1樓:匿名使用者
首先你先判斷取值範圍,α+π/6小於45°,那麼2α+π/6也是銳角sin(α+π/6)=1-(cos(α+π/6))^(1/2)=3/5
cos(2α+π/3)=2cos(α+π/6)^2-1=7/25sin(2α+π/3)=2sin(α+π/6)cos(α+π/6)=2*3/5*4*5=24/25
∴sin(2α+π/12)
=sin[(2α+π/3)-π/4]
=sin(2α+π/3)cosπ/4-cos(2α+π/3)sinπ/4
=24/25*(1/2)^(0.5)-7/25*(1/2)^0.5=50分之17倍根號2
2樓:皮皮鬼
若cos(α+π/6)=4/5
知sin(α+π/6)=3/5
則sin(2(α+π/6))=2sin(α+π/6)cos(α+π/6)=24/25
cos(2(α+π/6))=2cos(α+π/6)^2-1=7/25故sin(2α+π/12)
=sin(2(α+π/6)-π/4)
=sin(2(α+π/6)cosπ/4-cos(2(α+π/6)sinπ/4
=24/25×根2/2-7/25×根2/2=17根2/50
3樓:lll藍色楓林
α∈(0,π/2)
α+π/6∈(π/6,2π/3)
cos(α+π/6)=4/5>0
∴α+π/6∈(π/6,π/2)
∴2α+π/3∈(π/3,π)
cos(2α+π/3)
=2cos²(α+π/6)-1
=2*(4/5)²-1
=32/25-1
=7/25>0
∴2α+π/3∈(π/3,π/2)
sin(2α+π/3)=24/25
sin(2α+π/12)
=sin(2α+π/3-π/4)
=sin(2α+π/3)cosπ/4-cos(2α+π/3)sinπ/4
=24/25*√2/2-7/25*√2/2=24√2/50-7√2/50
=17√2/50
設a為銳角,若cos(a+π/6)=4/5,則sin(2a+π/12)值為
4樓:泣山彤
解:sin(2a+π/12)=sin【2(a+π/6) — π/4】=sin2(a+π/6)cos(π/4)—cos2(a+π/6)sin(π/4)
因為cos(a+π/6)=4/5,a為銳角,所以0
又 由二倍角公式可得,sin2(a+π/6)=24/25, cos2(a+π/6)=7/25, 所以sin(2a+π/12)=17√2/50. 這道題主要考查的是1.根據已知和未知角度拆分角度,便於運算; 2.二倍角公式的應用。 設a為銳角,若cos(a+π/6)=4/5.則sin(2a+π/12)=? 5樓:匿名使用者 ∵0<α<π/2 ∴π/6<α+π/6<2π/3 又cos(α+π/6)=4/5 ∴π/6<α+π/6<π/2 ∴sin(α+π/6)=3/5 ∴sin[2(α+π/6)]=2sin(α+π/6)cos(α+π/6)=2*3/5*4/5=24/25 cos[2(α+π/6)]=1-2sin²(α+π/6)=1-2*(3/5)²=7/25 又2α+π/12=2(α+π/6)-π/4 ∴sin(2α+π/12)=sin[2(α+π/6)-π/4]=sin[2(α+π/6)]cosπ/4-cos[2(α+π/6)]sinπ/4 =(24/25)*(√2/2)-(7/25)*(√2/2)=(17√2)/50 6樓:自我教 sin(2a+π/12)=sin【2(a+π/6) — π/4】=sin2(a+π/6)cos(π/4)—cos2(a+π/6)sin(π/4) 因為cos(a+π/6)=4/5,a為銳角,所以0 設a為銳角,若cos(a+π/6)=4/5,則sin(2a+π/12)的值為?
10 7樓:李永旭虛竹靖 解:sin(2a+π/12)=sin【2(a+π/6) — π/4】=sin2(a+π/6)cos(π/4)—cos2(a+π/6)sin(π/4) 因為cos(a+π/6)=4/5,a為銳角,所以0
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