cosx+π/6=4/5,求sin2x+π/12-爱科技问答

cosx+π/6=4/5,求sin2x+π/12

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avwuojdxne

cos(x+π/6)=4/5,求sin(2x+π/12)
解：∵cos(x+π/6)=4/5，∴x+π/6=arccos(4/5)，x=arccos(4/5)-π/6
于是sin(2x+π/12)=sin[2arccos(4/5)-π/3+π/12]=sin[2arccos(4/5)-π/4]
=sin[2arccos(4/5)]cos(π/4)-cos[2arccos(4/5)]sin(π/4)
=(√2/2)[2sinarccos(4/5)][cosarccos(4/5)]-(√2/2)[2cos?arccos(4/5)-1]
=(√2)[√(1-16/25)](4/5)-(√2)(4/5)?+√2/2
=(√2)[(3/5)(4/5)-(16/25)+1/2]
=(√2)[(12/25)-(16/25)+1/2]
=(17/50)√2.
lostcanvas

由cos(x + π/6) = 4/5，得x + π/6 = arccos(4/5)，故x = arccos(4/5) - π/6。
sin(2x + π/12) = sin。
化简后利用三角恒等式计算，结果为 7/25。
tancard

1
Q2865009

方法如下，请作参考：

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785779510a

解：
∵(17/12)π<(7/4)π ∴(5/3)π∴sin(x+π/4)<0 ∵cos(x+π/4)=3/5 ∴sin(x+π/4)=-4/5 ∴tan(x+π/4)=-4/3 ∴[tanx+tan(π/4)]/[1-tanxtan(π/4)]=-4/3 ∴(1+tanx)/(1-tanx)=-4/3 ∴3(tanx+1)=4(tanx-1) ∴3tanx+3=4tanx-4 ∴tanx=7 ∴sin(2x)+2(cosx)^2 =[sin(2x)+2(cosx)^2]/1 =[2sinxcosx+2(cosx)^2]/[(sinx)^2+(cosx)^2] =(2tanx+2)/[(tanx)^2+1] =(14+2)/(49+1) =8/25.

cosx+π/6=4/5,求sin2x+π/12