Đề kiểm tra Công thức lượng giác lớp 11 (có lời giải) - Đề 2

Chọn B

Ta có: $\cos (x + \frac{\pi }{4}) - \cos (x - \frac{\pi }{4}) = \, - \,2\sin \left( {\frac{{x + \frac{\pi }{4}\, + \,x - \frac{\pi }{4}}}{2}} \right).\sin \left( {\frac{{x + \frac{\pi }{4}\, - \,x + \frac{\pi }{4}}}{2}} \right)$

$= \, - \,2\sin x.\sin \frac{\pi }{4} = \, - \,\sqrt 2 \sin x$

Chọn C

Ta có: $\cos \frac{{37\pi }}{{12}} = \cos \left( {\frac{\pi }{2} + \frac{{7\pi }}{{12}} + 2\pi } \right) = \cos \left( {\frac{\pi }{2} + \frac{{7\pi }}{{12}}} \right)$

$$= \cos \frac{\pi }{2}\cos \frac{{7\pi }}{{12}} - \sin \frac{\pi }{2}\sin \frac{{7\pi }}{{12}} = - \frac{{\sqrt 6 + \sqrt 2 }}{4}$$

Chọn A

Ấn máy tính được đáp án$A$.

Chọn B

$\begin{array}{l}M = {\cos ^4}{15^{\rm{o}}} - {\sin ^4}{15^{\rm{o}}} = {\left( {{{\cos }^2}{{15}^{\rm{o}}}} \right)^2} - {\left( {{{\sin }^2}{{15}^{\rm{o}}}} \right)^2} = \left( {{{\cos }^2}{{15}^{\rm{o}}} - {{\sin }^2}{{15}^{\rm{o}}}} \right)\left( {{{\cos }^2}{{15}^{\rm{o}}} + {{\sin }^2}{{15}^{\rm{o}}}} \right)\\\,\,\,\,\,\,\, = {\cos ^2}{15^{\rm{o}}} - {\sin ^2}{15^{\rm{o}}} = \cos \left( {{{2.15}^{\rm{o}}}} \right) = \cos {30^{\rm{o}}} = \frac{{\sqrt 3 }}{2}.\end{array}$

Chọn B

Sử dụng máy tính dễ dàng có được đáp án $B$.

Chọn B

Ta có $\sin \alpha = \frac{{\sqrt 3 }}{2}$$\Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - \frac{3}{4} = \frac{1}{4} \Leftrightarrow \cos \alpha = \pm \frac{1}{2}$.

Từ $\frac{\pi }{2} < \alpha < \pi \Rightarrow \cos \alpha < 0$ nên $\cos \alpha = - \frac{1}{2}$.

Do đó $P = \cos \left( {2\alpha - \frac{\pi }{3}} \right) = \cos 2\alpha \cos \frac{\pi }{3} + \sin 2\alpha \sin \frac{\pi }{3}$

$= \left( {2{{\cos }^2}\alpha - 1} \right).\frac{1}{2} + 2\sin \alpha \cos \alpha .\frac{{\sqrt 3 }}{2} = {\cos ^2}\alpha - \frac{1}{2} + \sqrt 3 \sin \alpha \cos \alpha = - 1$.

Chọn C

$$\begin{array}{l}\frac{{si{n^2}2\alpha + 4si{n^4}\alpha - 4{{\sin }^2}\alpha .{{\cos }^2}\alpha }}{{4 - {{\sin }^2}2\alpha - 4{{\sin }^2}\alpha }} = \frac{{4si{n^4}\alpha }}{{4(1 - {{\sin }^2}\alpha ) - 4{{\sin }^2}\alpha .{{\cos }^2}\alpha }}\\ = \frac{{si{n^4}\alpha }}{{{{\cos }^2}\alpha (1 - {{\sin }^2}\alpha )}} = \frac{{si{n^4}\alpha }}{{{{\cos }^4}\alpha }} = {\tan ^4}a \Rightarrow {\rm{BT}} = {\tan ^4}\left( {\frac{\pi }{6}} \right) = \frac{1}{9}.\end{array}$$

Chọn D

Ta có :

$A={\sin }^{2}(a+b)-{\sin }^{2}a-{\sin }^{2}b$ $={\sin }^{2}(a+b)-\frac{1-\cos 2a}{2}-\frac{1-\cos 2b}{2}$

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

$= \cos \left( {a + b} \right)\left[ {\cos \left( {a - b} \right) - \cos \left( {a + b} \right)} \right]$$= 2\sin a\sin b\cos \left( {a + b} \right)$.

