên ${\text{cos}}\frac{\alpha }{2} = \sqrt {1 – {\text{si}}{{\text{n}}^2}\frac{\alpha }{2}} = \frac{1}{{\sqrt 5 }}$.

Biểu thức $A = {\text{tan}}\left( {\frac{\alpha }{2} – \frac{\pi }{4}} \right) = \frac{{{\text{tan}}\frac{\alpha }{2} – 1}}{{{\text{tan}}\frac{\alpha }{2} + 1}}$.

Do đó ${\text{tan}}\frac{\alpha }{2} = 2$.

Vậy biểu thức $A = \frac{{2 – 1}}{{2 + 1}} = \frac{1}{3}$.

Câu 30. Cho ${\text{cos}}x = \frac{1}{3}\left( { – \frac{\pi }{2} < x < 0} \right)$. Giá trị của ${\text{tan}}2x$ là
A. $\frac{{\sqrt 5 }}{2}$.
B. $\frac{{4\sqrt 2 }}{7}$.
C. $ – \frac{{\sqrt 5 }}{2}$.
D. $ – \frac{{4\sqrt 2 }}{7}$.

Chọn B

Lời giải

${\text{si}}{{\text{n}}^2}x = 1 – {\text{co}}{{\text{s}}^2}x = 1 – \frac{1}{9} = \frac{8}{9} \Rightarrow {\text{sin}}x = – \frac{{2\sqrt 2 }}{3}\left( {{\text{Vì}} – \frac{\pi }{2} < x < 0} \right){\text{.\;}}$

$ \Rightarrow {\text{tan}}x = – 2\sqrt 2 \Rightarrow {\text{tan}}2x = \frac{{2{\text{tan}}x}}{{1 – {\text{ta}}{{\text{n}}^2}x}} = \frac{{ – 4\sqrt 2 }}{{ – 7}} = \frac{{4\sqrt 2 }}{7}{\text{.\;}}$

Câu 31. Cho ${\text{cos}}x = 0$. Tính $A = {\text{si}}{{\text{n}}^2}\left( {x – \frac{\pi }{6}} \right) + {\text{si}}{{\text{n}}^2}\left( {x + \frac{\pi }{6}} \right)$.
A. $\frac{3}{2}$.
B. 2 .
C. 1 .
D. $\frac{1}{4}$.

Chọn A

Lời giải

Ta có ${\text{cos}}2x = 2{\text{co}}{{\text{s}}^2}x – 1 = – 1$. Sử dụng công thức hạ bậc và công thức biến đổi tổng thành tích ta được:

$A = \frac{{1 – {\text{cos}}\left( {2x – \frac{\pi }{3}} \right) + 1 – {\text{cos}}\left( {2x + \frac{\pi }{3}} \right)}}{2} = 1 – {\text{cos}}2x{\text{cos}}\frac{\pi }{3} = 1 + \frac{1}{2} = \frac{3}{2}$

Câu 32. Cho biết ${\text{cos}}\alpha = – \frac{2}{3}$. Giá trị của biểu thức $P = \frac{{{\text{cot}}\alpha + 3{\text{tan}}\alpha }}{{2{\text{cot}}\alpha + {\text{tan}}\alpha }}$ bằng bao nhiêu?
A. $P = \frac{{19}}{{13}}$.
B. $P = \frac{{25}}{{13}}$.
C. $P = – \frac{{25}}{{13}}$.
D. $P = – \frac{{19}}{{13}}$.

Chọn A

Lời giải

Ta có: ${\text{cos}}\alpha = – \frac{2}{3} \Rightarrow {\text{ta}}{{\text{n}}^2}\alpha = \frac{1}{{{\text{co}}{{\text{s}}^2}\alpha }} – 1 = \frac{1}{{{{\left( {\frac{{ – 2}}{3}} \right)}^2}}} – 1 = \frac{5}{4}$

$P = \frac{{{\text{cot}}\alpha + 3{\text{tan}}\alpha }}{{2{\text{cot}}\alpha + {\text{tan}}\alpha }} = \frac{{\frac{1}{{{\text{tan}}\alpha }} + 3{\text{tan}}\alpha }}{{\frac{2}{{{\text{tan}}\alpha }} + {\text{tan}}\alpha }} = \frac{{\frac{{1 + 3{\text{ta}}{{\text{n}}^2}\alpha }}{{{\text{tan}}\alpha }}}}{{\frac{{2 + {\text{ta}}{{\text{n}}^2}\alpha }}{{{\text{tan}}\alpha }}}} = \frac{{1 + 3{\text{ta}}{{\text{n}}^2}\alpha }}{{2 + {\text{ta}}{{\text{n}}^2}\alpha }} = \frac{{1 + 3 \cdot \frac{5}{4}}}{{2 + \frac{5}{4}}} = \frac{{19}}{{13}}$

