Đề kiểm tra Giá trị lượng giác của một góc từ 0 độ đến 180 độ lớp 10 (có lời giải) - Đề 2

Chọn B

\(\tan \alpha  - \cot \alpha  = 3 \Leftrightarrow {\left( {\tan \alpha  - \cot \alpha } \right)^2} = 9 \Leftrightarrow {\tan ^2}\alpha  + {\cot ^2}\alpha  - 2\tan \alpha .\cot \alpha  = 9\)

\( \Leftrightarrow {\tan ^2}\alpha  + {\cot ^2}\alpha  - 2 = 9 \Leftrightarrow {\tan ^2}\alpha  + {\cot ^2}\alpha  = 11\).

Chọn B

\(\begin{array}{l}\sin x + \cos x = m\,\, \Rightarrow \,{\left( {\sin x + \cos x} \right)^2}\, = {m^2}\\\, \Leftrightarrow \,\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sin x.\cos x = {m^2}\,\end{array}\)

\( \Leftrightarrow \,1 + 2\sin x.\cos x = {m^2} \Leftrightarrow \,\sin x.\cos x = \frac{{{m^2} - 1}}{2}\).

Vậy \(M = \frac{{{m^2} - 1}}{2}\).

a) Vì ${90}^{°}<\alpha <{180}^{°}$ nên \(\cos \alpha  < 0\).

b) Ta có: \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - {\left( {\frac{1}{3}} \right)^2} = \frac{8}{9} \Rightarrow \cos \alpha  =  - \frac{{2\sqrt 2 }}{3}{\rm{. }}\)

c) Do đó: \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} =  - \frac{1}{{2\sqrt 2 }}\).

Vì \(\tan \alpha  =  - \frac{3}{4} \Rightarrow \cot \alpha  = \frac{1}{{\tan \alpha }} =  - \frac{4}{3}\).

Theo giả thiết: ${90}^{°}<\alpha <{180}^{°}\Rightarrow \cos \alpha <0$

Ta có: \(\frac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha  = 1 + {\left( { - \frac{3}{4}} \right)^2} = \frac{{25}}{{16}} \Rightarrow {\cos ^2}\alpha  = \frac{{16}}{{25}} \Rightarrow \cos \alpha  =  - \frac{4}{5}\); \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \sin \alpha  = \tan \alpha  \cdot \cos \alpha  =  - \frac{3}{4} \cdot \left( { - \frac{4}{5}} \right) = \frac{3}{5}\).

Vì $0^{°}<\alpha <{90}^{°}$ nên \(\cos \alpha  > 0\).

Do đó \(\cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \frac{5}{{13}};\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{12}}{5};\cot \alpha  = \frac{5}{{12}}\).

a) \(A = 4 \cdot \frac{1}{2} + 3\sqrt 3  \cdot \frac{1}{{\sqrt 3 }} = 2 + 3 = 5\).

b) \(B = \frac{1}{{\sqrt 3 }} \cdot \frac{{\sqrt 3 }}{2} - 3\sqrt 2  \cdot \frac{{\sqrt 2 }}{2} + 1 = \frac{1}{2} - 3 + 1 =  - \frac{3}{2}\).

c) \(C = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - 2 = \frac{3}{4} + \frac{1}{3} - 2 =  - \frac{{11}}{{12}}\).

d) $D=2\sin {30}^{°}+2\sqrt{3}-\frac{1}{2\sqrt{2}}\cos {45}^{°}+3\tan {30}^{°}$ do $\left\{\begin{array}{l}\begin{array}{l}\sin {150}^{°}=\sin \left({180}^{°}-{30}^{°}\right)=\sin {30}^{°}\\ \cos {135}^{°}=\cos \left({180}^{°}-{45}^{°}\right)=-\cos {45}^{°}\\ \tan {150}^{°}=\tan \left({180}^{°}-{30}^{°}\right)=-\tan {30}^{°}\end{array}\end{array}\right.$

$D=2\cdot \frac{1}{2}+2\sqrt{3}-\frac{1}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{2}+3\cdot \frac{\sqrt{3}}{3}=1+2\sqrt{3}-\frac{1}{4}+\sqrt{3}=\frac{3}{4}+3\sqrt{3}.$

\(A = 3 \cdot 1 + 2 \cdot 1 - 3 \cdot \frac{1}{2} + 10 \cdot ( - 1) = - \frac{{13}}{2}\)

Ta có: \(7 = {\tan ^2}\alpha  + {\cot ^2}\alpha  = {(\tan \alpha  + \cot \alpha )^2} - 2 = {m^2} - 2\) \( \Rightarrow {m^2} = 9 \Rightarrow m =  \pm 3\).

\(A = \frac{{\sqrt 3 }}{2} + 2 \cdot \frac{{\sqrt 3 }}{2} - 3 \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 3 - 3\sqrt 2 }}{2}\).

$E={\sin }^2{5}^{°}+{\sin }^2{13}^{°}+{\sin }^2\left({90}^{°}-{13}^{°}\right)+{\sin }^2\left({90}^{°}-5^{°}\right)$

$={\sin }^2{5}^{°}+{\sin }^2{13}^0+{\cos }^2{13}^0+{\cos }^2{5}^0$

\( = \left( {{{\sin }^2}{5^0} + {{\cos }^2}{5^0}} \right) + \left( {{{\sin }^2}{{13}^0} + {{\cos }^2}{{13}^0}} \right) = 1 + 1 = 2\)

Vì \(\cot \alpha  = 2 \Rightarrow \sin \alpha  \ne 0\). Chia cả tử và mẫu cho \({\sin ^3}\alpha \) ta được:

\(B = \frac{{(\sin \alpha  + 2\cos \alpha )\frac{1}{{{{\sin }^3}\alpha }}}}{{\left( {{{\sin }^3}\alpha  - {{\cos }^3}\alpha } \right)\frac{1}{{{{\sin }^3}\alpha }}}} = \frac{{\frac{1}{{{{\sin }^2}\alpha }} + 2\cot \alpha  \cdot \frac{1}{{{{\sin }^2}\alpha }}}}{{1 - {{\cot }^3}\alpha }}\)\(\)

\( = \frac{{1 + {{\cot }^2}\alpha  + 2\cot \alpha \left( {1 + {{\cot }^2}\alpha } \right)}}{{1 - {{\cot }^3}\alpha }} = \frac{{2{{\cot }^3}\alpha  + {{\cot }^2}\alpha  + 2\cot \alpha  + 1}}{{1 - {{\cot }^3}\alpha }} =  - \frac{{25}}{7}{\rm{. }}\)

\({\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = \frac{{144}}{{169}} \Rightarrow \cos \alpha  =  \pm \frac{{12}}{{13}}\)

Do \(\alpha \) là góc tù nên \(\cos \alpha  < 0\), từ đó \(\cos \alpha  =  - \frac{{12}}{{13}}\)

Như vậy \(3\sin \alpha  + 2\cos \alpha  = 3 \cdot \frac{5}{{13}} + 2\left( { - \frac{{12}}{{13}}} \right) =  - \frac{9}{{13}}\)