Chọn C

$\cot x = \frac{3}{4} \Rightarrow \tan x = \frac{4}{3};\,\,\cot y = \frac{1}{7} \Rightarrow \tan y = 7$. $t = \tan \frac{x}{2} = 2$.

Chọn D

Ta có $\left\{ \begin{array}{l}\tan a + \tan b = 2\\\tan \left( {a + b} \right) = 4\end{array} \right.$.

từ $\tan \left( {a + b} \right) = 4 \Leftrightarrow \frac{{\tan a + \tan b}}{{1 - \tan a.\tan b}} = 4 \Rightarrow 2 = 4 - 4\tan a.\tan b \Rightarrow \tan a.\tan b = \frac{1}{2}$.

$\Rightarrow \tan a,\,\tan b$ theo thứ tự là nghiệm của phương trình ${X^2} - 2X + \frac{1}{2} = 0$.

$\Rightarrow \tan a = 1 - \frac{{\sqrt 2 }}{2},\,\tan b = 1 + \frac{{\sqrt 2 }}{2}$ hoặc ngược lại.

Chọn C

A. $\sin x.{\rm{co}}{{\rm{s}}^3}x - {\rm{cos }}x{\sin ^3}x = \sin x.{\rm{cos}}x({\rm{co}}{{\rm{s}}^2}x - {\sin ^2}x) = \frac{1}{2}\sin 2x.{\rm{cos2}}x = \frac{{\sin 4x}}{4}$.

B. ${\sin ^4}x + {\rm{co}}{{\rm{s}}^4}x = 1 - 2{\sin ^2}x.{\rm{co}}{{\rm{s}}^2}x = 1 - \frac{1}{2}{\sin ^2}2x = 1 - \frac{1}{2}(\frac{{1 - {\rm{cos}}4x}}{2}) = \frac{{3 + {\rm{cos}}4x}}{4}$.

C. $\frac{{1 + \sin x}}{{{\rm{cos }}x}} = \frac{{1 - {\rm{cos(}}\frac{\pi }{2}{\rm{ + x)}}}}{{\sin {\rm{(}}\frac{\pi }{2}{\rm{ + x)}}}} = \frac{{2{{\sin }^2}(\frac{\pi }{4}{\rm{ + }}\frac{{\rm{x}}}{2})}}{{2\sin {\rm{(}}\frac{\pi }{2}{\rm{ + x)cos}}(\frac{\pi }{4}{\rm{ + }}\frac{{\rm{x}}}{2})}} = \tan (\frac{\pi }{4} + \frac{x}{2})$.

D. ${\cot ^2}x + {\tan ^2}x = \frac{{{\rm{co}}{{\rm{s}}^2}x}}{{{{\sin }^2}x}} + \frac{{{{\sin }^2}x}}{{{\rm{co}}{{\rm{s}}^2}x}} = \frac{{{\rm{co}}{{\rm{s}}^4}x + {{\sin }^4}x}}{{{\rm{co}}{{\rm{s}}^2}x.{{\sin }^2}x}} = \frac{{\frac{{3 + {\rm{cos}}4x}}{4}}}{{\frac{{1 - {\rm{cos}}4x}}{8}}} = \frac{{2\cos 4x + 6}}{{1 - {\rm{cos}}4x}}$.

Chọn C

$\sin \left( {x + \frac{\pi }{6}} \right).\sin \left( {x - \frac{\pi }{6}} \right).\,{\rm{ }}\cos 2x = \frac{1}{2}\left( {\cos \frac{\pi }{3} - \cos 2x} \right){\rm{ }}\cos 2x$.

$= \frac{1}{4}\cos 2x - {\rm{ }}\frac{1}{2}{\cos ^2}2x{\rm{ }} = {\rm{ }}\frac{1}{4}\cos 2x - \frac{1}{4}\cos 4x - \frac{1}{4}.$

a) Đúng

b) Đúng

c) Sai

d) Sai

a) Vì $0 < x < \frac{\pi }{2}$ nên $\cos x > 0$.

