《單調(diào)性、最大值與最小值》三角函數(shù)PPT
第一部分內(nèi)容：課標(biāo)闡釋
1.理解正弦函數(shù)與余弦函數(shù)的單調(diào)性,會(huì)求函數(shù)的單調(diào)區(qū)間.
2.能夠利用三角函數(shù)單調(diào)性比較三角函數(shù)值的大小.
3.能夠結(jié)合三角函數(shù)的單調(diào)性求函數(shù)的最值和值域.
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單調(diào)性最大值與最小值PPT，第二部分內(nèi)容：自主預(yù)習(xí)
一、正弦函數(shù)與余弦函數(shù)的單調(diào)性
1.觀察正弦曲線,正弦函數(shù)在哪些區(qū)間上是增函數(shù)?在哪些區(qū)間上是減函數(shù)?如何將這些單調(diào)區(qū)間進(jìn)行整合?類似地,余弦函數(shù)在哪些區(qū)間上是增函數(shù)?在哪些區(qū)間上是減函數(shù)?怎樣整合這些區(qū)間?
提示:正弦函數(shù)在每一個(gè)閉區(qū)間["-" π/2+2kπ"," π/2+2kπ]
(k∈Z)上都是增函數(shù);在每一個(gè)閉區(qū)間[π/2+2kπ"," 3π/2+2kπ]
(k∈Z)上都是減函數(shù);余弦函數(shù)在每一個(gè)閉區(qū)間[-π+2kπ,2kπ](k∈Z)上都是增函數(shù);在每一個(gè)閉區(qū)間[2kπ,π+2kπ](k∈Z)上都是減函數(shù).
2.填空
(1)正弦函數(shù)y=sin x在每一個(gè)閉區(qū)間["-" π/2+2kπ"," π/2+2kπ](k∈Z)上都單調(diào)遞增;在每一個(gè)閉區(qū)間[π/2+2kπ"," 3π/2+2kπ](k∈Z)上都單調(diào)遞減;
(2)余弦函數(shù)y=cos x在每一個(gè)閉區(qū)間[-π+2kπ,2kπ](k∈Z)上都單調(diào)遞增;在每一個(gè)閉區(qū)間[2kπ,π+2kπ](k∈Z)上都單調(diào)遞減.
3.做一做
(1)函數(shù)y=sin 2x-1的單調(diào)遞增區(qū)間是___________;
(2)函數(shù)y=3-cos 2x的單調(diào)遞增區(qū)間是___________.
解析:(1)令-π/2+2kπ≤2x≤π/2+2kπ,k∈Z,
解得-π/4+kπ≤x≤π/4+kπ,k∈Z,故函數(shù)的單調(diào)遞增區(qū)間是["-" π/4+kπ"," π/4+kπ](k∈Z).
(2)函數(shù)y=3-cos 2x的單調(diào)遞增區(qū)間即為函數(shù)y=cos 2x的單調(diào)遞減區(qū)間,令2kπ≤2x≤π+2kπ,k∈Z,解得kπ≤x≤π/2+kπ,k∈Z,故函數(shù)的遞增區(qū)間是 kπ,π/2+kπ (k∈Z).
答案:(1)["-" π/4+kπ"," π/4+kπ](k∈Z)
(2)[kπ"," π/2+kπ](k∈Z)
二、正弦函數(shù)與余弦函數(shù)的最值和值域
1.觀察正弦曲線和余弦曲線,正、余弦函數(shù)是否存在最大值和最小值?若存在,其最大值和最小值分別為多少?當(dāng)自變量x分別取何值時(shí),正弦函數(shù)y=sin x取得最大值和最小值?余弦函數(shù)呢?
提示:正、余弦函數(shù)存在最大值1和最小值-1;正弦函數(shù)當(dāng)且僅當(dāng)x=2kπ+π/2(k∈Z)時(shí)取最大值1,當(dāng)且僅當(dāng)x=2kπ+3π/2(k∈Z)時(shí)取最小值-1;余弦函數(shù)當(dāng)且僅當(dāng)x=2kπ(k∈Z)時(shí)取最大值1,當(dāng)且僅當(dāng)x=π+2kπ(k∈Z)時(shí)取最小值-1.
2.填空
(1)正弦函數(shù)y=sin x當(dāng)且僅當(dāng)x=2kπ+π/2(k∈Z)時(shí)取最大值1;當(dāng)且僅當(dāng)x=2kπ+3π/2(k∈Z)時(shí)取最小值-1;
(2)余弦函數(shù)y=cos x當(dāng)且僅當(dāng)x=2kπ(k∈Z)時(shí)取最大值1;當(dāng)且僅當(dāng)x=2kπ+π(k∈Z)時(shí)取最小值-1.
(3)正弦函數(shù)y=sin x、余弦函數(shù)y=cos x的值域都是[-1,1].
3.做一做
(1)函數(shù)y=2-3sin x的最小值是___________;
(2)當(dāng)函數(shù)y=cos 取得最大值時(shí),x的值等于___________.