Câu 33. Cho ${\text{sin}}\alpha \cdot {\text{cos}}\left( {\alpha + \beta } \right) = {\text{sin}}\beta $ với $\alpha + \beta \ne \frac{\pi }{2} + k\pi ,\alpha \ne \frac{\pi }{2} + l\pi ,\left( {k,l \in \mathbb{Z}} \right)$. Ta có
A. ${\text{tan}}\left( {\alpha + \beta } \right) = 2{\text{cot}}\alpha $.
B. ${\text{tan}}\left( {\alpha + \beta } \right) = 2{\text{cot}}\beta $.
C. ${\text{tan}}\left( {\alpha + \beta } \right) = 2{\text{tan}}\beta $
.D. ${\text{tan}}\left( {\alpha + \beta } \right) = 2{\text{tan}}\alpha $.

Chọn D

Lời giải

Ta có ${\text{sin}}\alpha \cdot {\text{cos}}\left( {\alpha + \beta } \right) = {\text{sin}}\beta \Leftrightarrow \frac{1}{2}\left[ {{\text{sin}}\left( {2\alpha + \beta } \right) – {\text{sin}}\beta } \right] = {\text{sin}}\beta $

$ \Leftrightarrow {\text{sin}}\left[ {\left( {\alpha + \beta } \right) + \alpha } \right] = 3{\text{sin}}\beta \Leftrightarrow {\text{sin}}\left( {\alpha + \beta } \right){\text{cos}}\alpha + {\text{sin}}\alpha {\text{cos}}\left( {\alpha + \beta } \right) = 3{\text{sin}}\beta $

$ \Leftrightarrow \frac{{{\text{sin}}\left( {\alpha + \beta } \right)}}{{{\text{cos}}\left( {\alpha + \beta } \right)}}{\text{cos}}\alpha + {\text{sin}}\alpha = \frac{{3{\text{sin}}\beta }}{{{\text{cos}}\left( {\alpha + \beta } \right)}}($ vì ${\text{cos}}\left( {\alpha + \beta } \right) \ne 0)$

$ \Leftrightarrow \frac{{{\text{sin}}\left( {\alpha + \beta } \right)}}{{{\text{cos}}\left( {\alpha + \beta } \right)}} = \frac{{3{\text{sin}}\beta }}{{{\text{cos}}\alpha {\text{cos}}\left( {\alpha + \beta } \right)}} – \frac{{{\text{sin}}\alpha }}{{{\text{cos}}\alpha }}\left( {\text{*}} \right)($ vì ${\text{cos}}\alpha \ne 0)$

Mà $\frac{{{\text{sin}}\beta }}{{{\text{cos}}\left( {\alpha + \beta } \right)}} = {\text{sin}}\alpha $ (từ giả thiết), suy ra $\left( {\text{*}} \right) \Leftrightarrow {\text{tan}}\left( {\alpha + \beta } \right) = \frac{{3{\text{sin}}\alpha }}{{{\text{cos}}\alpha }} – \frac{{{\text{sin}}\alpha }}{{{\text{cos}}\alpha }} = 2{\text{tan}}\alpha $

Vậy ${\text{tan}}\left( {\alpha + \beta } \right) = 2{\text{tan}}\alpha $.

Câu 34. Biết rằng $\frac{1}{{{\text{co}}{{\text{s}}^2}x – {\text{si}}{{\text{n}}^2}x}} + \frac{{2 \cdot {\text{tan}}x}}{{1 – {\text{ta}}{{\text{n}}^2}x}} = \frac{{{\text{cos}}\left( {ax} \right)}}{{b – {\text{sin}}\left( {ax} \right)}}\left( {a,b \in \mathbb{R}} \right)$. Tính giá trị của biểu thức $P = a + b$.
A. $P = 4$.
B. $P = 1$.
C. $P = 2$.
D. $P = 3$.

Chọn D

Lời giải

Ta có: $\frac{1}{{{\text{co}}{{\text{s}}^2}x – {\text{si}}{{\text{n}}^2}x}} + \frac{{2 \cdot {\text{tan}}x}}{{1 – {\text{ta}}{{\text{n}}^2}x}} = \frac{1}{{{\text{cos}}2x}} + \frac{{\frac{{2{\text{sin}}x}}{{{\text{cos}}x}}}}{{1 – \frac{{{\text{si}}{{\text{n}}^2}x}}{{{\text{co}}{{\text{s}}^2}x}}}}$

$ = \frac{1}{{{\text{cos}}2x}} + \frac{{2{\text{sin}}x \cdot {\text{cos}}x}}{{{\text{co}}{{\text{s}}^2}x – {\text{si}}{{\text{n}}^2}x}} = \frac{1}{{{\text{cos}}2x}} + \frac{{{\text{sin}}2x}}{{{\text{cos}}2x}}$

$ = \frac{{1 + {\text{sin}}2x}}{{{\text{cos}}2x}} = \frac{{\left( {1 + {\text{sin}}2x} \right){\text{cos}}2x}}{{{\text{co}}{{\text{s}}^2}2x}} = \frac{{\left( {1 + {\text{sin}}2x} \right){\text{cos}}2x}}{{1 – {\text{si}}{{\text{n}}^2}2x}}$$ = \frac{{{\text{cos}}2x}}{{1 – {\text{sin}}2x}}$.

