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已知sin π 4 2A

sin(π/4+2a)sin(π/4-2a)=1/4 sin(π/4+2a)sin(π/2-π/4-2a)=1/4 sin(π/4+2a)sin[π/2-(π/4+2a)]=1/4 sin(π/4+2a)cos(π/4+2a)=1/4 2sin(π/4+2a)cos(π/4+2a)=1/2 sin(π/2+4a)=1/2 cos4a=1/2 a∈(π/4,π/2) 4a∈(π,2π) cos4a=1/2>0 所以4a∈(3π/2,2π) 即4a...

∵a∈(0.5π,π) ∴a+π/4∈(3π/4,5π/4) 又∵sin(a+π/4)=1/2 ∴a+π/4∈(3π/4,π) ∴a∈(π,3π/4) ∴a+π/4=5π/6 ∴a=7π/6 sin2a=sin7π/3=√3/2 cos2a=cos7π/3=1/2

解:∵sin(5π/2+α)=1/4 ==>sin(2π+π/2+α)=1/4 ==>sin(π/2+α)=1/4 ==>cosα=1/4 ∴cos2α=2cos²α-1 (应用倍角公式) =2*(1/4)²-1 =-7/8。

解: sin(π/2+a)=cosa=-3/5 ∵a∈(π/2,π), ∴sina>0 ∴sina=√(1-cos²a)=4/5, sin(π-2a) =sin2a =2sinacosa =2×4/5×(-3/5) =-24/25

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...

sin(π/6-a)=1/4 所以 cos(π/3-2a) =cos[2(π/6-a)] =1-2sin²(π/6-a) =1-2×(1/4)² =1-1/8 =7/8

cos(π/3+2a)=2cos(π/6+a)^2-1 =2sin[π/2-(π/6+a)]^2-1 =2sin(π/3-a)^2-1 = -7/8

解: sin(5/4π) =sin(π+π/4) =-sin(π/4) =-√2/2

sin(π/4-a)sin(π/4+a)=六分之根号二 2sin(π/4-a)sin(π/4+a)=√2/3 cos2a-cos(π/2)=√2/3 cos2a=√2/3 ∵0

a为第二象限,且sina=(√15)/4 那么cosa

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