Đề kiểm tra Giá trị lượng giác của một góc từ 0 độ đến 180 độ lớp 10 (có lời giải) - Đề 3

Lời giải

Chọn A

Ta có $\cos {45^{\rm{O}}} + \sin {45^{\rm{O}}} = \sqrt 2$.

Chọn A

Ta có $\sin {0^{\rm{O}}} + \cos {0^{\rm{O}}} = 1$.

Chọn B

Với $\alpha \in \left( {90^\circ ;180^\circ } \right)$, ta có $\sin \alpha > 0,\cos \alpha < 0$ suy ra: $\tan \alpha < 0,\cot \alpha < 0$

Vậy $\sin \alpha .\cot \alpha < 0$

Chọn C

$\cos {30}^{^{\circ}}+\sin {60}^{^{\circ}}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}$

Chọn A

$\tan {30}^{^{\circ}}+\cot {30}^{^{\circ}}=\frac{\sqrt{3}}{3}+\sqrt{3}=\frac{4\sqrt{3}}{3}$

Chọn A

Ta có $\cot \alpha = - \frac{1}{2} \Rightarrow \tan \alpha = - 2$.

$1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} \Leftrightarrow {\cos ^2}\alpha = \frac{1}{{1 + {{\tan }^2}\alpha }} = \frac{1}{{1 + {{\left( { - 2} \right)}^2}}} = \frac{1}{5}$.

Suy ra $\cos \alpha = \pm \frac{{\sqrt 5 }}{5}$.

Chọn B

$E = \frac{{\cot \alpha + 3\tan \alpha }}{{2\cot \alpha + \tan \alpha }} = \frac{{1 + 3{{\tan }^2}\alpha }}{{2 + {{\tan }^2}\alpha }} = \frac{{3\left( {{{\tan }^2}\alpha + 1} \right) - 2}}{{1 + \left( {1 + {{\tan }^2}\alpha } \right)}} = \frac{{\frac{3}{{{{\cos }^2}\alpha }} - 2}}{{\frac{1}{{{{\cos }^2}\alpha }} + 1}} = \frac{{3 - 2{{\cos }^2}\alpha }}{{1 + {{\cos }^2}\alpha }} = \frac{{19}}{{13}}$.

Chọn A

$G = \left[ {\left( {1 - {{\sin }^2}x} \right) - 1} \right]{\cot ^2}x + 1 = - {\sin ^2}x.{\cot ^2}x + 1 = 1 - {\cos ^2}x = {\sin ^2}x$.

Chọn B

Ta có: $\sin a + \cos a = \sqrt 2$$\Rightarrow 2 = {\left( {\sin a + \cos a} \right)^2}$$\Rightarrow \sin a.\cos a = \frac{1}{2}$.

${\sin ^4}a + {\cos ^4}a = \left( {{{\sin }^2}a + {{\cos }^2}a} \right) - 2{\sin ^2}a{\cos ^2}a = 1 - 2{\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}$.

Chọn B

$\begin{array}{l}{\tan ^2}x{\sin ^2}x - {\tan ^2}x + {\sin ^2}x = \\{\tan ^2}x\left( {{{\sin }^2}x - 1} \right) + {\sin ^2}x = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\left( { - {{\cos }^2}x} \right) + {\sin ^2}x = 0\end{array}$.

Chọn A

$\begin{matrix}E=\sin {36}^{^{\circ}}\cos {6}^{^{\circ}}\sin ({90}^{^{\circ}}+{36}^{^{\circ}})\cos ({90}^{^{\circ}}-{6}^{^{\circ}})= \\ \sin {36}^{^{\circ}}\cos {6}^{^{\circ}}-\cos {36}^{^{\circ}}\sin {6}^{^{\circ}}=\sin {30}^{^{\circ}}=\frac{1}{2}\end{matrix}$

Chọn C

$S={\sin }^{2}{2}^{^{\circ}}+{\sin }^{2}{4}^{^{\circ}}+{\sin }^{2}{6}^{^{\circ}}+...+{\sin }^{2}{84}^{^{\circ}}+{\sin }^{2}{86}^{^{\circ}}+{\sin }^{2}{88}^{^{\circ}}$

$=({\sin }^{2}{2}^{^{\circ}}+{\sin }^{2}{88}^{^{\circ}})+({\sin }^{2}{4}^{^{\circ}}+{\sin }^{2}{86}^{^{\circ}})+...+({\sin }^{2}{44}^{^{\circ}}+{\sin }^{2}{46}^{^{\circ}})$

$\ =({\sin }^{2}{2}^{^{\circ}}+{\cos }^{2}{2}^{^{\circ}})+({\sin }^{2}{4}^{^{\circ}}+{\cos }^{2}{4}^{^{\circ}})+...+({\sin }^{2}{44}^{^{\circ}}+{\cos }^{2}{44}^{^{\circ}})=22$

a) Sai

b) Đúng

c) Sai

d) Sai

Vì ${\sin ^2}\alpha + {\cos ^2}\alpha = 1 \Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}}$.

Mà ${90}^{^{\circ}}<α<{180}^{^{\circ}}$ nên $\cos \alpha < 0$.