Vậy $a = 2,b = 1$. Suy ra $P = a + b = 3$.

Câu 35. Cho ${\text{cos}}2\alpha = \frac{2}{3}$. Tính giá trị của biểu thức $P = {\text{cos}}\alpha \cdot {\text{cos}}3\alpha $.
A. $P = \frac{7}{{18}}$.
B. $P = \frac{7}{9}$.
C. $P = \frac{5}{9}$.
D. $\frac{5}{{18}}$.

Chọn D

Lời giải

Ta có $P = {\text{cos}}\alpha \cdot {\text{cos}}3\alpha = \frac{1}{2}\left( {{\text{cos}}2\alpha + {\text{cos}}4\alpha } \right) = \frac{1}{2}\left( {2{\text{co}}{{\text{s}}^2}2\alpha + {\text{cos}}2\alpha – 1} \right)$

$ = \frac{1}{2}\left[ {2{{\left( {\frac{2}{3}} \right)}^2} + \frac{2}{3} – 1} \right] = \frac{5}{{18}}$.

Câu 36. Cho ${\text{tan}}x = 2\left( {\pi < x < \frac{{3\pi }}{2}} \right)$. Giá trị của ${\text{sin}}\left( {x + \frac{\pi }{3}} \right)$ là
A. $\frac{{2 – \sqrt 3 }}{{2\sqrt 5 }}$.
B. $ – \frac{{2 + \sqrt 3 }}{{2\sqrt 5 }}$.
C. $\frac{{2 + \sqrt 3 }}{{2\sqrt 5 }}$.
D. $\frac{{ – 2 + \sqrt 3 }}{{2\sqrt 5 }}$.

Chọn B

Lời giải

$\pi < x < \frac{{3\pi }}{2}$ suy ra ${\text{sin}}x < 0,{\text{cos}}x < 0$.

Ta có: $1 + {\text{ta}}{{\text{n}}^2}x = \frac{1}{{{\text{co}}{{\text{s}}^2}x}} \Leftrightarrow {\text{co}}{{\text{s}}^2}x = \frac{1}{{1 + {\text{ta}}{{\text{n}}^2}x}} \Leftrightarrow {\text{co}}{{\text{s}}^2}x = \frac{1}{5} \Leftrightarrow {\text{cos}}x = \pm \frac{1}{{\sqrt 5 }}$

Do ${\text{cos}}x < 0$ nên nhận ${\text{cos}}x = – \frac{1}{{\sqrt 5 }}$.

${\text{tan}}x = \frac{{{\text{sin}}x}}{{{\text{cos}}x}} \Rightarrow {\text{sin}}x = {\text{tan}}x \cdot {\text{cos}}x = – \frac{2}{{\sqrt 5 }}$

${\text{sin}}\left( {x + \frac{\pi }{3}} \right) = {\text{sin}}x \cdot {\text{cos}}\frac{\pi }{3} + {\text{cos}}x \cdot {\text{sin}}\frac{\pi }{3} = \left( { – \frac{2}{{\sqrt 5 }}} \right) \cdot \frac{1}{2} + \left( { – \frac{1}{{\sqrt 5 }}} \right) \cdot \frac{{\sqrt 3 }}{2} = – \frac{{2 + \sqrt 3 }}{{2\sqrt 5 }}$

Câu 37. Tổng $A = {\text{tan}}{9^ \circ } + {\text{cot}}{9^ \circ } + {\text{tan}}{15^ \circ } + {\text{cot}}{15^ \circ } – {\text{tan}}{27^ \circ } – {\text{cot}}{27^ \circ }$ bằng:
A. 4 .
B. -4 .
C. 8 .
D. -8 .

Chọn C.

Lời giải

$A = {\text{tan}}{9^ \circ } + {\text{cot}}{9^ \circ } + {\text{tan}}{15^ \circ } + {\text{cot}}{15^ \circ } – {\text{tan}}{27^ \circ } – {\text{cot}}{27^ \circ }$

$\; = {\text{tan}}{9^ \circ } + {\text{cot}}{9^ \circ } – {\text{tan}}{27^ \circ } – {\text{cot}}{27^ \circ } + {\text{tan}}{15^ \circ } + {\text{cot}}{15^ \circ }$

$\; = {\text{tan}}{9^ \circ } + {\text{tan}}{81^ \circ } – {\text{tan}}{27^ \circ } – {\text{tan}}{63^ \circ } + {\text{tan}}{15^ \circ } + {\text{cot}}{15^ \circ }.$

