Đề kiểm tra Công thức lượng giác lớp 11 (có lời giải) - Đề 1

Chọn C

Có: $\sin {105^0} = \sin ({60^0} + {45^0})$$=$$\sin {60^0}.\cos {45^0} + \cos {60^0}.\sin {45^0}$.

$\Rightarrow$$\sin {105^0} = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 2 }}{2} + \frac{1}{2}.\frac{{\sqrt 2 }}{2}$$=$$\frac{{\sqrt 6 + \sqrt 2 }}{4}$.

Chọn A

Cách 1:$\tan {105^0}$$=$$\frac{{\sin {{105}^0}}}{{\cos {{105}^0}}}$$=$$\frac{{\frac{{\sqrt 6 + \sqrt 2 }}{4}}}{{ - \frac{{\sqrt 6 - \sqrt 2 }}{4}}}$$=$$- \frac{{\sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }}$$=$$- (2 + \sqrt 3 )$.

Cách 2:$\tan {105^0}$$=$$\tan ({45^0} + {60^0})$$=$$\frac{{\tan {{45}^0} + \tan {{60}^0}}}{{1 - \tan {{45}^0}\tan {{60}^0}}}$$=$$\frac{{1 + \sqrt 3 }}{{1 - \sqrt 3 }}$$=$$- (2 + \sqrt 3 )$.

Chọn B

Ta có: $\cos {54^0}\cos {4^0} - \cos {36^0}\cos {86^0} = \,\,\cos {54^0}\cos {4^0} - \sin {54^0}\sin {4^0} = \,\,\cos \left( {{{54}^0}\, + \,\,{4^0}} \right)\,\, = \,\,\cos {58^0}$

Chọn C

$\cos \frac{{37\pi }}{{12}}$$= \cos \left( {2\pi + \pi + \frac{\pi }{{12}}} \right)$$= \cos \left( {\pi + \frac{\pi }{{12}}} \right)$$= - \cos \left( {\frac{\pi }{{12}}} \right)$$= - \cos \left( {\frac{\pi }{3} - \frac{\pi }{4}} \right)$$= - \left( {\cos \frac{\pi }{3}.\cos \frac{\pi }{4} + \sin \frac{\pi }{3}.sin\frac{\pi }{4}} \right)$$= - \frac{{\sqrt 6 + \sqrt 2 }}{4}$.

Chọn B

Có $\cos 2a = 2{\cos ^2}a - 1$ nên Chọn B sai.

Chọn D

Ta có :

$$\cos \left( {a + b} \right).\cos \left( {a - b} \right) = \frac{1}{2}\left( {\cos 2a + \cos 2b} \right) = {\cos ^2}a + {\cos ^2}b - 1 = {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{4}} \right)^2} - 1 = - \frac{{119}}{{144}}$$

Chọn C

Ta có $a,\,b$ đều là các góc nhọn và dương.

$\sin a = \frac{8}{{17}} \Rightarrow \cos a = \sqrt {1 - \frac{{64}}{{289}}} = \frac{{15}}{{17}}$.

$\tan b = \frac{5}{{12}} \Rightarrow \cos b = \frac{1}{{\sqrt {1 + \frac{{25}}{{144}}} }} = \frac{{12}}{{13}} \Rightarrow \sin b = \tan b.\cos b = \frac{5}{{13}}$.

$\Rightarrow \sin \left( {a - b} \right) = \frac{8}{{17}}.\frac{{12}}{{13}} - \frac{{15}}{{17}}.\frac{5}{{13}} = \frac{{21}}{{221}}$.

Chọn A

$\tan x = 0.5 = \frac{1}{2},\,\sin y = \frac{3}{5}\,\,\left( {0 < y < {{90}^0}} \right) \Rightarrow \cos y = \frac{4}{5} \Rightarrow \tan y = \frac{3}{4}$.

$\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 - \tan x.\tan y}} = \frac{{\frac{1}{2} + \frac{3}{4}}}{{1 - \frac{1}{2}.\frac{3}{4}}} = 2$.

