Concept and properties of trigonometric functions Define sine, cosine, and tangent via the unit circle, graph them, understand their periodicity, monotonicity, parity, and extrema, and the effect of parameters in y=Asin(wx+phi).
Practice questions 1 已知sin ( α + π 12 ) = 5 5 \sin (α+ \dfrac {π}{12} )= \dfrac { \sqrt {5}}{5} sin ( α + 12 π ) = 5 5 sin ( π 3 − 2 α ) = ( ) \sin ( \dfrac {π}{3} -2α)= (\:\:\:\:) sin ( 3 π − 2 α ) = ( ) 4 5 \dfrac {4}{5} 5 4 2 函数y = 2 2 sin x − 1 y=\dfrac{2}{2\sin x-1} y = 2 sin x − 1 2 ( ) (\quad) ( ) [ − 2 3 , 2 ] \left[-\dfrac{2}{3},2\right] [ − 3 2 , 2 ] ( − ∞ , − 2 3 ] ∪ [ 2 , + ∞ ) \left(-∞,-\dfrac{2}{3}\right]∪[2,+∞) ( − ∞ , − 3 2 ] ∪ [ 2 , + ∞ ) 3 下列命题为真命题的是( ) (\quad) ( ) ∃ x 0 ∈ R \text{∃} x_{0}\in R ∃ x 0 ∈ R x 0 2 + 4 x 0 + 6 ⩽ 0 x_{0}^{2}\text{+}4x_{0}+6\leqslant 0 x 0 2 + 4 x 0 + 6 ⩽ 0 y = tan x y=\tan x y = tan x R R R 4 已知α是锐角，且cosα=0.9794，下列各值中，与α最接近的是（ ） A. 73°33′ B. 73°27′ C. 11°37′ D. 11°45′ 5 函数f ( x ) = sin x − cos x f(x)=\sin x-\cos x f ( x ) = sin x − cos x f ( x ) f(x) f ( x ) ( ) (\quad) ( ) 2 2 2 2 \sqrt{{2}} 2 0 0 0 − 2 -2 − 2 6 已知点P ( − 1 , 3 tan 5 π 6 ) P\left(-1,3\tan\dfrac{5\pi}{6}\right) P ( − 1 , 3 tan 6 5 π ) θ \theta θ cos θ \cos\theta cos θ ( ) (\quad) ( ) 1 2 \dfrac{1}{2} 2 1 3 2 \dfrac{\sqrt{3}}{2} 2 3 7 已知函数f ( x ) = cos ( ω x + φ ) ( ω > 1 , ω ∈ Z , ∣ φ ∣ < π 2 ) f(x)=\cos(ωx+φ)(ω>1,ω\in Z,|φ|< \dfrac {π}{2}) f ( x ) = cos ( ω x + φ ) ( ω > 1 , ω ∈ Z , ∣ φ ∣ < 2 π ) [ π 3 , 3 π 4 ] [\dfrac{π}{3},\dfrac{3π}{4}] [ 3 π , 4 3 π ] ( π 12 , 0 ) (\dfrac{π}{12},0) ( 12 π , 0 ) 8 选修4 − 4 4-4 4 − 4 x O y xOy x O y C 1 C_{1} C 1 { x = 3 cos α , y = sin α ( α \begin{cases}x= \sqrt{3}\cos α, \\ y=\sin α\end{cases}(α { x = 3 cos α , y = sin α ( α ) . ). ) . 9 已知 α ∈ ( 3 2 π , 2 π ) \alpha\in(\dfrac{3}{2}\pi,2\pi) α ∈ ( 2 3 π , 2 π ) α \alpha α P ( sin 1 , − cos 1 ) P(\sin 1,-\cos 1) P ( sin 1 , − cos 1 ) α = ( ) \alpha=(\quad) α = ( ) π + 2 \pi+2 π + 2 5 2 π − 2 \dfrac{5}{2}\pi-2 2 5 π − 2 10 函数f（x）=sin|ax+1|的图象恒过定点___，若函数y=f（x）的图象的对称轴为x=1，则非零实数a的值为___． 