Ta có

$A = {\text{tan}}{9^ \circ } + {\text{cot}}{9^ \circ } + {\text{tan}}{15^ \circ } + {\text{cot}}{15^ \circ } – {\text{tan}}{27^ \circ } – {\text{cot}}{27^ \circ }$
$\; = {\text{tan}}{9^ \circ } + {\text{cot}}{9^ \circ } – {\text{tan}}{27^ \circ } – {\text{cot}}{27^ \circ } + {\text{tan}}{15^ \circ } + {\text{cot}}{15^ \circ }$
$\; = {\text{tan}}{9^ \circ } + {\text{tan}}{81^ \circ } – {\text{tan}}{27^ \circ } – {\text{tan}}{63^ \circ } + {\text{tan}}{15^ \circ } + {\text{cot}}{15^ \circ }.$
Ta có
${\text{tan}}{9^ \circ } – {\text{tan}}{27^ \circ } + {\text{tan}}{81^ \circ } – {\text{tan}}{63^ \circ }$
$ = \frac{{ – {\text{sin}}{{18}^ \circ }}}{{{\text{cos}}{9^ \circ } \cdot {\text{cos}}{{27}^ \circ }}} + \frac{{{\text{sin}}{{18}^ \circ }}}{{{\text{cos}}{{81}^ \circ } \cdot {\text{cos}}{{63}^ \circ }}}$
$\; = {\text{sin}}{18^ \circ }\left( {\frac{{{\text{cos}}{9^ \circ } \cdot {\text{cos}}{{27}^ \circ } – {\text{cos}}{{81}^ \circ } \cdot {\text{cos}}{{63}^ \circ }}}{{{\text{cos}}{{81}^ \circ } \cdot {\text{cos}}{{63}^ \circ } \cdot {\text{cos}}{9^ \circ } \cdot {\text{cos}}{{27}^ \circ }}}} \right)$
$ = \frac{{{\text{sin}}{{18}^ \circ }\left( {{\text{cos}}{9^ \circ } \cdot {\text{cos}}{{27}^ \circ } – {\text{sin}}{9^ \circ } \cdot {\text{sin}}{{27}^ \circ }} \right)}}{{{\text{cos}}{{81}^ \circ } \cdot {\text{cos}}{{63}^ \circ } \cdot {\text{cos}}{9^ \circ } \cdot {\text{cos}}{{27}^ \circ }}}$
$ = \frac{{4{\text{sin}}{{18}^ \circ } \cdot {\text{cos}}{{36}^ \circ }}}{{\left( {{\text{cos}}{{72}^ \circ } + {\text{cos}}{{90}^ \circ }} \right)\left( {{\text{cos}}{{36}^ \circ } + {\text{cos}}{{90}^ \circ }} \right)}} = \frac{{4{\text{sin}}{{18}^ \circ }}}{{{\text{cos}}{{72}^ \circ }}} = 4$
Ta lại có
$\tan {15^0} + \cot {15^0} = \frac{{\sin {{15}^0}}}{{cos{{15}^0}}} + \frac{{cos{{15}^0}}}{{\sin {{15}^0}}}$
$ = \frac{{{{\sin }^2}{{15}^0} + co{s^2}{{15}^0}}}{{cos{{15}^0}.\sin {{15}^0}}} = \frac{1}{{\frac{1}{2}\sin {{30}^0}}} = 4$

Vậy $A = 8$.

Câu 38. Cho hai góc nhọn $a$ và $b$ với ${\text{sin}}a = \frac{1}{3},{\text{sin}}b = \frac{1}{2}$. Giá trị của ${\text{sin}}2\left( {a + b} \right)$ là:
A. $\frac{{2\sqrt 2 + 7\sqrt 3 }}{{18}}$.
B. $\frac{{3\sqrt 2 + 7\sqrt 3 }}{{18}}$.
C. $\frac{{4\sqrt 2 + 7\sqrt 3 }}{{18}}$.
D. $\frac{{5\sqrt 2 + 7\sqrt 3 }}{{18}}$.

Chọn C.

Lời giải

Ta có $\left\{ {\begin{array}{*{20}{l}}
{0 < a < \frac{\pi }{2}} \\
{{\text{sin}}a = \frac{1}{3}}
\end{array} \Rightarrow {\text{cos}}a = \frac{{2\sqrt 2 }}{3};\left\{ {\begin{array}{*{20}{l}}
{0 < b < \frac{\pi }{2}} \\
{{\text{sin}}b = \frac{1}{2}}
\end{array} \Rightarrow {\text{cos}}b = \frac{{\sqrt 3 }}{2}} \right.} \right.$.

