10 câu Trắc nghiệm Giá trị lượng giác của một góc bất kì từ 0o đến 150o có đáp án (Vận dụng)

Đáp án D

$$\begin{gathered} 7={\tan }^2\alpha +{\cot }^2\alpha \\ ={\left(\tan \alpha +\cot \alpha \right)}^2-2\tan \alpha \cot \alpha =m^2-2.1 \\ \Rightarrow m^2=9\iff m=\pm 3 \end{gathered}$$

Đáp án B

Ta có:

$$\begin{gathered} \cos \alpha +\sin \alpha =\frac{1}{3}\Rightarrow {\left(\cos \alpha +\sin \alpha \right)}^2=\frac{1}{9} \\ \iff 1+2\cos \alpha \sin \alpha =\frac{1}{9} \\ \iff \cos \alpha \sin \alpha =-\frac{4}{9} \end{gathered}$$

Ta có:

$$\begin{gathered} P=\sqrt{{\tan }^2\alpha +{\cot }^2\alpha } \\ =\sqrt{{\left(\tan \alpha +\cot \alpha \right)}^2-2\tan \alpha \cot \alpha } \\ =\sqrt{{\left(\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }\right)}^2-2} \\ =\sqrt{{\left(\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cos \alpha }\right)}^2-2} \\ =\sqrt{{\left(\frac{1}{\sin \alpha \cos \alpha }\right)}^2-2} \\ =\sqrt{{\left(-\frac{9}{4}\right)}^2-2}=\frac{7}{4} \end{gathered}$$

Đáp án D

Ta có:

$$\begin{gathered} E=2{\cos }^2\alpha +5\sin \alpha \cos \alpha +1 \\ ={\sin }^2\alpha \left(2.\frac{{\cos }^2\alpha }{{\sin }^2\alpha }+5\frac{\cos \alpha }{\sin \alpha }+\frac{1}{{\sin }^2\alpha }\right) \\ =\left(\frac{1}{{\cot }^2\alpha +1}\right)\left(2.{\cot }^2\alpha +5\cot \alpha +\frac{1}{\frac{1}{{\cot }^2\alpha +1}}\right) \\ =\left(\frac{1}{{\cot }^2\alpha +1}\right)\left(2.{\cot }^2\alpha +5\cot \alpha +{\cot }^2\alpha +1\right) \\ =\left(\frac{1}{{\cot }^2\alpha +1}\right)\left(3{\cot }^2\alpha +5\cot \alpha +1\right) \\ =\frac{1}{5^2+1}\left({3.5}^2+5.5+1\right)=\frac{101}{26} \end{gathered}$$

Đáp án B

Ta có:

$$\begin{gathered} E=\frac{\cot \alpha +3\tan \alpha }{2\cot \alpha +\tan \alpha }=\frac{1+3{\tan }^2\alpha }{2+{\tan }^2\alpha } \\ =\frac{3({\tan }^2\alpha +1)-2}{1+(1+{\tan }^2\alpha )}=\frac{\frac{3}{{\cos }^2\alpha }-2}{\frac{1}{{\cos }^2\alpha }+1} \\ =\frac{3-2{\cos }^2\alpha }{1+{\cos }^2\alpha }=\frac{19}{13} \end{gathered}$$

Đáp án A

Ta có:

$$\begin{gathered} 3\cos \alpha -\sin \alpha =1\iff 3\cos \alpha =\sin \alpha +1 \\ \Rightarrow 9{\cos }^2\alpha ={\left(\sin \alpha +1\right)}^2 \\ \iff 9{\cos }^2\alpha ={\sin }^2\alpha +2\sin \alpha +1 \\ \iff 9\left(1-{\sin }^2\alpha \right)={\sin }^2\alpha +2\sin \alpha +1 \\ \iff 10{\sin }^2\alpha +2\sin \alpha -8=0 \\ \iff \left[\begin{array}{c}\sin \alpha =-1\\ \sin \alpha =\frac{4}{5}\end{array}\right. \end{gathered}$$

• $\sin \alpha =-1$ không thỏa mãn vì $0^0<\alpha <{90}^0$
• $\sin \alpha =\frac{4}{5}\Rightarrow \cos \alpha =\frac{3}{5}\Rightarrow \tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{4}{3}$