11 已知sin θ + cos θ = 1 5 , θ ∈ ( 0, π ) \sin\theta{+}\cos\theta{=}\dfrac{1}{5}{,}\theta{∈}(0{,}\pi) sin θ + cos θ = 5 1 , θ ∈ ( 0 , π ) tan θ \tan\theta tan θ 1 − 2 sin θ cos θ cos 2 θ − sin 2 θ \dfrac{1-2\sin \theta \cos \theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta } cos 2 θ − sin 2 θ 1 − 2 sin θ cos θ 12 点A ( sin 2018 0 , tan 117 0 ) A\left(\sin {2018}^{0},\tan {117}^{0}\right) A ( sin 2018 0 , tan 117 0 ) 13 已知sin ( α − π 6 ) = 1 3 \sin(α-\dfrac{π}{6})=\dfrac{1}{3} sin ( α − 6 π ) = 3 1 cos ( π 3 − 2 α ) = ( ) \cos(\dfrac{π}{3}-2α)=(\quad) cos ( 3 π − 2 α ) = ( ) − 7 9 -\dfrac{7}{9} − 9 7 − 2 3 -\dfrac{2}{3} − 3 2 14 已知角α α α sin 2 α < 0 \sin 2α < 0 sin 2 α < 0 sin α − cos α < 0 \sin α-\cos α < 0 sin α − cos α < 0 α α α ( ) () ( ) 15 与-457°角的终边相同的角的集合是（ ） A. {α|α=475°+k•360°，k∈Z} B. α|α=97°+k•360°，k∈Z} C. α|α=263°+k•360°，k∈Z} D. α|α=-263°+k•360°，k∈Z} 16 已知f ( x ) = sin ( 2 x + φ ) f(x)=\sin (2x+φ) f ( x ) = sin ( 2 x + φ ) f ( π 3 ) = 0 f( \dfrac {π}{3})=0 f ( 3 π ) = 0 f ( x ) f(x) f ( x ) ( ) () ( ) x = π 3 x= \dfrac {π}{3} x = 3 π x = 2 π 3 x= \dfrac {2π}{3} x = 3 2 π 17 已知函数f ( x ) = 3 sin ( ω x − π 6 ) + b ( ω > 0 ) f(x)= \sqrt{3}\sin (ωx- \dfrac{π}{6})+b(ω > 0) f ( x ) = 3 sin ( ω x − 6 π ) + b ( ω > 0 ) π 4 \dfrac{π}{4} 4 π x ∈ [ 0 , π 4 ] x∈[0, \dfrac{π}{4}] x ∈ [ 0 , 4 π ] 18 已知函数f ( x ) = sin ω x + 3 cos ω x f(x)=\sin ωx+ \sqrt {3}\cos ωx f ( x ) = sin ω x + 3 cos ω x π π π x ∈ R x∈R x ∈ R ω > 0 ω > 0 ω > 0 ω ω ω f ( θ 2 + π 12 ) = 6 5 f( \dfrac {θ}{2}+ \dfrac {π}{12})= \dfrac {6}{5} f ( 2 θ + 12 π ) = 5 6 19 要得到函数y = sin 2 x y=\sin 2x y = sin 2 x y = sin ( 2 x − π 3 ) y=\sin (2x- \dfrac {π}{3}) y = sin ( 2 x − 3 π ) ( ) () ( ) π 6 \dfrac {π}{6} 6 π π 6 \dfrac {π}{6} 6 π 20 已知y = f ( x + π 6 ) y=f(x+\dfrac{π}{6}) y = f ( x + 6 π ) π π π f ( x ) = f(x)= f ( x ) = . ( .( . ( ) . ). ) . 