Chọn B

$$\begin{array}{l}\frac{{3 - 4\cos 2\alpha + cos4\alpha }}{{3 + 4\cos 2\alpha + cos4\alpha }} = \frac{{3 - 4\left( {1 - 2{{\sin }^2}\alpha } \right) + 2{{\left( {1 - 2si{n^2}\alpha } \right)}^2} - 1}}{{3 + 4\left( {2{{\cos }^2}\alpha - 1} \right) + 2{{\left( {2co{s^2}\alpha - 1} \right)}^2} - 1}}\\ = \frac{{8{{\sin }^2}a - 8{{\sin }^2}\alpha + 8{{\sin }^4}\alpha }}{{8{{\cos }^2}a - 8{{\cos }^2}\alpha + 8{{\cos }^4}\alpha }} = {\tan ^4}\alpha .\end{array}$$

Chọn D

Đặt $t = \tan \frac{x}{2} = \frac{1}{2}$ nên $\sin x = \frac{{2t}}{{1 + {t^2}}} = \frac{{2\frac{1}{2}}}{{1 + \frac{1}{4}}} = \frac{4}{5}$, $\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}} = \frac{{1 - \frac{1}{4}}}{{1 + \frac{1}{4}}} = \frac{3}{5}$.

Vậy $\frac{{\sin x}}{{2 - 3\cos x}} = \frac{{\frac{4}{5}}}{{2 - \frac{9}{5}}} = 4$.

Chọn C

$\cos 10^\circ .\cos 30^\circ .\cos 50^\circ .\cos 70^\circ = \cos 10^\circ .\cos 30^\circ .\frac{1}{2}\left( {\cos {{120}^{\rm{o}}} + \cos {{20}^{\rm{o}}}} \right)$$= \frac{{\sqrt 3 }}{4}\left( { - \frac{{\cos 10^\circ }}{2} + \frac{{\cos 30^\circ + \cos 10^\circ }}{2}} \right)$$= \frac{{\sqrt 3 }}{4}.\frac{1}{4} = \frac{{\sqrt 3 }}{{16}}$.

Chọn D

Vì $A,\,\,B,\,\,C$ là các góc của tam giác $ABC$ nên $A + B + C = {180^o} \Rightarrow C = {180^o} - \left( {A + B} \right).$

Do đó $C$ và $\left( {A + B} \right)$ là 2 góc bù nhau.

$$\Rightarrow \sin C = \sin \left( {A + B} \right);\,\,\cos C = - \cos \left( {a + b} \right);\,\,\tan C = - \tan \left( {A + B} \right);\,\,\cot C = \cot \left( {A + B} \right).$$

a) Đúng

b) Sai

c) Sai

d) Đúng

Vì $\pi < \alpha < \frac{{3\pi }}{2}$ nên $\sin \alpha < 0,\cos \alpha < 0$.

Ta có: $\cos 2\alpha = - \frac{1}{4} \Rightarrow 1 - 2{\sin ^2}\alpha = - \frac{1}{4} \Rightarrow {\sin ^2}\alpha = \frac{5}{8} \Rightarrow \sin \alpha = - \frac{{\sqrt {10} }}{4}$;

$\begin{array}{l}\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - \frac{{10}}{{16}}} = - \frac{{\sqrt 6 }}{4};\\\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\sqrt {15} }}{3}\\\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{{\sqrt {15} }}{5}\end{array}$

a) Sai

b) Sai

c) Đúng

d) Đúng

a) $\cos 3x + \cos x = 2\cos 2x \cdot \cos x$

b) $\sin 3x + \sin 2x = 2\sin \frac{{5x}}{2}\cos \frac{x}{2}$;

c) $\cos 4x - \cos x = - 2\sin \frac{{5x}}{2}\sin \frac{{3x}}{2}$

d) $\sin 5x - \sin x = 2\cos 3x\sin 2x$

a) Đúng

b) Đúng

c) Sai

d) Sai

$\begin{array}{l}{\rm{ V\`i }}\frac{\pi }{2} < \alpha < \pi \Rightarrow \cos \alpha < 0.{\rm{ }}\\{\rm{Ta c\'o }}\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \frac{4}{5}\\ \Rightarrow \tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{3}{4}{\rm{. }}\end{array}$

Ta có: $\tan \left( {\alpha + \frac{\pi }{3}} \right) = \frac{{\tan \alpha + \tan \frac{\pi }{3}}}{{1 - \tan \alpha \tan \frac{\pi }{3}}} = \frac{{\tan \alpha + \sqrt 3 }}{{1 - \sqrt 3 \tan \alpha }}$.

