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【問題思考】

2.填空:對于余弦函數(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法是否正確,正確的打“√”,錯誤的打“×”.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.對于sin

x=a(x∈R),-1≤a≤1,這個方程的解可表示成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辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯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辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析反思感悟cos

x=a(-1≤a≤1),當(dāng)x∈[0,π]時,則x=arccos

a,當(dāng)x∈R時,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辨析反思感悟cosx=a(-1≤a≤探究一探究二探究三易錯辨析變式訓(xùn)練1已知cosx=-0.345.(1)當(dāng)x∈[0,π]時,求x;(2)當(dāng)x∈R時,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,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,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辨析變式訓(xùn)練1已知cosx=-0.3探究一探究二探究三易錯辨析已知正切值求角

探究一探究二探究三易錯辨析已知正切值求角探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析反思感悟?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辨析變式訓(xùn)練2已知tanx=2,且x∈[3π,4π],求x.(用符號表示)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辨析易錯點: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辨析易錯點:因忽視角的范圍而致誤探究一探究二探究三易錯辨析糾錯心得arcsin

x,arccos

x,arctan

x都是有范圍的,忽略它們的范圍是求解問題出錯的根源.探究一探究二探究三易錯辨析糾錯心得arcsinx,arcc探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析高中人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【問題思考】

2.填空:對于余弦函數(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法是否正確,正確的打“√”,錯誤的打“×”.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.對于sin

x=a(x∈R),-1≤a≤1,這個方程的解可表示成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辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯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辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析反思感悟cos

x=a(-1≤a≤1),當(dāng)x∈[0,π]時,則x=arccos

a,當(dāng)x∈R時,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辨析反思感悟cosx=a(-1≤a≤探究一探究二探究三易錯辨析變式訓(xùn)練1已知cosx=-0.345.(1)當(dāng)x∈[0,π]時,求x;(2)當(dāng)x∈R時,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,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,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辨析變式訓(xùn)練1已知cosx=-0.3探究一探究二探究三易錯辨析已知正切值求角

探究一探究二探究三易錯辨析已知正切值求角探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析反思感悟?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辨析變式訓(xùn)練2已知tanx=2,且x∈[3π,4π],求x.(用符號表示)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辨析易錯點: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辨析易錯點:因忽視角的范圍而致誤探究一探究二探究三易錯辨析糾錯心得arcsin

x,arccos

x,arctan

x都是有范圍的,忽略它們的范圍是求解問題出錯的根源.探究一探究二探究三易錯辨析糾錯心得arcsinx,arcc探究一探究二探究三易錯辨析探究一探究二探究三易錯辨析高中人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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