高中人B數(shù)學(xué)必修四課件：133-已知三角函數(shù)值求角-

1.3.3

已知三角函數(shù)值求角1.3.3已知三角函數(shù)值求角高中人B數(shù)學(xué)必修四課件：133-已知三角函數(shù)值求角-一二三一、已知正弦值,求角【問(wèn)題思考】

一二三一、已知正弦值,求角一二三一二三一二三二、已知余弦值,求角【問(wèn)題思考】

2.填空:對(duì)于余弦函數(shù)y=cosx,如果已知函數(shù)值y(y∈[-1,1]),那么在[0,π]上有唯一的x值和它對(duì)應(yīng),記作x=arccosy(-1≤y≤1,0≤x≤π).

一二三二、已知余弦值,求角2.填空:一二三一二三一二三三、已知正切值,求角【問(wèn)題思考】

1.已知tanx=-1,求x.一二三三、已知正切值,求角思考辨析判斷下列說(shuō)法是否正確,正確的打“√”,錯(cuò)誤的打“×”.答案:(1)×

(2)×

(3)×

(4)√思考辨析答案:(1)×(2)×(3)×(4)√探究一探究二探究三易錯(cuò)辨析已知正弦值求角

分析:借助正弦函數(shù)的圖象及所給角的范圍求解.探究一探究二探究三易錯(cuò)辨析已知正弦值求角分析:借助正弦函數(shù)探究一探究二探究三易錯(cuò)辨析反思感悟給值求角,由于范圍不同,所得的角可能不同,一定要注意范圍條件的約束作用.對(duì)于sin

x=a(x∈R),-1≤a≤1,這個(gè)方程的解可表示成x=2kπ+arcsin

a或x=2kπ+π-arcsin

a(k∈Z).從而方程的解集為{x|x=kπ+(-1)karcsin

a,k∈Z}.探究一探究二探究三易錯(cuò)辨析反思感悟給值求角,由于范圍不同,所探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析已知余弦值求角

分析:借助余弦函數(shù)的圖象及所給角的范圍求解即可.探究一探究二探究三易錯(cuò)辨析已知余弦值求角分析:借助余弦函數(shù)探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析反思感悟cos

x=a(-1≤a≤1),當(dāng)x∈[0,π]時(shí),則x=arccos

a,當(dāng)x∈R時(shí),可先求得[0,2π]內(nèi)的所有解,再利用周期性可求得{x|x=2kπ±arccos

a,k∈Z}.探究一探究二探究三易錯(cuò)辨析反思感悟cosx=a(-1≤a≤探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練1已知cosx=-0.345.(1)當(dāng)x∈[0,π]時(shí),求x;(2)當(dāng)x∈R時(shí),求x的取值集合.解:(1)∵cos

x=-0.345,且x∈[0,π],∴x=arccos(-0.345)=π-arccos

0.345.(2)當(dāng)x∈R時(shí),先求出[0,2π]上的解.∵cos

α=-0.345,∴α是第二或第三象限的角,由(1)知x1=π-arccos

0.345為第二象限的角,∵cos(π+arccos

0.345)=-0.345且π+arccos

0.345∈

,∴x2=π+arccos

0.345,∴當(dāng)x=2kπ+x1或2kπ+x2,k∈Z時(shí),cos

x=-0.345.即所求x的集合為{x|x=2kπ±arccos(-0.345),k∈Z}.探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練1已知cosx=-0.3探究一探究二探究三易錯(cuò)辨析已知正切值求角

探究一探究二探究三易錯(cuò)辨析已知正切值求角探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析反思感悟?qū)τ谝阎兄登蠼怯腥缦乱?guī)律:探究一探究二探究三易錯(cuò)辨析反思感悟?qū)τ谝阎兄登蠼怯腥缦乱?guī)探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練2已知tanx=2,且x∈[3π,4π],求x.(用符號(hào)表示)解:∵3π≤x≤4π,∴x-3π=arctan

2,∴x=3π+arctan

2.探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練2已知tanx=2,且x探究一探究二探究三易錯(cuò)辨析易錯(cuò)點(diǎn):因忽視角的范圍而致誤【典例】

求函數(shù)y=(arcsinx)2+arcsinx-1的最值.探究一探究二探究三易錯(cuò)辨析易錯(cuò)點(diǎn):因忽視角的范圍而致誤探究一探究二探究三易錯(cuò)辨析糾錯(cuò)心得arcsin

x,arccos

x,arctan

x都是有范圍的,忽略它們的范圍是求解問(wèn)題出錯(cuò)的根源.探究一探究二探究三易錯(cuò)辨析糾錯(cuò)心得arcsinx,arcc探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析高中人B數(shù)學(xué)必修四課件：133-已知三角函數(shù)值求角-答案:C4.滿足等式sin(2x+45°)=cos(30°-x)的最小正角x是

.

解析:sin(2x+45°)=sin(60°+x),要使x>0,且最小,則2x+45°=60°+x,所以x=15°.答案:15°5.若arccos(2x-1)有意義,則x的取值范圍是

.