${\text{sin}}2\left( {a + b} \right) = 2{\text{sin}}\left( {a + b} \right) \cdot {\text{cos}}\left( {a + b} \right) = 2\left( {{\text{sin}}a \cdot {\text{cos}}b + {\text{sin}}b \cdot {\text{cos}}a} \right)\left( {{\text{cos}}a \cdot {\text{cos}}b + {\text{sin}}a \cdot {\text{sin}}b} \right)$ $ = \frac{{4\sqrt 2 + 7\sqrt 3 }}{{18}}$

Câu 39. Biểu thức $A = \frac{{2{\text{co}}{{\text{s}}^2}2\alpha + \sqrt 3 {\text{sin}}4\alpha – 1}}{{2{\text{si}}{{\text{n}}^2}2\alpha + \sqrt 3 {\text{sin}}4\alpha – 1}}$ có kết quả rút gọn là:
A. $\frac{{{\text{cos}}\left( {4\alpha + {{30}^ \circ }} \right)}}{{{\text{cos}}\left( {4\alpha – {{30}^ \circ }} \right)}}$.
B. $\frac{{{\text{cos}}\left( {4\alpha – {{30}^ \circ }} \right)}}{{{\text{cos}}\left( {4\alpha + {{30}^ \circ }} \right)}}$.
C. $\frac{{{\text{sin}}\left( {4\alpha + {{30}^ \circ }} \right)}}{{{\text{sin}}\left( {4\alpha – {{30}^ \circ }} \right)}}$.
D. $\frac{{{\text{sin}}\left( {4\alpha – {{30}^ \circ }} \right)}}{{{\text{sin}}\left( {4\alpha + {{30}^ \circ }} \right)}}$.

Chọn C.

Lời giải

Ta có :

$A = \frac{{2{\text{co}}{{\text{s}}^2}2\alpha + \sqrt 3 {\text{sin}}4\alpha – 1}}{{2{\text{si}}{{\text{n}}^2}2\alpha + \sqrt 3 {\text{sin}}4\alpha – 1}} = \frac{{{\text{cos}}4\alpha + \sqrt 3 {\text{sin}}4\alpha }}{{\sqrt 3 {\text{sin}}4\alpha – {\text{cos}}4\alpha }} = \frac{{{\text{sin}}\left( {4\alpha + {{30}^ \circ }} \right)}}{{{\text{sin}}\left( {4\alpha – {{30}^ \circ }} \right)}}$

Câu 40. Kết quả nào sau đây ${\mathbf{SAI}}$ ?
A. ${\text{sin}}{33^ \circ } + {\text{cos}}{60^ \circ } = {\text{cos}}{3^ \circ }$.
B. $\frac{{{\text{sin}}{9^ \circ }}}{{{\text{sin}}{{48}^ \circ }}} = \frac{{{\text{sin}}{{12}^ \circ }}}{{{\text{sin}}{{81}^ \circ }}}$.
C. ${\text{cos}}{20^ \circ } + 2{\text{si}}{{\text{n}}^2}{55^ \circ } = 1 + \sqrt 2 {\text{sin}}{65^ \circ }$.
D. $\frac{1}{{{\text{cos}}{{290}^ \circ }}} + \frac{1}{{\sqrt 3 {\text{sin}}{{250}^ \circ }}} = \frac{4}{{\sqrt 3 }}$.

Chọn A.

Lời giải

Ta có : $\frac{{{\text{sin}}{9^ \circ }}}{{{\text{sin}}{{48}^ \circ }}} = \frac{{{\text{sin}}{{12}^ \circ }}}{{{\text{sin}}{{81}^ \circ }}} \Leftrightarrow {\text{sin}}{9^ \circ } \cdot {\text{sin}}{81^ \circ } – {\text{sin}}{12^ \circ } \cdot {\text{sin}}{48^ \circ } = 0$

$\; \Leftrightarrow \frac{1}{2}\left( {{\text{cos}}{{72}^ \circ } – {\text{cos}}{{90}^ \circ }} \right) – \frac{1}{2}\left( {{\text{cos}}{{36}^ \circ } – {\text{cos}}{{60}^ \circ }} \right) = 0 \Leftrightarrow 2{\text{cos}}{72^ \circ } – 2{\text{cos}}{36^ \circ } + 1 = 0$

Suy ra B đúng.

Tương tự, ta cũng chứng minh được các biểu thức ở ${\text{C}}$ và ${\text{D}}$ đúng. Biểu thức ở đáp án ${\text{A}}$ sai.

Câu 41. Nếu $5{\text{sin}}\alpha = 3{\text{sin}}\left( {\alpha + 2\beta } \right)$ thì:
A. ${\text{tan}}\left( {\alpha + \beta } \right) = 2{\text{tan}}\beta $.
B. ${\text{tan}}\left( {\alpha + \beta } \right) = 3{\text{tan}}\beta $.
C. ${\text{tan}}\left( {\alpha + \beta } \right) = 4{\text{tan}}\beta $.
D. ${\text{tan}}\left( {\alpha + \beta } \right) = 5{\text{tan}}\beta $.

Chọn C.

