作业帮sin(π/4+θ)*sin(π/4-θ)=2/5,则cos2θ=
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sin(π/4+θ)*sin(π/4-θ)=2/5,则cos2θ=
sin(π/4+θ)*sin(π/4-θ)=2/5,则cos2θ=
sin(π/4+θ)*sin(π/4-θ)=2/5,则cos2θ=
因为
π/4-θ=π/2-(π/4+θ)
所以sin(π/4-θ)=sin[π/2-(π/4+θ)]=cos(π/4+θ)
所以
sin(π/4+θ)*sin(π/4-θ)
=sin(π/4+θ)*cos(π/4+θ)
=1/2sin(π+2θ）
=-1/2sin2θ=2/5
所以sin2θ=-4/5
cos2θ=3/5或者-3/5
sin(π/4-θ)=cos[π/2-(π/4-θ)]=cos(π/4+θ)
所以已知条件就变为：sin(π/4+θ)cos(π/4+θ)=2/5
那么2sin(π/4+θ)cos(π/4+θ)=4/5
即sin[2(π/4+θ)]=4/5，即sin(π/2+2θ)=4/5
所以cos2θ=sin(π/2+2θ)=4/5
化简求值sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)
sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)化简求值
已知sin(π+θ)
sin(x+π/4)
sin(阿尔法-π/4)
sin(-π/4)求值
sin(π/4+θ)*sin(π/4-θ)=2/5,则cos2θ=
2(sinθ+cosθ)= 2√2sin（θ+π/4)为什么?; （-π/2
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（cosθ＋sinθ）＝√2sin（θ＋π／4）
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2sin(x+π/4)sin(x-π/4)=?
sin(x+π/4)sin(x-π/4)等于什么?