解析:要使arccos(2x-1)有意義,則需-1≤2x-1≤1,即0≤x≤1,故x∈[0,1].答案:[0,1]答案: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已知三角函數(shù)值求角1.3.3已知三角函數(shù)值求角高中人B數(shù)學(xué)必修四課件：133-已知三角函數(shù)值求角-一二三一、已知正弦值,求角【問(wèn)題思考】

一二三一、已知正弦值,求角一二三一二三一二三二、已知余弦值,求角【問(wèn)題思考】

2.填空:對(duì)于余弦函數(shù)y=cosx,如果已知函數(shù)值y(y∈[-1,1]),那么在[0,π]上有唯一的x值和它對(duì)應(yīng),記作x=arccosy(-1≤y≤1,0≤x≤π).

一二三二、已知余弦值,求角2.填空:一二三一二三一二三三、已知正切值,求角【問(wèn)題思考】

1.已知tanx=-1,求x.一二三三、已知正切值,求角思考辨析判斷下列說(shuō)法是否正確,正確的打“√”,錯(cuò)誤的打“×”.答案:(1)×

(2)×

(3)×

(4)√思考辨析答案:(1)×(2)×(3)×(4)√探究一探究二探究三易錯(cuò)辨析已知正弦值求角

分析:借助正弦函數(shù)的圖象及所給角的范圍求解.探究一探究二探究三易錯(cuò)辨析已知正弦值求角分析:借助正弦函數(shù)探究一探究二探究三易錯(cuò)辨析反思感悟給值求角,由于范圍不同,所得的角可能不同,一定要注意范圍條件的約束作用.對(duì)于sin

x=a(x∈R),-1≤a≤1,這個(gè)方程的解可表示成x=2kπ+arcsin

a或x=2kπ+π-arcsin

a(k∈Z).從而方程的解集為{x|x=kπ+(-1)karcsin

a,k∈Z}.探究一探究二探究三易錯(cuò)辨析反思感悟給值求角,由于范圍不同,所探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析已知余弦值求角

分析:借助余弦函數(shù)的圖象及所給角的范圍求解即可.探究一探究二探究三易錯(cuò)辨析已知余弦值求角分析:借助余弦函數(shù)探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析反思感悟cos

x=a(-1≤a≤1),當(dāng)x∈[0,π]時(shí),則x=arccos

a,當(dāng)x∈R時(shí),可先求得[0,2π]內(nèi)的所有解,再利用周期性可求得{x|x=2kπ±arccos

a,k∈Z}.探究一探究二探究三易錯(cuò)辨析反思感悟cosx=a(-1≤a≤探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練1已知cosx=-0.345.(1)當(dāng)x∈[0,π]時(shí),求x;(2)當(dāng)x∈R時(shí),求x的取值集合.解:(1)∵cos

x=-0.345,且x∈[0,π],∴x=arccos(-0.345)=π-arccos

0.345.(2)當(dāng)x∈R時(shí),先求出[0,2π]上的解.∵cos

α=-0.345,∴α是第二或第三象限的角,由(1)知x1=π-arccos

0.345為第二象限的角,∵cos(π+arccos

0.345)=-0.345且π+arccos

0.345∈

,∴x2=π+arccos

0.345,∴當(dāng)x=2kπ+x1或2kπ+x2,k∈Z時(shí),cos

x=-0.345.即所求x的集合為{x|x=2kπ±arccos(-0.345),k∈Z}.探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練1已知cosx=-0.3探究一探究二探究三易錯(cuò)辨析已知正切值求角

探究一探究二探究三易錯(cuò)辨析已知正切值求角探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析反思感悟?qū)τ谝阎兄登蠼怯腥缦乱?guī)律:探究一探究二探究三易錯(cuò)辨析反思感悟?qū)τ谝阎兄登蠼怯腥缦乱?guī)探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練2已知tanx=2,且x∈[3π,4π],求x.(用符號(hào)表示)解:∵3π≤x≤4π,∴x-3π=arctan

2,∴x=3π+arctan

2.探究一探究二探究三易錯(cuò)辨析變式訓(xùn)練2已知tanx=2,且x探究一探究二探究三易錯(cuò)辨析易錯(cuò)點(diǎn):因忽視角的范圍而致誤【典例】

求函數(shù)y=(arcsinx)2+arcsinx-1的最值.探究一探究二探究三易錯(cuò)辨析易錯(cuò)點(diǎn):因忽視角的范圍而致誤探究一探究二探究三易錯(cuò)辨析糾錯(cuò)心得arcsin

x,arccos

x,arctan

x都是有范圍的,忽略它們的范圍是求解問(wèn)題出錯(cuò)的根源.探究一探究二探究三易錯(cuò)辨析糾錯(cuò)心得arcsinx,arcc探究一探究二探究三易錯(cuò)辨析探究一探究二探究三易錯(cuò)辨析高中人B數(shù)學(xué)必修四課件：133-已知三角函數(shù)值求角-答案:C4.滿足等式sin(2x+45°)=cos(30°-x)的最小正角x是

.

解析:sin(2x+45°)=sin(60°+x),要使x>0,且最小,則2x+45°=60°+x,所以x=15°.答案:15°5.若arccos(2x-1)有意義,則x的取值范圍是

.

解析:要使arccos(2x-1)有意義,則需-1≤2x-1≤1,即0≤x≤1,故x∈[0,1].答案:[0,1]答案: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