14 câu Trắc nghiệm Ôn tập Toán 10 Chương 6 có đáp án

Đáp án A

Ta có:

$\cot \alpha =3\Rightarrow \frac{\cos \alpha }{\sin \alpha }=3\iff \cos \alpha =3\sin \alpha$

Thay vào biểu thức đề bài, ta được:

$$\begin{gathered} \frac{3\sin \alpha -2.3\sin \alpha }{12{\sin }^3\alpha +4{\left(3\sin \alpha \right)}^3}=\frac{-3\sin \alpha }{120{\sin }^3\alpha } \\ =-\frac{1}{40}.\frac{1}{{\sin }^2\alpha }=-\frac{1}{40}\left(1+{\cot }^2\alpha \right) \\ =-\frac{1}{40}\left(1+3^2\right)=-\frac{1}{4} \end{gathered}$$

Đáp án D

Ta có:

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

Đặt $\tan \frac{\alpha }{2}=t$ thì $\cos \alpha =\frac{1-t^2}{1+t^2}$

Với t = 2 $\Rightarrow \cos \alpha =\frac{1-4}{1+4}=\frac{-3}{5}$

Suy ra $B=\frac{1+5\left(\frac{-3}{5}\right)}{3-2\left(\frac{-3}{5}\right)}=\frac{-2}{\frac{21}{5}}=\frac{-10}{21}$

Đáp án B

$$\begin{gathered} A={\tan }^2\frac{\pi }{24}+{\cot }^2\frac{\pi }{24} \\ =\frac{1}{{\cos }^2\frac{\pi }{24}}-1+\frac{1}{{\sin }^2\frac{\pi }{24}}-1 \\ =\frac{1}{{\cos }^2\frac{\pi }{24}.{\sin }^2\frac{\pi }{24}}-2 \\ =\frac{4}{{\sin }^2\frac{\pi }{12}}-2=\frac{8}{1-\cos \frac{\pi }{6}}-2 \\ =\frac{8}{1-\frac{\sqrt{3}}{2}}-2=\frac{16}{2-\sqrt{3}}-2 \\ =\frac{16-4+2\sqrt{3}}{2-\sqrt{3}}=\frac{12+2\sqrt{3}}{2-\sqrt{3}} \end{gathered}$$

Đáp án B

Ta có: $\sin a-\cos a=\frac{3}{4}$ suy ra ${(\sin a-\cos a)}^2=\frac{9}{16}$

$$\begin{gathered} \iff {\sin }^{2}a+{\cos }^{2}a-2\sin a\cos a=\frac{9}{16} \\ \iff 1-\sin 2a=\frac{9}{16}\iff \sin 2a=1-\frac{9}{16} \\ \iff \sin 2a=\frac{7}{16} \end{gathered}$$

Đáp án B

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

nên A sai

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

nên B đúng

${\left(\sin a-\cos a\right)}^2=1-2\sin a.\cos a=1-\sin 2a$nên C sai

${\left({\sin }^{2}a+{\cos }^{2}a\right)}^3=1$ và $1+2{\sin }^{4}a.{\cos }^{4}a=1+2.{\left(\frac{1}{2}\sin 2a\right)}^4=1+\frac{1}{8}{\sin }^{4}2a$ nên D sai

Đáp án A

$$\begin{gathered} G={\cos }^2\frac{\pi }{6}+{\cos }^2\frac{2\pi }{6}+\mathrm{...}+{\cos }^2\frac{5\pi }{6}+{\cos }^2\pi \\ =\left({\cos }^2\frac{\pi }{6}+{\cos }^2\frac{2\pi }{6}\right)+\left({\cos }^2\frac{4\pi }{6}+{\cos }^2\frac{5\pi }{6}\right)+\left({\cos }^2\frac{\pi }{2}+{\cos }^2\pi \right) \\ =\left({\cos }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}\right)+\left({\cos }^2\frac{2\pi }{6}+{\cos }^2\frac{\pi }{6}\right)+1 \\ =2\left({\cos }^2\frac{\pi }{6}+{\cos }^2\frac{2\pi }{6}\right)+1 \\ =2\left({\cos }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{6}\right)+1=3 \end{gathered}$$