Ta có: $\sin x = \frac{1}{{\sqrt 3 }} \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x} = \sqrt {1 - \frac{1}{3}} = \frac{{\sqrt 6 }}{3}$.

$\cos \left( {x + \frac{\pi }{3}} \right) = \cos x\cos \frac{\pi }{3} - \sin x\sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3} \cdot \frac{1}{2} - \frac{1}{{\sqrt 3 }} \cdot \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 6 - 3}}{6}{\rm{. }}$

a) Sai

b) Đúng

c) Sai

d) Đúng

Vì $\pi < x < \frac{{3\pi }}{2}$ nên $\sin x < 0$.

Ta có: $\cos x = - \frac{{12}}{{13}} \Rightarrow \sin x = - \sqrt {1 - {{\cos }^2}x} = - \sqrt {1 - {{\left( { - \frac{{12}}{{13}}} \right)}^2}} = - \frac{5}{{13}}$.

$\sin \left( {\frac{\pi }{3} - x} \right) = \sin \frac{\pi }{3}\cos x - \cos \frac{\pi }{3}\sin x = \frac{{\sqrt 3 }}{2}\left( { - \frac{{12}}{{13}}} \right) - \frac{1}{2}\left( { - \frac{5}{{13}}} \right) = \frac{{5 - 12\sqrt 3 }}{{26}}$

a) Đúng

b) Đúng

c) Đúng

d) Sai

Vì $\frac{\pi }{2} < \alpha < \frac{{3\pi }}{2}$ nên $\cos \alpha < 0$. Ta có hệ: $\left\{ {\begin{array}{llllllllllllllllllll}{{{\sin }^2}\alpha + {{\cos }^2}\alpha = 1}\\{2\sin \alpha \cos \alpha = - \frac{4}{5}}\end{array}} \right.$

$\Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{\frac{4}{{25 c o s{ ^2}\alpha }} + c o s{ ^2}\alpha = 1}\\{ s i n \alpha = - \frac{2}{{5 c o s \alpha }}}\end{array} \Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{25{{\cos }^4}\alpha - 25{{\cos }^2}\alpha + 4 = 0}\\{\sin \alpha = - \frac{2}{{5\cos \alpha }}}\end{array}} \right.} \right.$

$\Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\cos { ^2}\alpha = \frac{4}{5}\\\cos { ^2}\alpha = \frac{1}{5}\end{array} \right.\\\sin \alpha = - \frac{2}{{5 \cos \alpha }}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\cos \alpha = \frac{{ - 2}}{{\sqrt 5 }}\\\cos \alpha = \frac{{ - 1}}{{\sqrt 5 }}\end{array} \right.\\\sin \alpha = - \frac{2}{{5 \cos \alpha }}\end{array} \right. \Rightarrow \left[ {\begin{array}{llllllllllllllllllll}{\cos \alpha = \frac{{ - 2}}{{\sqrt 5 }},\sin \alpha = \frac{1}{{\sqrt 5 }}}\\{\cos \alpha = \frac{{ - 1}}{{\sqrt 5 }},\sin \alpha = \frac{2}{{\sqrt 5 }}}\end{array}} \right.$

a) Đúng

b) Đúng

c) Sai

d) Sai

$\sin \alpha = \frac{2}{3},\frac{\pi }{2} < \alpha < \pi$

Ta có: ${\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{2}{3}} \right)^2} = \frac{5}{9} \Rightarrow \cos \alpha = \pm \frac{{\sqrt 5 }}{3}$

Vì $\frac{\pi }{2} < \alpha < \pi$ nên $\cos \alpha = - \frac{{\sqrt 5 }}{3}$

$\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{{2\sqrt 5 }}{5}$.

