13 câu Trắc nghiệm Giá trị có lượng giác của một cung có đáp án

Đáp án D

$$\begin{gathered} P=3{\sin }^{2}x+4{\cos }^{2}x=3({\sin }^{2}x+{\cos }^{2}x)+{\cos }^{2}x \\ =3+{\left(\frac{1}{2}\right)}^2=\frac{13}{4} \end{gathered}$$

Đáp án A

$$\begin{gathered} A=\frac{cos{750}^0+\sin {420}^0}{\sin \left(-{330}^0\right)-cos\left(-{390}^0\right)} \\ =\frac{cos\left({30}^0+{2.360}^0\right)+\sin \left({60}^0+{360}^0\right)}{-\sin \left(-{30}^0+{360}^0\right)-cos\left({30}^0+{360}^0\right)} \\ =\frac{cos{30}^0+\sin {60}^0}{\sin {30}^0-cos{30}^0} \\ =\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{\frac{1}{2}-\frac{\sqrt{3}}{2}}=\frac{2\sqrt{3}}{1-\sqrt{3}}=-3-\sqrt{3} \end{gathered}$$

Đáp án C

Ta có:

$$\begin{gathered} C=2\left({\sin }^{4}x+co{s}^{4}x+{\sin }^{2}xco{s}^{2}x\right){}^2-\left({\sin }^{8}x+co{s}^{8}x\right) \\ =2{\left[{\left({\sin }^{2}x+co{s}^{2}x\right)}^2-{\sin }^{2}xco{s}^{2}x\right]}^2-\left[{\left({\sin }^{4}x+co{s}^{4}x\right)}^2-2{\sin }^{4}xco{s}^{4}x\right] \\ =2{\left[1-{\sin }^{2}xco{s}^{2}x\right]}^2-\left[{\left({\sin }^{2}x+co{s}^{2}x\right)}^2-2{\sin }^{2}xco{s}^{2}x\right]{}^2+2{\sin }^{4}xco{s}^{4}x \\ =2{\left[1-{\sin }^{2}xco{s}^{2}x\right]}^2-\left[1-2{\sin }^{2}x.co{s}^{2}x\right]{}^2+2{\sin }^{4}xco{s}^{4}x \\ =2\left(1-2{\sin }^{2}xco{s}^{2}x+{\sin }^{4}xco{s}^{4}x\right)-\left(1-4{\sin }^{2}xco{s}^{2}x+4{\sin }^{4}xco{s}^{4}x\right)+2{\sin }^{4}xco{s}^{4}x \\ =1 \end{gathered}$$

Đáp án B

$$\begin{gathered} A=a{\cos }^{2}x+2b\sin x.cosx+c{\sin }^{2}x \\ \iff \frac{A}{co{s}^{2}x}=a+2b\tan x+c{\tan }^{2}x \\ \iff A\left(1+{\tan }^{2}x\right)=a+2b\tan x+c{\tan }^{2}x \\ \iff A\left(1+{\left(\frac{2b}{a-c}\right)}^2\right)=a+2b\frac{2b}{a-c}+c{\left(\frac{2b}{a-c}\right)}^2 \\ \iff A\frac{{\left(a-c\right)}^2+{\left(2b\right)}^2}{{\left(a-c\right)}^2}=\frac{a{\left(a-c\right)}^2+4b^2\left(a-c\right)+c4b^2}{{\left(a-c\right)}^2} \\ \iff A\frac{{\left(a-c\right)}^2+{\left(2b\right)}^2}{{\left(a-c\right)}^2}=\frac{a{\left(a-c\right)}^2+4b^{2}a}{{\left(a-c\right)}^2}=\frac{a\left({\left(a-c\right)}^2+4b^2\right)}{{\left(a-c\right)}^2} \\ \iff A=a \end{gathered}$$

Đáp án B

Vì $\cos {90}^{\circ } = 0$ nên $A = \cos {235}^{\circ }. \sin {60}^{\circ }. \tan {125}^{\circ }. \cos {90}^{\circ }=0$

Đáp án A

$$\begin{gathered} A=\frac{{\tan }^{2}a-{\sin }^{2}a}{{\cot }^{2}a-co{s}^{2}a} \\ \iff A=\frac{{\sin }^{2}a\left(\frac{1}{co{s}^{2}a}-1\right)}{co{s}^{2}a\left(\frac{1}{{\sin }^{2}a}-1\right)} \\ =\frac{{\tan }^{2}a.{\tan }^{2}a}{{\cot }^{2}a}={\tan }^{6}a \end{gathered}$$

