《習(xí)題課 函數(shù)y=Asin(ωx+φ)的性質(zhì)及其應(yīng)用》三角函數(shù)PPT
第一部分內(nèi)容：課標(biāo)闡釋
1.了解函數(shù)y=Asin(ωx+φ)中,參數(shù)A,ω,φ的物理意義.
2.能夠根據(jù)y=Asin(ωx+φ)的圖象確定其解析式.
3.掌握函數(shù)y=Asin(ωx+φ)的性質(zhì),能夠利用性質(zhì)解決相關(guān)問題.
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習(xí)題課函數(shù)y=Asin(ωx+φ)的性質(zhì)及其應(yīng)用PPT，第二部分內(nèi)容：自主預(yù)習(xí)
函數(shù)y=Asin(ωx+φ)的性質(zhì)
1.對(duì)于正弦函數(shù)y=sin x,我們研究過其定義域、值域、周期性、奇偶性、對(duì)稱性、單調(diào)區(qū)間等,那么對(duì)于形如y=Asin(ωx+φ)的函數(shù),例如:函數(shù)y=3sin(2x"-" π/4),其定義域、值域、周期性、奇偶性、對(duì)稱軸、對(duì)稱中心、單調(diào)區(qū)間如何求解呢?
提示:以正弦函數(shù)的性質(zhì)為基礎(chǔ),充分利用整體代換方法研究函數(shù)y=Asin(ωx+φ)的各種性質(zhì).
2.函數(shù)y=Asin(ωx+φ)(A>0)的性質(zhì)
3.做一做
(1)函數(shù)f(x)=sin(3x"-" π/3)-4的值域?yàn)? )
A.[-1,1] B.[-4,4] C.[-5,5] D.[-5,-3]
(2)若函數(shù)f(x)=-2sin(4x+φ)(0<φ<2π)是一個(gè)奇函數(shù),則φ的值等于( )
A.π/2 B.π/8 C.π D.π/4
(3)若函數(shù)y=1/3sin(ωx+π/6)(ω>0)的最小正周期是4π,則其圖象的一條對(duì)稱軸為( )
A.x=-4π/3 B.x=-π/3
C.x=π/2 D.x=5π/3
解析:(1)因?yàn)?1≤sin(3x"-" π/3)≤1,
所以-5≤sin(3x"-" π/3)-4≤-3,
即函數(shù)值域?yàn)閇-5,-3],選D.
(2)依題意有φ=kπ,k∈Z,而0<φ<2π,
所以φ=π,故選C.
(3)依題意有2π/ω=4π,
所以ω=1/2,即y=1/3sin(1/2 x+π/6),
而當(dāng)x=-4π/3時(shí),函數(shù)取得最小值-1/3,故x=-4π/3是其圖象的一條對(duì)稱軸.選A.
答案:(1)D (2)C (3)A
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習(xí)題課函數(shù)y=Asin(ωx+φ)的性質(zhì)及其應(yīng)用PPT，第三部分內(nèi)容：探究學(xué)習(xí)
三角函數(shù)圖象變換的應(yīng)用
例1將函數(shù)y=sin(2x+φ)的圖象沿x軸向左平移π/8個(gè)單位后,得到一個(gè)偶函數(shù)的圖象,則φ的一個(gè)可能取值為( )
A.3π/4 B.π/4 C.0 D.-π/4
解析:把函數(shù)y=sin(2x+φ)的圖象向左平移π/8個(gè)單位后,得到的圖象的解析式是y=sin(2x+π/4+φ),該函數(shù)是偶函數(shù)的條件是π/4+φ=kπ+π/2,k∈Z,根據(jù)選項(xiàng)檢驗(yàn)可知φ的一個(gè)可能取值為π/4.
答案:B
反思感悟 函數(shù)y=Asin(ωx+φ)的奇偶性:
(1)當(dāng)φ=kπ(k∈Z)時(shí),函數(shù)是奇函數(shù);
(2)當(dāng)φ=kπ+π/2(k∈Z)時(shí),函數(shù)是偶函數(shù);
(3)當(dāng)φ≠kπ,且φ≠kπ+π/2(k∈Z)時(shí),函數(shù)是非奇非偶函數(shù).
延伸探究 本例中,若將函數(shù)y=sin(2x+φ)的圖象向右平移π/6個(gè)單位,得到的圖象關(guān)于直線x=π/4對(duì)稱,則φ的最小正值等于________.
解析:函數(shù)y=sin(2x+φ)的圖象向右平移π/6個(gè)單位,得到y(tǒng)=sin(2x"-" π/3+φ)的圖象,該圖象關(guān)于直線x=π/4對(duì)稱,則有2×π/4-π/3+φ=kπ+π/2(k∈Z),則φ=kπ+π/3(k∈Z),因此φ的最小正值等于π/3.
答案:π/3
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習(xí)題課函數(shù)y=Asin(ωx+φ)的性質(zhì)及其應(yīng)用PPT，第四部分內(nèi)容：規(guī)范解答
函數(shù)y=Asin(ωx+φ)的性質(zhì)及應(yīng)用
典例 設(shè)函數(shù)f(x)=sin(2x+φ)(-π<φ<0),f(x)圖象的一條對(duì)稱軸是直線x=π/8 .
(1)求函數(shù)y=f(x)的單調(diào)增區(qū)間;
(2)畫出函數(shù)y=f(x)在區(qū)間[0,π]上的圖象.
【規(guī)范展示】 (1)∵x=π/8是函數(shù)y=f(x)的圖象的對(duì)稱軸,∴sin(2×π/8+φ)=±1.
∴π/4+φ=kπ+π/2(k∈Z),φ=kπ+π/4(k∈Z).
∵-π<φ<0,∴φ=-3π/4.∴y=sin(2x"-" 3π/4).
由題意得2kπ-π/2≤2x-3π/4≤2kπ+π/2(k∈Z),
∴kπ+π/8≤x≤kπ+5π/8(k∈Z).
即函數(shù)y=sin(2x"-" 3π/4)的單調(diào)增區(qū)間為[kπ+π/8 "," kπ+5π/8](k∈Z).
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習(xí)題課函數(shù)y=Asin(ωx+φ)的性質(zhì)及其應(yīng)用PPT，第五部分內(nèi)容：隨堂演練
1.如圖,是函數(shù)f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<π)的圖象,則其解析式為( )
A.f(x)=5sin(4/3 x+π/3)
B.f(x)=5sin(2/3 x+π/3)
C.f(x)=5sin(2/3 x+π/6)
D.f(x)=5sin(2/3 x"-" π/3)
解析:由題圖知,A=5,由T/2=5π/2-π=3π/2,知T=3π,
∴ω=2π/T=2/3,則y=5sin(2/3 x+φ).
由圖象知最高點(diǎn)坐標(biāo)為(π/4 "," 5),
將其代入y=5sin(2/3 x+φ),得5sin(π/6+φ)=5,
∴π/6+φ=2kπ+π/2(k∈Z).
解得φ=2kπ+π/3(k∈Z).
∵|φ|<π,∴φ=π/3,∴y=5sin(2/3 x+π/3).
答案:B
2.在同一平面直角坐標(biāo)系中,函數(shù)y=cos x/2+3π/2 (x∈[0,2π])的圖象和直線y=1/2的交點(diǎn)個(gè)數(shù)是( )
A.0 B.1 C.2 D.4
解析:作出函數(shù)y=cos x/2+3/2π ,x∈[0,2π]的圖象及y=1/2的圖象可得,應(yīng)選C.
答案:C
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