解析:(1)因?yàn)閥=sin x的最大值為1,所以y=2-3sin x的最小值是-1.
(2)當(dāng) =2kπ,k∈Z,即x=4kπ,k∈Z時(shí),函數(shù)y=cos 取得最大值.
答案:(1)-1 (2)4kπ(k∈Z)
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單調(diào)性最大值與最小值PPT，第三部分內(nèi)容：探究學(xué)習(xí)
求三角函數(shù)的單調(diào)區(qū)間
例1求下列函數(shù)的單調(diào)遞減區(qū)間:
(1)y=1/2cos(2x+π/3);
(2)y=2sin(π/4 "-" x).
分析:(1)可采用整體換元法并結(jié)合正弦函數(shù)、余弦函數(shù)的單調(diào)區(qū)間求解;(2)可先將自變量x的系數(shù)轉(zhuǎn)化為正數(shù)再求單調(diào)區(qū)間.
解:(1)令z=2x+π/3,而函數(shù)y=cos z的單調(diào)遞減區(qū)間是[2kπ,2kπ+π](k∈Z).
∴當(dāng)原函數(shù)單調(diào)遞減時(shí),可得2kπ≤2x+π/3≤2kπ+π(k∈Z),
解得kπ-π/6≤x≤kπ+π/3(k∈Z).
∴原函數(shù)的單調(diào)遞減區(qū)間是[kπ"-" π/6 "," kπ+π/3](k∈Z).
反思感悟 與正弦函數(shù)、余弦函數(shù)有關(guān)的單調(diào)區(qū)間的求解技巧:
(1)結(jié)合正弦、余弦函數(shù)的圖象,熟記它們的單調(diào)區(qū)間;
(2)確定函數(shù)y=Asin(ωx+φ)(A>0,ω>0)單調(diào)區(qū)間的方法:采用“換元”法整體代換,將ωx+φ看作一個(gè)整體,可令“z=ωx+φ”,即通過求y=Asin z的單調(diào)區(qū)間求出原函數(shù)的單調(diào)區(qū)間.若ω<0,則可利用誘導(dǎo)公式將x的系數(shù)轉(zhuǎn)變?yōu)檎龜?shù).
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單調(diào)性最大值與最小值PPT，第四部分內(nèi)容：思維辨析
求三角函數(shù)最值時(shí)忽視分類討論或忽略定義域致誤
1.忽視分類討論
典例1已知函數(shù)y=2asin(2x"-" π/3)+b的定義域?yàn)?0,π/2 ,函數(shù)的最大值為1,最小值為-5,求a和b的值.
錯(cuò)解∵0≤x≤π/2,∴-π/3≤2x-π/3≤2π/3.
∴-√3/2≤sin(2x"-" π/3)≤1.
則{■(2a+b=1"," @"-" √3 a+b="-" 5"," )┤解得{■(a=12"-" 6√3 "," @b="-" 23+12√3 "." )┤
錯(cuò)解錯(cuò)在什么地方?你能發(fā)現(xiàn)嗎?怎樣避免這類錯(cuò)誤呢?
提示:錯(cuò)解中默認(rèn)為a>0,忽視了對(duì)a<0這一情況的討論,導(dǎo)致丟解.
正解:∵0≤x≤π/2,∴-π/3≤2x-π/3≤2π/3.
∴-√3/2≤sin(2x"-" π/3)≤1.
若a>0,則{■(2a+b=1"," @"-" √3 a+b="-" 5"," )┤解得{■(a=12"-" 6√3 "," @b="-" 23+12√3 "." )┤
若a<0,則{■(2a+b="-" 5"," @"-" √3 a+b=1"," )┤解得{■(a="-" 12+6√3 "," @b=19"-" 12√3 "." )┤
防范措施 形如y=Asin(ωx+φ)+B或y=Acos(ωx+φ)+B的函數(shù),其最值與參數(shù)A的正負(fù)有關(guān),因此在解決這類問題時(shí),要注意對(duì)A分A>0和A<0兩種情況進(jìn)行分類討論.
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單調(diào)性最大值與最小值PPT，第五部分內(nèi)容：隨堂演練
1.函數(shù)y=-cos x在區(qū)間["-" π/2 "," π/2]上是( )
A.增函數(shù) B.減函數(shù)
C.先減后增函數(shù) D.先增后減函數(shù)
解析:結(jié)合函數(shù)在["-" π/2 "," π/2]上的圖象可知C正確.
答案:C
2.函數(shù)y=2-sin x的最大值及取最大值時(shí)x的值為( )
A.ymax=3,x=π/2
B.ymax=1,x=π/2+2kπ(k∈Z)
C.ymax=3,x=-π/2+2kπ(k∈Z)
D.ymax=3,x=π/2+2kπ(k∈Z)
解析:因?yàn)閥=2-sin x,所以當(dāng)sin x=-1時(shí),ymax=3,此時(shí)x=-π/2+2kπ(k∈Z).
答案:C
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