Đáp án C

$$\begin{gathered} {\sin }^{2}a.{\tan }^{2}a+4{\sin }^{2}a-{\tan }^{2}a+3co{s}^{2}a \\ ={\sin }^{2}a\left(\frac{1}{co{s}^{2}a}-1\right)+4{\sin }^{2}a-{\tan }^{2}a+3co{s}^{2}a \\ =\frac{{\sin }^{2}a}{co{s}^{2}a}-{\sin }^{2}a+4{\sin }^{2}a-{\tan }^{2}a+3co{s}^{2}a \\ =3{\sin }^{2}a+3co{s}^{2}a \\ =3 \end{gathered}$$

Đáp án A

$$\begin{gathered} S=\cos ({90}^{\circ }-x) \sin ({180}^{\circ }-x)-\sin ({90}^{\circ }-x) \cos ({180}^{\circ }–x) \\ = \sin x.\sin x - \cos x.(-\cos x) \\ = {\sin }^{2}x + {\cos }^{2}x = 1 \end{gathered}$$

Đáp án C

$$\begin{gathered} A=\frac{\sin \left(-{234}^0\right)-cos{216}^0}{\sin {144}^0-cos{126}^0}.\tan {36}^0 \\ =\frac{-\sin \left({180}^0+{54}^0\right)-cos\left({180}^0+{36}^0\right)}{\sin \left({180}^0-{36}^0\right)-cos\left({180}^0-{54}^0\right)}.\tan {36}^0 \\ =\frac{\sin {54}^0+cos{36}^0}{\sin {36}^0+cos{54}^0}.\tan {36}^0 \\ =\frac{cos{36}^0+cos{36}^0}{\sin {36}^0+\sin {36}^0}.\tan {36}^0 \\ =\cot {36}^0.\tan {36}^0=1 \end{gathered}$$

Đáp án A

$$\begin{gathered} A=\frac{\sin \left(-{328}^0\right).\sin {958}^0}{\cot {572}^0}-\frac{cos\left(-{508}^0\right).cos\left(-{1022}^0\right)}{\tan \left(-{212}^0\right)} \\ A=-\frac{\sin {32}^0.\sin {58}^0}{\cot {32}^0}-\frac{cos{32}^0.cos{58}^0}{\tan \left(-{212}^0\right)} \\ A=-\frac{\sin {32}^0.cos{32}^0}{\cot {32}^0}-\frac{cos{32}^0.\sin {32}^0}{\tan {32}^0} \\ A=-{\sin }^2{32}^0-co{s}^2{32}^0=-1 \end{gathered}$$

Đáp án C

$$\begin{gathered} A=\frac{\sin {155}^0.cos{115}^0+\cot {42}^0.\cot {48}^0}{\cot {55}^0.\cot \left(-{145}^0\right)+\tan {17}^0.\cot {17}^0} \\ \iff A=\frac{\sin {25}^0.\left(-\sin {25}^0\right)+\cot {42}^0.\tan {42}^0}{\cot {55}^0.\tan {55}^0+1} \\ \iff A=\frac{-{\sin }^2{25}^0+1}{2} \\ \iff A=\frac{co{s}^2{25}^0}{2} \end{gathered}$$

Đáp án D

A = cos(α + 26π) − cos(α − 7π)   − cos(α − 1,5π) − cos(α + 2003$\frac{\pi }{2}$) + cos(α − 1,5π).cot(α − 8π)

= cos(α + 13.2π) − cos(α – π − 2.3π)  − cos(α − $\frac{\pi }{2}$ − π) − cos(α − $\frac{\pi }{2}$ + 1002π) + cos(α − $\frac{\pi }{2}$ − π).cot(α − 4.2π)

= cosα − cos(α − π) + sinα  − cos(α − $\frac{\pi }{2}$ ) − cos(α − $\frac{\pi }{2}$ ).cotα

= cosα + cosα + sinα  − sinα − sinα.cotα  = cosα

Đáp án D

Do ABC là tam giác nên $A + B + C = {180}^{\circ }$ và $\frac{A+C}{2}={90}^0-\frac{B}{2}$

Khi đó: $\sin \frac{A+C}{2}=\sin \left({90}^0-\frac{B}{2}\right)=cos\frac{B}{2}$ nên A đúng

$cos\frac{A+C}{2}=cos\left({90}^0-\frac{B}{2}\right)=\sin \frac{B}{2}$ nên B đúng

$\sin \left(A+B\right)=\sin \left({180}^0-C\right)=\sin C$ nên C đúng

$cos\left(A+B\right)=cos\left({180}^0-C\right)=-\cos C$ nên D sai