Lời giải

Ta có :

$5{\text{sin}}\alpha = 3{\text{sin}}\left( {\alpha + 2\beta } \right) \Leftrightarrow 5{\text{sin}}\left[ {\left( {\alpha + \beta } \right) – \beta } \right] = 3{\text{sin}}\left[ {\left( {\alpha + \beta } \right) + \beta } \right]$

$ \Leftrightarrow 5{\text{sin}}\left( {\alpha + \beta } \right){\text{cos}}\beta – 5{\text{cos}}\left( {\alpha + \beta } \right){\text{sin}}\beta = 3{\text{sin}}\left( {\alpha + \beta } \right){\text{cos}}\beta + 3{\text{cos}}\left( {\alpha + \beta } \right){\text{sin}}\beta $

$ \Leftrightarrow 2{\text{sin}}\left( {\alpha + \beta } \right){\text{cos}}\beta = 8{\text{cos}}\left( {\alpha + \beta } \right){\text{sin}}\beta \Leftrightarrow \frac{{{\text{sin}}\left( {\alpha + \beta } \right)}}{{{\text{cos}}\left( {\alpha + \beta } \right)}} = 4\frac{{{\text{sin}}\beta }}{{{\text{cos}}\beta }} \Leftrightarrow {\text{tan}}\left( {\alpha + \beta } \right) = 4{\text{tan}}\beta $

Câu 42. Cho biểu thức $A = {\text{si}}{{\text{n}}^2}\left( {a + b} \right) – {\text{si}}{{\text{n}}^2}a – {\text{si}}{{\text{n}}^2}b$. Hãy chọn kết quả đúng:
A. $A = 2{\text{cos}}a \cdot {\text{sin}}b \cdot {\text{sin}}\left( {a + b} \right)$.
B. $A = 2{\text{sin}}a \cdot {\text{cos}}b \cdot {\text{cos}}\left( {a + b} \right)$.
C. $A = 2{\text{cos}}a \cdot {\text{cos}}b \cdot {\text{cos}}\left( {a + b} \right)$.
D. $A = 2{\text{sin}}a \cdot {\text{sin}}b \cdot {\text{cos}}\left( {a + b} \right)$.

Chọn D.

Lời giải

Ta có :

$A = {\text{si}}{{\text{n}}^2}\left( {a + b} \right) – {\text{si}}{{\text{n}}^2}a – {\text{si}}{{\text{n}}^2}b = {\text{si}}{{\text{n}}^2}\left( {a + b} \right) – \frac{{1 – {\text{cos}}2a}}{2} – \frac{{1 – {\text{cos}}2b}}{2}$

$ = {\text{si}}{{\text{n}}^2}\left( {a + b} \right) – 1 + \frac{1}{2}\left( {{\text{cos}}2a + {\text{cos}}2b} \right) = – {\text{co}}{{\text{s}}^2}\left( {a + b} \right) + {\text{cos}}\left( {a + b} \right){\text{cos}}\left( {a – b} \right)$

$ = {\text{cos}}\left( {a + b} \right)\left[ {{\text{cos}}\left( {a – b} \right) – {\text{cos}}\left( {a + b} \right)} \right] = 2{\text{sin}}a{\text{sin}}b{\text{cos}}\left( {a + b} \right)$

Câu 43. Xác định hệ thức ${\mathbf{SAI}}$ trong các hệ thức sau:
A. ${\text{cos}}{40^ \circ } + {\text{tan}}\alpha \cdot {\text{sin}}{40^ \circ } = \frac{{{\text{cos}}\left( {{{40}^ \circ } – \alpha } \right)}}{{{\text{cos}}\alpha }}$.
B. ${\text{sin}}{15^ \circ } + {\text{tan}}{30^ \circ } \cdot {\text{cos}}{15^ \circ } = \frac{{\sqrt 6 }}{3}$.
C. ${\text{co}}{{\text{s}}^2}x – 2{\text{cos}}a \cdot {\text{cos}}x \cdot {\text{cos}}\left( {a + x} \right) + {\text{co}}{{\text{s}}^2}\left( {a + x} \right) = {\text{si}}{{\text{n}}^2}a$.
D. ${\text{si}}{{\text{n}}^2}x + 2{\text{sin}}\left( {a – x} \right) \cdot {\text{sin}}x \cdot {\text{cos}}a + {\text{si}}{{\text{n}}^2}\left( {a – x} \right) = {\text{co}}{{\text{s}}^2}a$.

Chọn D.

Lời giải

Ta có :

${\text{cos}}{40^ \circ } + {\text{tan}}\alpha \cdot {\text{sin}}{40^ \circ } = {\text{cos}}{40^ \circ } + \frac{{{\text{sin}}\alpha }}{{{\text{cos}}\alpha }} \cdot {\text{sin}}{40^ \circ } = \frac{{{\text{cos}}{{40}^ \circ }{\text{cos}}\alpha + {\text{sin}}{{40}^ \circ }{\text{sin}}\alpha }}{{{\text{cos}}\alpha }} = \frac{{{\text{cos}}\left( {{{40}^ \circ } – \alpha } \right)}}{{{\text{cos}}\alpha }} \cdot {\text{A}}$ đúng.