Đáp án A

$$\begin{gathered} E=\sin \frac{\pi }{5}+\sin \frac{2\pi }{5}+\mathrm{...}+\sin \frac{9\pi }{5} \\ =\left(\sin \frac{\pi }{5}+\sin \frac{9\pi }{5}\right)+\left(\sin \frac{2\pi }{5}+\sin \frac{8\pi }{5}\right)+\mathrm{...}+\left(\sin \frac{4\pi }{5}+\sin \frac{6\pi }{5}\right)+\sin \frac{5\pi }{5} \\ =\left[\sin \frac{\pi }{5}+\sin \left(2\pi -\frac{\pi }{5}\right)\right]+\left[\sin \frac{2\pi }{5}+\sin \left(2\pi -\frac{2\pi }{5}\right)\right]+\mathrm{...}+\left[\sin \frac{4\pi }{5}+\sin \left(2\pi -\frac{4\pi }{5}\right)\right]+\sin \pi \\ =\left[\sin \frac{\pi }{5}+\sin \left(-\frac{\pi }{5}\right)\right]+\left[\sin \frac{2\pi }{5}+\sin \left(-\frac{2\pi }{5}\right)\right]+\mathrm{...}+\left[\sin \frac{4\pi }{5}+\sin \left(-\frac{4\pi }{5}\right)\right]+0 \\ =\left(\sin \frac{\pi }{5}-\sin \frac{\pi }{5}\right)+\left(\sin \frac{2\pi }{5}-\sin \frac{2\pi }{5}\right)+\mathrm{...}+\left(\sin \frac{4\pi }{5}-\sin \frac{4\pi }{5}\right) \\ = 0 + 0 + \dots + 0 = 0 \end{gathered}$$

Đáp án A

$$\begin{gathered} {\sin }^{8}x+{\cos }^{8}x={\left({\sin }^{4}x+{\cos }^{4}x\right)}^2-2{\sin }^{4}x.{\cos }^{4}x \\ =\left(1-2{\sin }^{2}x.{\cos }^{2}x\right)-\frac{1}{8}{\sin }^{4}2x \\ ={\left(1-\frac{1}{2}{\sin }^{2}2x.\right)}^2-\frac{1}{8}{\sin }^{4}2x=1-{\sin }^{2}2x+\frac{1}{8}{\sin }^{4}2x \\ =1-\frac{1-\cos 4x}{2}+\frac{1}{8}{\left(\frac{1-\cos 4x}{2}\right)}^2 \\ =1-\frac{1-\cos 4x}{2}+\frac{1}{32}{\left(1-2\cos 4x+\frac{1+\cos 8x}{2}\right)}^2 \\ =\frac{35}{64}+\frac{7}{16}\cos 4x+\frac{1}{64}\cos 8x \\ \Rightarrow a=35,b=7,c=1\Rightarrow a-5b+c=1 \end{gathered}$$

Đáp án C

Ta có:

$$\begin{gathered} 0<\alpha <{90}^0\iff 0<\frac{\alpha }{2}<{45}^0\Rightarrow 0<\sin \frac{\alpha }{2}<\frac{\sqrt{2}}{2} \\ \iff 0<\sqrt{\frac{x-1}{2x}}<\frac{\sqrt{2}}{2}\iff x>0 \\ {\sin }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}=1\Rightarrow \cos \frac{\alpha }{2}=\sqrt{1-{\sin }^2\frac{\alpha }{2}} \\ vì 0<\frac{\alpha }{2}<{45}^0 \\ \iff \cos \frac{\alpha }{2}=\sqrt{\frac{x+1}{2x}}\Rightarrow \tan \frac{\alpha }{2}=\sqrt{\frac{x-1}{x+1}} \\ \tan \alpha =\frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}=\frac{2\sqrt{\frac{x-1}{x+1}}}{1-\frac{x-1}{x+1}}=\sqrt{x^2-1} \\ \Rightarrow \cot \alpha =\frac{1}{\tan \alpha }=\frac{1}{\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}}{x^2-1} \end{gathered}$$

Đáp án C

Ta có:

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

$$\begin{gathered} \sin A+\sin B+\sin C=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2} \\ =2\sin \left(\frac{\pi }{2}-\frac{C}{2}\right)\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2} \\ =2\cos \frac{C}{2}\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2} \\ =2\cos \frac{C}{2}\left(\cos \frac{A-B}{2}+\sin \left(\frac{\pi }{2}-\frac{A+B}{2}\right)\right) \\ =2\cos \frac{C}{2}\left(\cos \frac{A-B}{2}+\cos \frac{A+B}{2}\right) \\ =2\cos \frac{C}{2}.2\cos \frac{A}{2}\cos \frac{B}{2} \\ =4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} \\ \Rightarrow \frac{\cos A+\cos B+\cos C+1}{\sin A+\sin B+\sin C} \\ =\frac{4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}=\tan \frac{C}{2} \\ \Rightarrow \tan \frac{C}{2}=1\iff \frac{C}{2}={45}^0\iff C={90}^0 \end{gathered}$$

$\Rightarrow \Delta ABC$ vuông tại C

Đáp án C

$$\begin{gathered} +)\sin \text{x}{\cos }^{3}x-\cos x{\sin }^{3}x=\mathrm{s}\text{inx}\frac{3\cos x+\cos 3x}{4}-\cos x.\frac{3\sin x-\sin 3x}{4} \\ =\frac{3}{4}\sin \text{x}\cos x+\frac{1}{4}\sin x\cos 3x-\frac{3}{4}\sin \text{x}\cos x+\frac{1}{4}\sin 3x\cos x \\ =\frac{1}{4}\left(\sin x\cos 3x+\sin 3x\cos x\right) \\ =\frac{1}{4}\sin \left(x+3x\right)=\frac{\sin 4x}{4} \end{gathered}$$