Do đó $\cos \alpha = - \sqrt {\frac{{16}}{{25}}} = - \frac{4}{5};\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{3}{4}$.

a) Đúng

b) Đúng

c) Sai

d) Sai

a) $\sin {30}^{^{\circ}}=\cos {60}^{^{\circ}}$

b) $\sin {150}^{^{\circ}}=\sin ({180}^{^{\circ}}-{30}^{^{\circ}})=\sin {30}^{^{\circ}}$;

c) $\cos {40}^{^{\circ}}=\cos ({180}^{^{\circ}}-{140}^{^{\circ}})=-\cos {140}^{^{\circ}}$;

d) $\tan {25}^{^{\circ}}=\tan ({180}^{^{\circ}}-{155}^{^{\circ}})=-\tan {155}^{^{\circ}}$.

a) Đúng

b) Sai

c) Đúng

d) Đúng

a) Ta có: $\sin \alpha = \frac{1}{3} \Rightarrow {\sin ^2}\alpha = \frac{1}{9}$. Vì ${\sin ^2}\alpha + {\cos ^2}\alpha = 1$ nên ${\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - \frac{1}{9} = \frac{8}{9}$

b) $A = \frac{1}{9} + 3 \cdot \frac{8}{9} = \frac{{25}}{9}$.

c) $B = 5 \cdot \frac{1}{9} - \frac{8}{9} = - \frac{1}{3}$

d) $C = \sqrt {\frac{1}{9} + 3 \cdot \frac{8}{9}} + \sqrt {\frac{8}{9} - 7 \cdot \frac{1}{9}} = \frac{5}{3} + \frac{1}{3} = 2$.

a) Đúng

b) Đúng

c) Đúng

d) Đúng

a)${\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} = 1$

b) $1 + 2\sin \alpha \cdot \cos \alpha = {\sin ^2}\alpha + {\cos ^2}\alpha + 2\sin \alpha \cdot \cos \alpha = {(\sin \alpha + \cos \alpha )^2}$.

c) $1 - 2\sin \alpha \cdot \cos \alpha = {\sin ^2}\alpha + {\cos ^2}\alpha - 2\sin \alpha \cdot \cos \alpha = {(\sin \alpha - \cos \alpha )^2}$.

d) Ta có: ${\sin ^4}\alpha + {\cos ^4}\alpha + 2{\sin ^2}\alpha \cdot {\cos ^2}\alpha = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} = 1$.

Do đó: $1 - 2{\sin ^2}\alpha {\cos ^2}\alpha = {\sin ^4}\alpha + {\cos ^4}\alpha$.

$F=1-2{\sin }^{2}({90}^{^{\circ}}-{35}^{^{\circ}})+4{\cos }^{2}{60}^{^{\circ}}-2{\sin }^{2}{35}^{^{\circ}}+\tan ({90}^{^{\circ}}-{35}^{^{\circ}})\tan {35}^{^{\circ}}$

$F=1-2{\cos }^{2}{35}^{^{\circ}}+4{\cos }^{2}{60}^{^{\circ}}-2{\sin }^{2}{35}^{^{\circ}}+\cot {35}^{^{\circ}}\cdot \tan {55}^{^{\circ}}$

$=1-2({\sin }^{2}{35}^{^{\circ}}+{\cos }^{2}{35}^{^{\circ}})+4{\cos }^{2}{60}^{^{\circ}}+\tan {35}^{^{\circ}}\cot {35}^{^{\circ}}=1-2\cdot 1+4\cdot {(\frac{1}{2})}^{2}+1=1$

Vì ${30}^{^{\circ}}$ và ${60}^{^{\circ}}$ là hai góc phụ nhau nên $\{\begin{matrix}\sin {30}^{^{\circ}}=\cos {60}^{^{\circ}} \\ \sin {60}^{^{\circ}}=\cos {30}^{^{\circ}}\end{matrix}$

$P=\cos {30}^{^{\circ}}\cdot \cos {60}^{^{\circ}}-\sin {30}^{^{\circ}}\cdot \sin {60}^{^{\circ}}=\cos {30}^{^{\circ}}\cdot \cos {60}^{^{\circ}}-\cos {60}^{^{\circ}}\cdot \cos {30}^{^{\circ}}=0.\$

$\sin A\cdot \cos (B+C)+\cos A\cdot \sin (B+C)=\sin A\cdot \cos ({180}^{^{\circ}}-A)+\cos A\cdot \sin ({180}^{^{\circ}}-A)$

= - \sin A \cdot \cos A + \cos A \cdot \sin A = 0\).

$A=2({\sin }^{2}{10}^{^{\circ}}+{\sin }^{2}{20}^{^{\circ}}+\ldots +{\sin }^{2}{80}^{^{\circ}})+1$

$\begin{matrix}=2[({\sin }^{2}{10}^{^{\circ}}+{\sin }^{2}{80}^{^{\circ}})+({\sin }^{2}{20}^{^{\circ}}+{\sin }^{2}{70}^{^{\circ}})+({\sin }^{2}{30}^{^{\circ}}+{\sin }^{2}{60}^{^{\circ}})+({\sin }^{2}{40}^{^{\circ}}+{\sin }^{2}{50}^{^{\circ}})]+1 \\ =2[({\sin }^{2}{10}^{^{\circ}}+{\cos }^{2}{10}^{^{\circ}})+({\sin }^{2}{20}^{^{\circ}}+{\cos }^{2}{20}^{^{\circ}})+({\sin }^{2}{30}^{^{\circ}}+{\cos }^{2}{30}^{^{\circ}})+({\sin }^{2}{40}^{^{\circ}}+{\cos }^{2}{40}^{^{\circ}})]+1 \\ =2.4+1=9\end{matrix}$