$\cos \left( {\frac{\pi }{3} + \alpha } \right) = \cos \frac{\pi }{3}\cos \alpha - \sin \frac{\pi }{3}\sin \alpha = \frac{1}{2} \cdot \left( {\frac{{ - \sqrt 5 }}{3}} \right) - \frac{{\sqrt 3 }}{2} \cdot \frac{2}{3} = \frac{{ - \sqrt 5 - 2\sqrt 3 }}{6}$

$\cos (\frac{π}{4}-α)=\cos \frac{π}{4}\cos α+\sin \frac{π}{4}\sin α=\frac{\sqrt{2}}{2}\cdot (\frac{-\sqrt{5}}{3})+\frac{\sqrt{2}}{2}\cdot \frac{2}{3}=\frac{-\sqrt{10}+2\sqrt{2}}{6}$

Ta có: ${\sin ^2}a = 1 - {\cos ^2}a = \frac{8}{9};{\sin ^2}b = 1 - {\cos ^2}b = \frac{{15}}{{16}}$.

$\begin{array}{llllllllllllllllllll}P&{ = (\cos a\cos b - \sin a\sin b)(\cos a\cos b + \sin a\sin b)}\\{}&{ = {{(\cos a\cos b)}^2} - {{(\sin a\sin b)}^2} = {{\cos }^2}a{{\cos }^2}b - {{\sin }^2}a{{\sin }^2}b}\\{}&{ = \frac{1}{9} \cdot \frac{1}{{16}} - \frac{8}{9} \cdot \frac{{15}}{{16}} = - \frac{{119}}{{144}}}\end{array}$

Ta có: ${\rm{cos}}\alpha = \frac{3}{5}$ và $0 \le \alpha \le {45^0} \Rightarrow \sin \alpha = \frac{4}{5} \Rightarrow \sin 2\alpha = 2\sin \alpha .{\rm{cos}}\alpha {\rm{ = }}\frac{{24}}{{25}}.$

Khi đó $d = \frac{{{v_0}^2\sin 2\alpha }}{g} = \frac{{{{15}^2}.\frac{{24}}{{25}}}}{{10}} = \frac{{108}}{5}$.

Vậy $d = \frac{{108}}{5}\,{\rm{cm}}$.

Ta có:

$\begin{array}{l}\tan \alpha = \tan (\widehat {BAD} - \widehat {CAD})\\ = \frac{{\tan \widehat {BAD} - \tan \widehat {CAD}}}{{1 + \tan \widehat {BAD}\tan \widehat {CAD}}} = \frac{{\frac{{15}}{{12}} - \frac{9}{{12}}}}{{1 + \frac{{15}}{{12}} \cdot \frac{9}{{12}}}} = \frac{8}{{31}}.\end{array}$

Vì vậy $α\approx 14,{47}^{^{\circ}}$

$\begin{array}{llllllllllllllllllll}B&{ = \frac{{\sin 2x + 2\sin 3x + \sin 4x}}{{\cos 3x + 2\cos 4x + \cos 5x}} = \frac{{2\sin 3x\cos x + 2\sin 3x}}{{2\cos 4x\cos x + 2\cos 4x}} = \frac{{2\sin 3x(\cos x + 1)}}{{2\cos 4x(\cos x + 1)}} = \frac{{\sin 3x}}{{\cos 4x}}}\\{}&{}\end{array}$

Gọi $\widehat {BOA} = \alpha$ $\left( {0^\circ < \alpha < 90^\circ } \right)$

Vẽ $BB' \bot AH$, $CC' \bot AH$$\left( {B' \in AH,\,C' \in AH} \right)$

Khi đó ta có $\sin \alpha = \frac{{27}}{{60}} = \frac{9}{{20}} \Rightarrow {\rm{cos}}\alpha = \frac{{\sqrt {319} }}{{20}}$.

Lại có $\widehat {COA} = 2\alpha \Rightarrow \sin \widehat {COA} = \sin 2\alpha = 2\sin \alpha .{\rm{cos}}\alpha = 2.\frac{9}{{20}}.\frac{{\sqrt {319} }}{{20}} = \frac{{9\sqrt {319} }}{{200}}$.

Trong tam giác $COC'$ ta có $\frac{{CC'}}{{OC}} = \sin 2\alpha \Rightarrow CC' = OC.\sin 2\alpha = 60.\frac{{9\sqrt {319} }}{{200}} \approx 48.2$.

Vậy khoảng cách từ $C$đến $AH$khoảng $48{,}2\,{\rm{cm}}$.