Đáp án C

Ta có:

$$\begin{gathered} 2\cos \alpha +\sqrt{2}\sin \alpha =2\iff \sqrt{2}\sin \alpha =2-2\cos \alpha \\ \iff 2{\sin }^2\alpha ={\left(2-2\cos \alpha \right)}^2 \\ \iff 2{\sin }^2\alpha =4-8\cos \alpha +4{\cos }^2\alpha \\ \iff 2(1-{\cos }^2\alpha )=4-8\cos \alpha +4{\cos }^2\alpha \\ \iff 6{\cos }^2\alpha -8\cos \alpha +2=0 \\ \iff \left[\begin{array}{c}\cos \alpha =1\\ \cos \alpha =\frac{1}{3}\end{array}\right. \end{gathered}$$

• $\cos \alpha =1$ không thỏa mãn vì $0^0<\alpha <{90}^0$
• $\cos \alpha =\frac{1}{3}\Rightarrow \sin \alpha =\frac{2\sqrt{2}}{3}\Rightarrow \cot \alpha =\frac{\cos \alpha }{\sin \alpha }=\frac{\sqrt{2}}{4}$

Đáp án B

Ta có:

$$\begin{gathered} \sin a+\cos a=\sqrt{2}\Rightarrow 2={\left(\sin a+\cos a\right)}^2 \\ \iff 2={\sin }^{2}a+2\sin a.\cos a+{\cos }^{2}a \\ \iff 2=1+2\sin a.\cos a \\ \iff 1=2\sin a.\cos a \\ \iff \sin a.\cos a=\frac{1}{2} \end{gathered}$$

Do đó: 

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

Đáp án B

Ta có:

$$\begin{gathered} \sin a-\cos a=\frac{1}{\sqrt{5}}\Rightarrow {\left(\sin a-\cos a\right)}^2=\frac{1}{5} \\ \iff 1-2\sin a\cos a=\frac{1}{5} \\ \iff \sin a\cos a=\frac{2}{5} \end{gathered}$$

Ta có:

$$\begin{gathered} P=\sqrt{{\sin }^{4}a+{\cos }^{4}a} \\ =\sqrt{{\left({\sin }^{2}a+{\cos }^{2}a\right)}^2-2{\sin }^{2}a{\cos }^{2}a} \\ =\sqrt{1-2{\left(\sin a\cos a\right)}^2}=\frac{\sqrt{17}}{5} \end{gathered}$$

Đáp án A

$$\begin{gathered} {\sin }^{4}a+{\cos }^{4}a=1-2{\sin }^{2}a{\cos }^{2}a \\ \begin{array}{l}{\sin }^{6}a+{\cos }^{6}a={\left({\sin }^{2}a\right)}^3+{\left(co{s}^{2}a\right)}^3\\ =\left({\sin }^{2}a+co{s}^{2}a\right)\left({\left({\sin }^{2}a\right)}^2-{\sin }^{2}a{\cos }^{2}a+{\left({\cos }^{2}a\right)}^2\right)\\ ={\left({\sin }^{2}a+{\cos }^{2}a\right)}^2-3{\sin }^{2}a{\cos }^{2}a\\ =1-3{\sin }^{2}a{\cos }^{2}a\end{array} \\ f\left(a\right)=3\left(1-2{\sin }^{2}a{\cos }^{2}a\right)-2\left(1-3{\sin }^{2}a{\cos }^{2}a\right)=1 \end{gathered}$$

Đáp án C

Hai góc ${15}^{\circ }$ và ${75}^{\circ }$ phụ nhau nên $\sin {75}^{\circ }=\cos {15}^{\circ }$

Hai góc ${20}^{\circ }$ và ${110}^{\circ }$ hơn kém nhau ${90}^{\circ }$ nên $\cos {110}^{\circ }=-\sin {20}^{\circ }$.

Do đó:

$$\begin{gathered} S={\sin }^2{15}^0+{\cos }^2{20}^0+{\sin }^2{75}^0+{\cos }^2{110}^0 \\ ={\sin }^2{15}^0+{\cos }^2{20}^0+{\cos }^2{15}^0+{(-\sin {20}^0)}^2 \\ =({\sin }^2{15}^0+{\cos }^2{15}^0)({\sin }^2{20}^0+{\cos }^2{20}^0) \\ =2 \end{gathered}$$