Suy ra:$\tan \left( {\alpha + \frac{\pi }{3}} \right) = \frac{{ - \frac{3}{4} + \sqrt 3 }}{{1 - \sqrt 3 \left( { - \frac{3}{4}} \right)}} = \frac{{ - 3 + 4\sqrt 3 }}{{4 + 3\sqrt 3 }} = \frac{{48 - 25\sqrt 3 }}{{11}}$.

a) Sai

b) Sai

c) Đúng

d) Đúng

$\begin{array}{l}\frac{\pi }{2} < x < \pi \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \sin \frac{x}{2} > 0.\\\sin \frac{x}{2} = \sqrt {\frac{{1 - \cos x}}{2}} = \sqrt {\frac{{1 - \frac{1}{5}}}{2}} = \frac{{\sqrt {10} }}{5}\end{array}$.

$\begin{array}{l}\frac{\pi }{2} < x < \pi \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \cos \frac{x}{2} > 0.\\\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} = \sqrt {\frac{{1 + \frac{1}{5}}}{2}} = \frac{{\sqrt {15} }}{5}.\\\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\frac{{\sqrt {10} }}{5}}}{{\frac{{\sqrt {15} }}{5}}} = \frac{{\sqrt 6 }}{3}.\\\cot \frac{x}{2} = \frac{{\cos \frac{x}{2}}}{{\sin \frac{x}{2}}} = \frac{{\frac{{\sqrt {15} }}{5}}}{{\frac{{\sqrt {10} }}{5}}} = \frac{{\sqrt 6 }}{2}.\end{array}$

Cho $\sin x = \frac{1}{5},\frac{\pi }{2} < x < \pi$. Tính $\cot 2x$.

$\begin{array}{l}\frac{\pi }{2} < x < \pi \Rightarrow \frac{{2\pi }}{2} < 2x < 2\pi \Rightarrow \pi < 2x < 2\pi \Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{\cos 2x > 0}\\{\tan 2x < 0}\end{array}} \right.\\\cos 2x = 1 - 2{\sin ^2}x = 1 - 2 \cdot \frac{1}{{25}} = \frac{{23}}{{25}}\\\frac{\pi }{2} < x < \pi \Rightarrow \cos x < 0\\\sin x = \frac{1}{5} \Rightarrow \cos x = - \sqrt {1 - {{\sin }^2}x} = - \frac{{2\sqrt 6 }}{5}.\\\sin 2x = 2\sin x \cdot \cos x = 2 \cdot \frac{1}{5} \cdot \left( { - \frac{{2\sqrt 6 }}{5}} \right) = - \frac{{4\sqrt 6 }}{5}\\\cot 2x = \frac{{\cos 2x}}{{\sin 2x}} = \frac{{\frac{{23}}{{25}}}}{{ - \frac{{4\sqrt 6 }}{5}}} = - \frac{{23\sqrt 6 }}{{120}}\end{array}$

Do $\frac{\pi }{2} < \alpha ,\beta < \pi \Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{\cos \alpha < 0}\\{\sin \beta > 0}\end{array}} \right.$.

Ta có $\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3};\sin \beta = \sqrt {1 - {{\cos }^2}\beta } = \sqrt {1 - \frac{4}{9}} = \frac{{\sqrt 5 }}{3}$.

Suy ra $\sin (\alpha + \beta ) = \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) + \left( { - \frac{{2\sqrt 2 }}{3}} \right) \cdot \frac{{\sqrt 5 }}{3} = - \frac{{2 + 2\sqrt {10} }}{9}$.

Vậy $\sin (\alpha + \beta ) = - \frac{{2 + 2\sqrt {10} }}{9}$.