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

$$\begin{gathered} +){\cot }^{2}x+{\tan }^{2}x=\frac{{\cos }^{2}x}{{\sin }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}=\frac{{\cos }^{4}x+{\sin }^{4}x}{{\sin }^{2}x{\cos }^{2}x} \\ =\frac{\frac{3+\cos 4x}{4}}{\frac{1}{4}{\sin }^{2}2x}=\frac{3+\cos 4x}{\frac{1}{2}\left(1-\cos 4x\right)}=\frac{2\cos 4x+6}{1-\cos 4x} \end{gathered}$$

Đáp án C

$$\begin{gathered} \sin a-\cos a=\frac{1}{5}\Rightarrow {\left(\sin a-\cos a\right)}^2=\frac{1}{25} \\ \iff {\sin }^{2}a-2\sin a\cos a+{\cos }^{2}a=\frac{1}{25} \\ \iff 1-\sin 2a=\frac{1}{25}\iff \sin 2a=\frac{24}{25} \end{gathered}$$

Ta có:

$$\begin{gathered} {\sin }^{2}2a+{\cos }^{2}2a=1\Rightarrow {\left(\frac{24}{25}\right)}^2+{\cos }^{2}2a=1 \\ \iff {\cos }^{2}2a=\frac{49}{625}\iff \cos 2a=\pm \frac{7}{25} \\ Mà {135}^0<a<{180}^0\iff {270}^0<2a<{360}^0 \\ \Rightarrow \cos 2a>0\Rightarrow \cos 2a=\frac{7}{25} \\ \tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{\frac{24}{25}}{\frac{7}{25}}=\frac{24}{7} \end{gathered}$$

Đáp án A

$$\begin{gathered} \frac{2{\cos }^{2}x-1}{4\tan \left(\frac{\pi }{4}-x\right){\sin }^2\left(\frac{\pi }{4}+x\right)}=\frac{\cos 2x}{4.\frac{\sin \left(\frac{\pi }{4}-x\right)}{\cos \left(\frac{\pi }{4}-x\right)}.\frac{1-\cos \left(\frac{\pi }{2}+2x\right)}{2}} \\ =\frac{\cos 2x}{2.\frac{\sqrt{2}\left(\cos x-\mathrm{s}\text{inx}\right)}{\sqrt{2}\left(\cos x+\mathrm{s}\text{inx}\right)}.\left(1+\mathrm{s}\text{in2x}\right)} \\ =\frac{\cos 2x}{2.\frac{\left(\cos x-\mathrm{s}\text{inx}\right)}{\left(\cos x+\mathrm{s}\text{inx}\right)}.{\left(\sin x+\cos \text{x}\right)}^2} \\ =\frac{\cos 2x}{2\left(\cos x-\mathrm{s}\text{inx}\right)\left(\mathrm{s}\text{inx}+\cos x\right)} \\ =\frac{\cos 2x}{2\left({\cos }^{2}x-{\sin }^{2}x\right)}=\frac{\cos 2x}{2\cos 2x}=\frac{1}{2} \end{gathered}$$

Đáp án B

$$\begin{gathered} B={\sin }^3\frac{a}{3}+3{\sin }^3\frac{a}{3^2}+3^2{\sin }^3\frac{a}{3^3}+\mathrm{...}+3^{n-1}{\sin }^3\frac{a}{3^n} \\ =\frac{3\sin \frac{a}{3}-\sin a}{4}+3.\frac{3\sin \frac{a}{3^2}-\sin \frac{a}{3}}{4}+3^2.\frac{3\sin \frac{a}{3^3}-\sin \frac{a}{3^2}}{4}+\mathrm{...}+3^{n-1}.\frac{3\sin \frac{a}{3^n}-\sin \frac{a}{3^{n-1}}}{4} \\ =\frac{1}{4}.\left(-\sin a+3\sin \frac{a}{3}-3\sin \frac{a}{3}+3^2\sin \frac{a}{3^2}-3^2\sin \frac{a}{3^2}+3^3\sin \frac{a}{3^3}-\mathrm{...}-3^{n-1}\sin \frac{a}{3^{n-1}}+3^n\sin \frac{a}{3^n}\right) \\ =\frac{1}{4}\left(3^n\sin \frac{a}{3^n}-\sin a\right) \\ =\frac{3^{\mathrm{n}}\sin \frac{\mathrm{a}}{3^{\mathrm{n}}}-\mathrm{sina}}{4} \end{gathered}$$