21 求函数f ( x ) = tan x f(x)=\tan x f ( x ) = tan x x ∈ [ − π 4 , π 3 ] x\in[-\dfrac{\pi}{4},\dfrac{\pi}{3}] x ∈ [ − 4 π , 3 π ] 22 在∆ A B C ∆ABC ∆ A BC a a a b b b c c c A A A B B B C C C b cos C = ( 2 a + c ) x o s ( π − B ) b\cos C=(2a+c)xos(π-B) b cos C = ( 2 a + c ) x os ( π − B ) B B B ; ( 2 ) ;(2) ; ( 2 ) b = 13 b= \sqrt{13} b = 13 23 已知α α α sin ( π + α ) = − 3 5 \sin (π+α)=- \dfrac {3}{5} sin ( π + α ) = − 5 3 tan 2 α \tan 2α tan 2 α 24 (1)若π 2 < α < π \dfrac{π}{2} < α < π 2 π < α < π sin α = 4 5 \sin α= \dfrac{4}{5} sin α = 5 4 sin ( 2 π − α ) tan ( π + α ) cos ( − π + α ) sin ( π 2 − α ) cos ( π 2 + α ) \dfrac{\sin (2π-α)\tan (π+α)\cos (-π+α)}{\sin ( \dfrac{π}{2}-α)\cos ( \dfrac{π}{2}+α)} sin ( 2 π − α ) cos ( 2 π + α ) sin ( 2 π − α ) tan ( π + α ) cos ( − π + α ) 25 已知命题p ： sin x = 1 2 p：\sin x= \dfrac {1}{2} p ： sin x = 2 1 q ： x = π 6 + 2 k π , k ∈ Z q：x= \dfrac {π}{6}+2kπ,k∈Z q ： x = 6 π + 2 kπ , k ∈ Z p p p q q q ( ) () ( ) 26 已知sinα+cosα＞1，则α是第 象限角． 27 将函数y = 2 sin ( 2 x + π 6 ) y\mathrm{\text{=}}2\mathrm{\sin}(2x\mathrm{\text{+}}\dfrac{\pi}{6}) y = 2 sin ( 2 x + 6 π ) 1 4 \dfrac{1}{4} 4 1 ( ) () ( ) 28 把函数y = sin ( 2 x + 4 π 3 ) y=\sin (2x+ \dfrac {4π}{3}) y = sin ( 2 x + 3 4 π ) φ ( φ > 0 ) φ(φ > 0) φ ( φ > 0 ) y y y φ φ φ 29 若α α α sin 2 α \sin\:2α sin 2 α sin α 2 \sin\dfrac{\alpha}{2} sin 2 α tan 2 α \tan\:2α tan 2 α tan α 2 \tan\dfrac{\alpha}{2} tan 2 α ( ) (\quad) ( ) 0 0\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: 0 30 将函数f ( x ) = cos ( 2 x − π 4 ) f(x)=\cos(2x-\dfrac{π}{4}) f ( x ) = cos ( 2 x − 4 π ) π 8 \dfrac{π}{8} 8 π g ( x ) g(x) g ( x ) g ( x ) ( ) g(x)(\quad) g ( x ) ( ) ( 0 , π 4 ) (0,\dfrac{π}{4}) ( 0 , 4 π ) 31 已知函数f ( x ) = sin ( ω x + φ ) ( ω > 0 , ∣ φ ∣ < π 2 ) f\left(x\right)=\sin\left(\omega x+\varphi\right)(\omega>0,\left|\varphi\right|< \dfrac{\pi}{2}) f ( x ) = sin ( ω x + φ ) ( ω > 0 , ∣ φ ∣ < 2 π ) π \pi π 32 已知函数y = cos x y=\cos x y = cos x y = sin ( 2 x + φ ) ( 0 ⩽ φ ⩽ π ) y=\sin (2x+φ)(0\leqslant φ\leqslant π) y = sin ( 2 x + φ ) ( 0 ⩽ φ ⩽ π ) π 3 \dfrac {π}{3} 3 π φ = ( ) φ=() φ = ( ) π 6 \dfrac {π}{6} 6 π 33 关于函数f ( x ) = sin ( 2 x − π 3 ) f(x)=\sin(2x-\dfrac{π}{3}) f ( x ) = sin ( 2 x − 3 π ) ( ) (\quad) ( ) y = sin 2 x y=\sin 2x y = sin 2 x π 6 \dfrac{π}{6} 6 π f ( x ) f(x) f ( x ) 34 若将函数y = cos 2 x y=\cos 2x y = cos 2 x π 12 \dfrac{\pi }{12} 12 π ( ) () ( ) x = k π 2 − π 6 ( k ∈ Z ) x=\dfrac{k\pi }{2}-\dfrac{\pi }{6}\left( k\in Z \right) x = 2 kπ − 6 π ( k ∈ Z ) 35 下列各角中，与60 ° 60° 60° ( ) (\quad) ( ) − 300 ° -300° − 300° − 60 ° -60° − 60° 150 ° 150° 150° 240 ° 240° 240° 36 已知函数f ( x ) = 2 sin ω x cos ω x + 2 3 sin 2 ω x − 3 ( ω > 0 ) f(x)=2\sin ωx\cos ωx+2 \sqrt {3}\sin ^{2}ωx- \sqrt {3}(ω > 0) f ( x ) = 2 sin ω x cos ω x + 2 3 sin 2 ω x − 3 ( ω > 0 ) π π π ω ω ω f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x ) 37 将函数y = sin x y=\sin x y = sin x π 10 \dfrac {\pi }{10} 10 π 2 2 2 ( ( ( ) ) ) ( ) () ( ) y = sin ( 2 x − π 10 ) y=\sin (2x- \dfrac {\pi }{10}) y = sin ( 2 x − 10 π ) 38 已知a < 0 , a\text{ < }0, a < 0 , f ( x ) = a cos x + 1 + sin x + 1 − sin x f\left(x\right)=a\cos x+\sqrt{1+\sin x}+\sqrt{1-\sin x} f ( x ) = a cos x + 1 + sin x + 1 − sin x x ∈ [ − π 2 , π 2 ] . ( 1 ) x{\in}\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].(1) x ∈ [ − 2 π , 2 π ] . ( 1 ) 39 在平面直角坐标系中，角α \alpha α O x Ox O x P ( 1 , m ) ( m < 0 ) P(1,m)(m< 0) P ( 1 , m ) ( m < 0 ) ( ) (\quad) ( ) sin α + cos α \sin\alpha+\cos\alpha sin α + cos α cos α − sin α \cos\alpha-\sin\alpha cos α − sin α 40 下列叙述正确的是 ( ) (\quad) ( ) y = tan x y=\tan x y = tan x R R\:\:\:\:\:\:\:\:\:\:\:\:\:\: R y = x y=\sqrt{x} y = x R R R y = 1 x y=\dfrac{1}{x} y = x 1 41 tan 5 π 6 \tan \dfrac {5π}{6} tan 6 5 π ( ) (\:\:\:\:) ( ) − 1 2 - \dfrac {1}{2} − 2 1 − 3 3 - \dfrac { \sqrt {3}}{3} − 3 3 − 3 2 - \dfrac { \sqrt {3}}{2} − 2 3 42 已知函数g ( x ) = cos x + sin x g(x)=\cos x+\sin x g ( x ) = cos x + sin x h ( x ) = sin ( x + π 2 ) + sin ( x + π ) . h(x)=\sin(x+\dfrac{π}{2})+\sin(x+π). h ( x ) = sin ( x + 2 π ) + sin ( x + π ) . f ( x ) = g ( x − π 6 ) h ( x − π 6 ) f(x)=g(x-\dfrac{π}{6})h(x-\dfrac{π}{6}) f ( x ) = g ( x − 6 π ) h ( x − 6 π ) f ( x ) f(x) f ( x ) 43 已知sin ( α − π ) + cos ( π − α ) sin ( π 2 − α ) + cos ( π 2 + α ) = 5 \dfrac{\sin(α-π)+\cos(π-α)}{\sin(\dfrac{π}{2}-α)+\cos(\dfrac{π}{2}+α)}=5 sin ( 2 π − α ) + cos ( 2 π + α ) sin ( α − π ) + cos ( π − α ) = 5 tan α \tan α tan α ( ) (\quad) ( ) − 2 -2 − 2 2 2 2 44 已知a → = ( 2 λ sin x , sin x + cos x ) \overrightarrow{a}=(2λ\sin x,\sin x+\cos x) a = ( 2 λ sin x , sin x + cos x ) b → = ( 3 cos x , λ ( sin x − cos x ) ) ( λ > 0 ) \overrightarrow{b}=( \sqrt {3}\cos x,λ(\sin x-\cos x))(λ > 0) b = ( 3 cos x , λ ( sin x − cos x )) ( λ > 0 ) 45 设复数z z z z ‾ \overline{z} z 2 z + z ‾ = 3 + i 2z+\overline{z}=3+i 2 z + z = 3 + i ω = sin θ − i cos θ \omega=\sin\theta-i\cos\theta ω = sin θ − i cos θ z − ω z-\omega z − ω O M → . ( 1 ) \overrightarrow{OM}.(1) OM . ( 1 ) 46 如果函数y = 3 cos ( 2 x + φ ) y=3\cos (2x+φ) y = 3 cos ( 2 x + φ ) ( 4 π 3 , 0 ) ( \dfrac {4π}{3},0) ( 3 4 π , 0 ) ∣ φ ∣ |φ| ∣ φ ∣ ( ) () ( ) π 6 \dfrac {π}{6} 6 π π 4 \dfrac {π}{4} 4 π 47 己知数列{ a n } \{a_{n}\} { a n } a 2 a 3 a 4 = − a 7 2 = − 64 a_{2}a_{3}a_{4}=-a_{7}^{2}=-64 a 2 a 3 a 4 = − a 7 2 = − 64 tan ( a 4 a 6 3 ⋅ π ) = ( ) \tan (\dfrac{{{a}_{4}}{{a}_{6}}}{3}\cdot \mathrm{ }\!\!\pi\!\!{ })=() tan ( 3 a 4 a 6 ⋅ π ) = ( ) 48 在△ A B C \triangle ABC △ A BC S S S △ A B C \triangle ABC △ A BC S = 1 2 ( b 2 + c 2 − a 2 ) S= \dfrac {1}{2}(b^{2}+c^{2}-a^{2}) S = 2 1 ( b 2 + c 2 − a 2 ) tan B + tan C − 2 tan B tan C = ( ) \tan B+\tan C-2\tan B\tan C=() tan B + tan C − 2 tan B tan C = ( ) 49 在三角形A B C ABC A BC A B = 5 AB=5 A B = 5 A C = 3 AC=3 A C = 3 B C = 7 BC=7 BC = 7 ∠ B A C ∠BAC ∠ B A C ( ) () ( ) 2 π 3 \dfrac {2π}{3} 3 2 π 5 π 6 \dfrac {5π}{6} 6 5 π 3 π 4 \dfrac {3π}{4} 4 3 π 50 若角α α α P ( a , 2 a ) ( a < 0 ) P(a,2a)(a < 0) P ( a , 2 a ) ( a < 0 ) cos α = \cos α= cos α = 51 下列各组中，终边相同的角是( ) A. 3 π 5 \dfrac{3\pi }{5} 5 3 π 2 k π − 3 π 5 ( k ∈ Z ) 2k\pi -\dfrac{3\pi }{5}(k\in Z) 2 kπ − 5 3 π ( k ∈ Z ) − π 5 -\dfrac{\pi }{5} − 5 π 26 π 5 \dfrac{26\pi }{5} 5 26 π 52 下列几个命题： ①函数y = x 2 - 1 + 1 - x 2 y\text{=}\sqrt{x^{2}\text{-}1}\text{+}\sqrt{1\text{-}x^{2}} y = x 2 - 1 + 1 - x 2 { a > 0 Δ = b 2 - 4 a c ⩽ 0 \left\{\begin{matrix}a\text{ > }0\\ {\Delta=}b^{2}\text{-}4ac{\leqslant}0\\\end{matrix}\right. { a > 0 Δ = b 2 - 4 a c ⩽ 0 53 在锐角△ A B C \triangle ABC △ A BC A A A B B B C C C a a a b b b c c c a = b cos C + 3 3 c sin B a=b\cos C+ \dfrac { \sqrt {3}}{3}c\sin B a = b cos C + 3 3 c sin B a = 2 a=2 a = 2 54 已知a = 1.6 0.8 a=1.6^{0.8} a = 1. 6 0.8 b = log 3 10 b=\log_{3}10 b = log 3 10 c = sin 1 − cos 1 c=\sin 1-\cos 1 c = sin 1 − cos 1 a a a b b b c c c ( ) (\quad) ( ) a > b > c a>b>c a > b > c c > b > a c>b>a c > b > a 55 在平面直角坐标系x O y xOy x O y O O O x x x C 1 {{C}_{1}} C 1 x 2 2 + y 2 = 1 \dfrac{{{x}^{2}}}{2}+{{y}^{2}}=1 2 x 2 + y 2 = 1 C 2 {{C}_{2}} C 2 56 已知函数f ( x ) = sin ( 2 x + φ ) + b ( 0 < φ < π ) f(x)=\sin(2x+φ)+b(0< φ< π) f ( x ) = sin ( 2 x + φ ) + b ( 0 < φ < π ) π 12 \dfrac{π}{12} 12 π 3 π 4 \dfrac{3π}{4} 4 3 π b = ( ) b=(\quad) b = ( ) 1 1 1 3 2 \dfrac{\sqrt{3}}{2} 2 3 57 已知函数f ( x ) = 2 cos ( π 2 − x ) cos ( 2 π − x ) . ( 1 ) f(x)=2\cos(\dfrac{π}{2}-x)\cos(2π-x).(1) f ( x ) = 2 cos ( 2 π − x ) cos ( 2 π − x ) . ( 1 ) f ( x ) f(x) f ( x ) x ∈ [ 0 , π 2 ] x\in[0,\dfrac{π}{2}] x ∈ [ 0 , 2 π ] g ( x ) = f ( x ) + cos 2 x g(x)=f(x)+\cos 2x g ( x ) = f ( x ) + cos 2 x 58 化简sin 50 ∘ cos 38 ∘ + cos 50 ∘ cos 128 ∘ \sin 50^{\circ}\cos 38^{\circ}+\cos 50^{\circ}\cos 128^{\circ} sin 5 0 ∘ cos 3 8 ∘ + cos 5 0 ∘ cos 12 8 ∘ 59 f ( x ) = 2 cos 2 x − 2 a cos x − 1 − 2 a f(x)=2\cos ^{2}x-2a\cos x-1-2a f ( x ) = 2 cos 2 x − 2 a cos x − 1 − 2 a g ( a ) g(a) g ( a ) a ∈ R ( 1 ) a∈R(1) a ∈ R ( 1 ) g ( a ) g(a) g ( a ) g ( a ) = 1 2 g(a)= \dfrac {1}{2} g ( a ) = 2 1 a a a f ( x ) f(x) f ( x ) 60 cos ( − 120 o ) = ( ) \cos (-120^{o})=() cos ( − 12 0 o ) = ( ) 1 2 \dfrac {1}{2} 2 1 − 1 2 - \dfrac {1}{2} − 2 1 3 2 \dfrac { \sqrt {3}}{2} 2 3 − 3 2 - \dfrac { \sqrt {3}}{2} − 2 3