Ta có: $C = \tan 3x - \tan x - \frac{{2\sin x}}{{\cos 3x}} = \frac{{\sin 3x}}{{\cos 3x}} - \frac{{\sin x}}{{\cos x}} - \frac{{2\sin x}}{{\cos 3x}}$

$\begin{array}{l} = \frac{{\sin 3x\cos x - \sin x\cos 3x}}{{\cos 3x\cos x}} - \frac{{2\sin x}}{{\cos 3x}} = \frac{{\sin 2x}}{{\cos 3x\cos x}} - \frac{{2\sin x}}{{\cos 3x}}\\ = \frac{{2\sin x\cos x}}{{\cos 3x\cos x}} - \frac{{2\sin x}}{{\cos 3x}} = \frac{{2\sin x}}{{\cos 3x}} - \frac{{2\sin x}}{{\cos 3x}} = 0.\end{array}$

Kẻ $AM \bot CK,{\rm{ }}BN \bot CK$(hình vẽ) ta có:

$$\begin{array}{llllllllllllllllllll}{{\rm{BN = AM = HK = 20 }}\left( {\rm{m}} \right){\rm{;}}}\\{{\rm{CN = CK -- NK = CK -- BH = 32 -- 24 = 8 }}\left( {\rm{m}} \right){\rm{;}}}\\{{\rm{MN = AB = BH -- AH = 24 -- 6 = 18 }}\left( {\rm{m}} \right){\rm{;}}}\\{{\rm{CM = CN + MN = 8 + 18 = 26 }}\left( {\rm{m}} \right){\rm{.}}}\end{array}$$

Đặt $\widehat {BCN} = \alpha ,\widehat {ACM} = \beta$.

Xét ${\rm{\Delta BCN}}$ vuông tại ${\rm{N}}$ có: ${\rm{tan}}\alpha = \frac{{BN}}{{CN}} = \frac{{20}}{8} = \frac{5}{2}$;

Xét ${\rm{\Delta }}ACM$ vuông tại M có: ${\rm{tan}}\beta = \frac{{AM}}{{CM}} = \frac{{20}}{{26}} = \frac{{10}}{{13}}$;

Ta có: ${\rm{tan}}\widehat {ACB} = {\rm{tan}}\left( {\widehat {BCN} - \widehat {ACM}} \right) = {\rm{tan}}\left( {\alpha - \beta } \right)$

$\Rightarrow {\rm{tan}}\widehat {ACB} = \frac{{{\rm{tan}}\alpha - {\rm{tan}}\beta }}{{1 + {\rm{tan}}\alpha {\rm{tan}}\beta }} = \frac{{\frac{5}{2} - \frac{{10}}{{13}}}}{{1 + \frac{5}{2} \cdot \frac{{10}}{{13}}}} = \frac{{45}}{{76}}$.

Xét tam giác ABC vuông tại B có: ${\rm{tan}}\widehat {BAC} = \frac{3}{4}$

Ta lại có: $\widehat {BAD} = \widehat {BAC} + \widehat {CAD}$

$\Rightarrow {\rm{tan}}\widehat {{\rm{BAD}}} = {\rm{tan}}\left( {\widehat {{\rm{BAC}}} + \widehat {{\rm{CAD}}}} \right) = {\rm{tan}}\left( {\widehat {{\rm{BAC}}} + {{30}^ \circ }} \right) = \frac{{{\rm{tan}}\widehat {{\rm{BAC}}} + {\rm{tan}}{{30}^ \circ }}}{{1 - {\rm{tan}}\widehat {{\rm{BAC}}} \cdot {\rm{tan}}{{30}^ \circ }}}$$= \frac{{\frac{3}{4} + \frac{{\sqrt 3 }}{3}}}{{1 - \frac{3}{4} \cdot \frac{{\sqrt 3 }}{3}}} = \frac{{48 + 25\sqrt 3 }}{{39}} \approx 2{,}34.$

Xét tam giác ABD vuông tại B có:

$\begin{array}{rrrrrrrrrrrrrrrrrrrr}{}&{{\rm{tan}}\widehat {BAD} = \frac{{BD}}{{AB}} \Rightarrow BD = {\rm{tan}}\widehat {BAD} \cdot AB = 2{,}34.4 \approx 9{,}36.}\\{}&{\; \Rightarrow CD = BD - BC \approx 9{,}36 - 3 = 6{,}36.}\end{array}$

Tại $\alpha = \frac{\pi }{2}$ thì ${\rm{H}}$ trùng ${\rm{I}},{\rm{M}}$ trùng ${\rm{O}}$ nên ${\rm{MH}} = {\rm{OI}}$ do đó ${\rm{OM}} = {\rm{IH}}$.

Xét tam giác AHI vuông tại H có: ${\rm{IH}} = {\rm{cos}}a.{\rm{IA}} = 8{\rm{cos}}\alpha$.