${\text{co}}{{\text{s}}^2}x – 2{\text{cos}}a \cdot {\text{cos}}x \cdot {\text{cos}}\left( {a + x} \right) + {\text{co}}{{\text{s}}^2}\left( {a + x} \right) = {\text{co}}{{\text{s}}^2}x + {\text{cos}}\left( {a + x} \right)\left[ { – 2{\text{cos}}a{\text{cos}}x + {\text{cos}}\left( {a + x} \right)} \right]$ $ = {\text{co}}{{\text{s}}^2}x – {\text{cos}}\left( {a + x} \right){\text{cos}}\left( {a – x} \right)$

$ = {\text{co}}{{\text{s}}^2}x – \frac{1}{2}\left( {{\text{cos}}2a + {\text{cos}}2x} \right) = {\text{co}}{{\text{s}}^2}x – {\text{co}}{{\text{s}}^2}a – {\text{co}}{{\text{s}}^2}x + 1 = {\text{si}}{{\text{n}}^2}a \cdot {\text{C đúng}}$

${\text{si}}{{\text{n}}^2}x + 2{\text{sin}}\left( {a – x} \right) \cdot {\text{sin}}x \cdot {\text{cos}}a + {\text{si}}{{\text{n}}^2}\left( {a – x} \right)$

$ = {\text{si}}{{\text{n}}^2}x + {\text{sin}}\left( {a – x} \right)\left( {2{\text{sin}}x{\text{cos}}a + {\text{sin}}\left( {a – x} \right)} \right)$

$ = {\text{si}}{{\text{n}}^2}x + {\text{sin}}\left( {a – x} \right){\text{sin}}\left( {a + x} \right) = {\text{si}}{{\text{n}}^2}x + \frac{1}{2}\left( {{\text{cos}}2x – {\text{cos}}2a} \right)$

$ = {\text{si}}{{\text{n}}^2}x – {\text{co}}{{\text{s}}^2}a – {\text{si}}{{\text{n}}^2}x + 1 = {\text{si}}{{\text{n}}^2}a \cdot$

D sai.

Câu 44. Giá trị nhỏ nhất của ${\text{si}}{{\text{n}}^6}x + {\text{co}}{{\text{s}}^6}x$ là
A. 0 .
B. $\frac{1}{2}$.
C. $\frac{1}{4}$.
D. $\frac{1}{8}$.

Chọn C

Lời giải

Ta có ${\text{si}}{{\text{n}}^6}x + {\text{co}}{{\text{s}}^6}x = {\left( {{\text{si}}{{\text{n}}^2}x + {\text{co}}{{\text{s}}^2}x} \right)^3} – 3{\text{si}}{{\text{n}}^2}x{\text{co}}{{\text{s}}^2}x\left( {{\text{si}}{{\text{n}}^2}x + {\text{co}}{{\text{s}}^2}x} \right) = 1 – \frac{3}{4}{\text{si}}{{\text{n}}^2}2x \geqslant 1 – \frac{3}{4} = \frac{1}{4}$.
Dấu “=” xảy ra khi và chỉ khi ${\text{si}}{{\text{n}}^2}2x = 1 \Leftrightarrow {\text{cos}}2x = 0 \Leftrightarrow 2x = \frac{\pi }{2} + k\pi \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\left( {k \in \mathbb{Z}} \right)$.

Câu 45. Giá trị lớn nhất của $M = {\text{si}}{{\text{n}}^4}x + {\text{co}}{{\text{s}}^4}x$ bằng:
A. 4 .
B. 1 .
C. 2 .
D. 3 .

Chọn B

Lời giải

Ta có $M = 1 – \frac{1}{2}{\text{si}}{{\text{n}}^2}2x$

Vì $0 \leqslant {\text{si}}{{\text{n}}^2}x \leqslant 1$

$ \Leftrightarrow – \frac{1}{2} \leqslant – \frac{1}{2}{\text{si}}{{\text{n}}^2}2x \leqslant 0$

$ \Leftrightarrow \frac{1}{2} \leqslant 1 – \frac{1}{2}{\text{si}}{{\text{n}}^2}2x \leqslant 1$

Nên giá trị lớn nhất là 1 .

Câu 46. Cho $M = 3{\text{sin}}x + 4{\text{cos}}x$. Chọn khẳng định đúng.
A. $ – 5 \leqslant M \leqslant 5$.
B. $M > 5$.
C. $M \geqslant 5$.
D. $M \leqslant 5$.

Chọn A

Lời giải

$M = 5\left( {\frac{3}{5}{\text{sin}}x + \frac{4}{5}{\text{cos}}x} \right) = 5{\text{sin}}\left( {x + a} \right)$ với ${\text{cos}}a = \frac{3}{5};{\text{sin}}a = \frac{4}{5}$.

