《習(xí)題課 三角恒等變換的應(yīng)用》三角函數(shù)PPT
第一部分內(nèi)容：課標(biāo)闡釋
1.能夠運(yùn)用三角函數(shù)公式對(duì)函數(shù)解析式進(jìn)行化簡(jiǎn),以研究函數(shù)的性質(zhì).
2.能夠運(yùn)用三角函數(shù)公式解決求值與化簡(jiǎn)問題.
3.掌握三角恒等變換在實(shí)際問題中的應(yīng)用.
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習(xí)題課三角恒等變換的應(yīng)用PPT，第二部分內(nèi)容：自主預(yù)習(xí)
一、降冪和升冪公式
1.填空
(1)降冪公式:sin2α=(1"-" cos2α)/2,cos2α=(1+cos2α)/2,sin αcos α=1/2sin 2α.
(2)升冪公式:1+cos α=2cos2α/2,1-cos α=2sin2α/2.
2.做一做
(1)函數(shù)y=sin(2x+π/3)cos(2x+π/3)的最小正周期為( )
A.2π B.π C.π/2 D.π/4
(2)函數(shù)f(x)=cos2(x+π/4),x∈R,則f(x)( )
A.是奇函數(shù)
B.是偶函數(shù)
C.既是奇函數(shù),也是偶函數(shù)
D.既不是奇函數(shù),也不是偶函數(shù)
解析:(1)因?yàn)閥=sin(2x+π/3)cos(2x+π/3)
=1/2sin(4x+2π/3),所以最小正周期為2π/4=π/2.
(2)∵f(x)=(1+cos(2x+π/2))/2=1/2-1/2sin 2x,x∈R,
∴f(-x)=1/2-1/2sin 2(-x)=1/2+1/2sin 2x.
∴f("-" π/4)=1/2+1/2sin π/2=1,
f(π/4)=1/2-1/2sin π/2=0,
∴f("-" π/4)≠f(π/4),f("-" π/4)≠-f(π/4).
∴f(x)既不是奇函數(shù),也不是偶函數(shù).
答案:(1)C (2)D
二、輔助角公式
1.填空
asin x+bcos x=√(a^2+b^2 )sin(x+φ)
("其中" sinφ=b/√(a^2+b^2 ) "," cosφ=a/√(a^2+b^2 ) "," tanφ=b/a).
2.做一做
(1)若函數(shù)f(x)=sin x+3cos x,x∈R,則f(x)的值域是( )
A.[1,3] B.[1,2]
C.[-√10,√10] D.[0,√10]
(2)函數(shù)f(x)=sin x-√3cos x(x∈[-π,0])的單調(diào)遞增區(qū)間是( )
A.["-" π",-" 5π/6] B.["-" 5π/6 ",-" π/6] C.["-" π/3 "," 0] D.["-" π/6 "," 0]
解析:(1)因?yàn)閒(x)=sin x+3cos x=√10sin(x+φ)(其中tan φ=3),所以函數(shù)f(x)=sin x+3cos x,x∈R,則f(x)的值域是[-√10,√10].
(2)f(x)=2(1/2 sinx"-" √3/2 cosx)=2sin(x"-" π/3),令2kπ-π/2≤x-π/3≤2kπ+π/2,k∈Z,得2kπ-π/6≤x≤2kπ+5π/6,k∈Z.
又x∈[-π,0],
∴x∈["-" π/6 "," 0].
答案:(1)C (2)D
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習(xí)題課三角恒等變換的應(yīng)用PPT，第三部分內(nèi)容：規(guī)范解答
三角恒等變換與三角函數(shù)性質(zhì)的綜合應(yīng)用
典例 已知函數(shù)f(x)=sin(π/2 "-" x)sin x-√3cos2x.
(1)求f(x)的最小正周期和最大值;
(2)討論f(x)在[π/6 "," 2π/3]上的單調(diào)性.
【審題策略】 先利用三角恒等變換將函數(shù)f(x)的解析式化成f(x)=Asin(ωx+φ)+k的形式,然后確定其性質(zhì).
【規(guī)范展示】 解:(1)f(x)=sin(π/2 "-" x)sin x-√3cos2x=cos xsin x-√3/2(1+cos 2x)
=1/2sin 2x-√3/2cos 2x-√3/2=sin(2x"-" π/3)-√3/2,
因此f(x)的最小正周期為π,最大值為(2"-" √3)/2.
(2)當(dāng)x∈[π/6 "," 2π/3]時(shí),0≤2x-π/3≤π,從而當(dāng)0≤2x-π/3≤π/2,即π/6≤x≤5π/12時(shí),f(x)單調(diào)遞增;當(dāng)π/2<2x-π/3≤π,即5π/12<x≤2π/3時(shí),f(x)單調(diào)遞減.綜上可知,f(x)在[π/6 "," 5π/12]上單調(diào)遞增;在[5π/12 "," 2π/3]上單調(diào)遞減.
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習(xí)題課三角恒等變換的應(yīng)用PPT，第四部分內(nèi)容：隨堂演練
1.(多選題)已知函數(shù)f(x)=sin(x+π/6)cos(x+π/6),則下列判斷不正確的是( )
A.f(x)的最小正周期為π/2
B.f(x"-" π/6)是奇函數(shù)
C.f(x)的一個(gè)對(duì)稱中心為(π/6 "," 0)
D.f(x)的一條對(duì)稱軸為x=π/6
解析:因?yàn)閒(x)=sin(x+π/6)cos(x+π/6)=1/2sin(2x+π/3),所以f(x"-" π/6)=1/2sin 2x,所以f(x"-" π/6)是奇函數(shù).
答案:ACD
2.函數(shù)y=cos(x+π/2)+sin(π/3 "-" x)具有性質(zhì)( )
A.最大值為1,圖象關(guān)于點(diǎn)(π/6 "," 0)對(duì)稱
B.最大值為√3,圖象關(guān)于點(diǎn)(π/6 "," 0)對(duì)稱
C.最大值為1,圖象關(guān)于直線x=π/6對(duì)稱
D.最大值為√3,圖象關(guān)于直線x=π/6對(duì)稱
解析:∵y=-sin x+√3/2cos x-1/2sin x
=-√3 (√3/2 sinx"-" 1/2 cosx)=-√3sin(x"-" π/6),
∴最大值為√3,圖象關(guān)于點(diǎn)(π/6 "," 0)對(duì)稱.
答案:B
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