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:

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