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已知sin(π/4+α)sin(π/4%α)=1/6 α∈(π/2,π),求tAn4α

sin(π/6-α)=1/4, 则cos(π/3-2α)=cos[2(π/6-α)] =1-2 sin²(π/6-α) =7/8. 所以cos(2α-π/3) =7/8. sin(π/6+2α)= sin((2α-π/3)+ π/2) = cos(2α-π/3)==7/8. 如果你认可我的回答,敬请及时采纳, ~如果你认可我的回答,请及时点击【采纳为满意回...

解:∵α∈(π/2,π) ∴cosα

sin(π/4+2a) * sin(π/4-2a) =1/4 则(1/2)[cos4a-cos(π/2)]=1/4 cos4a=1/2 a∈(π/4,π/2) 2a∈(π/2,π) 可见cos2a0 由cos4a=2cos²2a-1=1/2 cos²2a=3/4 cos2a=-√3/2 (1) 又cos4a=1-2sin²2a=1/2 sin²2a=1/4 sin2a=1/2 (2) 2sin^a+t...

解: ∵π/3<a<π ∴π/2<a+π/6<7π/6 ∵sin(a+π/6)=1/3, ∴π/2<a+π/6<π, ∴cos(a+π)=-√[1-sin²(a+π/6)]=-2√2/3, sin(π/12-a) =sin[π/4-(a+π/6)] =ain(π/4)cos(a+π/6)-cos(π/4)sin(a+π/6) =√2/2×(-2√2/9)-√2/2×1/3 =-2/9-√2/6 =-(4...

-√6/3 根据公式:(cosx-sinx)²=1-2sinxcosx=1-1/3=2/3 所以 cosx-sinx=+-√6/3 因为 π/4

由已知得cosa√3/2+sina*1/2+sina=4/5*√3,cosa*1/2+sina*√3/2=4/5,cosasinπ/6+sinacosπ/6=4/5,sin(a+π/6)=4/5

(1)∵f(x)=sin(π4+x)sin(π4-x)+3sinxcosx=12cos2x+32sin2x…(2分)=sin(2x+π6),…(4分)∴f(π6)=1.…(6分)(2)由f(A2)=sin(A+π6)=1,而0<A<π可得:A+π6=π2,即A=π3.(8分)∴sinB+sinC=sinB+sin(2π3-B)=32sinB+32cosB=3si...

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cos[π/2-(π/6+α)]=sin(π/6+α)=1/3 所以cos(π/3-α)=1/3 利用倍角公式 则cos2(π/3-α)=cos(2π/3-2α)=2[cos(π/3-α)]^2-1=2*(1/9)-1=-7/9

(1)、由sin(π-a)=4/5可以知道, sina=sin(π-a)=4/5 且a属于(0,π/2), 故cosa也大于0,又(cosa)^2+(sina)^2=1, 解得cosa=3/5, 所以sin2a - cos^2 a/2=2sinacosa - (0.5+cosa) =24/25 - 0.5 -3/5 = -0.14 (2)、由cosa=3/5可知, f(x)=5/6* co...

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