Ta có: $ – 1 \leqslant {\text{sin}}\left( {x + a} \right) \leqslant 1$

$ \Leftrightarrow – 5 \leqslant 5{\text{sin}}\left( {x + a} \right) \leqslant 5$

Câu 47. Giá trị lớn nhất của $M = {\text{si}}{{\text{n}}^6}x – {\text{co}}{{\text{s}}^6}x$ bằng:
A. 2
B. 3
C. 0 .
D. 1 .

Chọn D

Lời giải

Ta có.

$M = \left( {{\text{si}}{{\text{n}}^2}x – {\text{co}}{{\text{s}}^2}x} \right)\left( {{\text{si}}{{\text{n}}^4}x + {\text{si}}{{\text{n}}^2}x{\text{co}}{{\text{s}}^2}x + {\text{co}}{{\text{s}}^4}x} \right)$

$ = – {\text{cos}}2x\left( {1 – {\text{si}}{{\text{n}}^2}x{\text{co}}{{\text{s}}^2}x} \right)$

$\; = – {\text{cos}}2x\left( {1 – \frac{1}{4}{\text{si}}{{\text{n}}^2}2x} \right)$

$\; = – {\text{cos}}2x\left( {\frac{3}{4} + \frac{1}{4}{\text{co}}{{\text{s}}^2}2x} \right) \leqslant \frac{3}{4} + \frac{1}{4}{\text{co}}{{\text{s}}^2}2x \leqslant \frac{3}{4} + \frac{1}{4} = 1$ Do $\left( {{\text{cos}}2x \leqslant 1} \right)$

Nên giá trị lớn nhất là 1 .

Câu 48. Cho biểu thức $M = \frac{{1 + {\text{tan}}{x^3}}}{{{{(1 + {\text{tan}}x)}^3}}},\left( {x \ne – \frac{\pi }{4} + k\pi ,x \ne \frac{\pi }{2} + k\pi ,k \in \mathbb{Z}} \right)$, mệnh đề nào trong các mệnh đề sau đúng?
A. $M \leqslant 1$.
B. $M \geqslant \frac{1}{4}$.
C. $\frac{1}{4} \leqslant M \leqslant 1$.
D. $M < 1$.

Chọn B

Lời giải

Đặt $t = {\text{tan}}x,t \in \mathbb{R} \setminus \left\{ { – 1} \right\}$.

Ta có: $M = \frac{{1 + {t^3}}}{{{{(1 + t)}^3}}} = \frac{{{t^2} – t + 1}}{{{t^2} + 2t + 1}} \Rightarrow \left( {M – 1} \right){t^2} + \left( {2M + 1} \right)t + M – 1 = 0.\left( {\text{*}} \right)$.

Với $M = 1$ thì $\left( {{\;^{\text{*}}}} \right)$ có nghiệm $t = 0$.

Với $M \ne 1$ để $\left( {\text{*}} \right)$ có nghiệm khác -1 thì.

Câu 49. Cho $M = 6{\text{co}}{{\text{s}}^2}x + 5{\text{si}}{{\text{n}}^2}x$. Khi đó giá trị lớn nhất của $M$ là
A. 11 .
B. 1 .
C. 5 .
D. 6 .

Chọn D

Lời giải

$M = 6\left( {1 – {\text{si}}{{\text{n}}^2}x} \right) + 5{\text{si}}{{\text{n}}^2}x = 6 – {\text{si}}{{\text{n}}^2}x$

Ta có: $0 \leqslant {\text{si}}{{\text{n}}^2}x \leqslant 1,\forall x \in R$

$ \Leftrightarrow 0 \geqslant – {\text{si}}{{\text{n}}^2}x \geqslant – 1,\forall x \in R$

$ \Leftrightarrow 6 \geqslant 6 – {\text{si}}{{\text{n}}^2}x \geqslant 5,\forall x \in R$.

Gía trị lớn nhất là 6 .

Câu 50. Giá trị lớn nhất của biểu thức $M = 7{\text{co}}{{\text{s}}^2}x – 2{\text{si}}{{\text{n}}^2}x$ là
A. -2 .
B. 5 .
C. 7 .
D. 16 .

Chọn C

Lời giải

$M = 7\left( {1 – {\text{si}}{{\text{n}}^2}x} \right) – 2{\text{si}}{{\text{n}}^2}x = 7 – 9{\text{si}}{{\text{n}}^2}x$

Ta có: $0 \leqslant {\text{si}}{{\text{n}}^2}x \leqslant 1$

$ \Leftrightarrow 0 \geqslant – 9{\text{si}}{{\text{n}}^2}x \geqslant – 9,\forall x \in R$

$ \Leftrightarrow 7 \geqslant 7 – 2{\text{si}}{{\text{n}}^2}x \geqslant – 2$.

Gía trị lớn nhất là 7 .