判断无穷数列 sin(π/2)+sin(2π/2)+sin(3π/2)+sin(4π/2)+… 的敛散性.
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α属于(0,π/2)且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin
数列求和 sin²1°+sin²2°+sin²3°+.+sin²88°+sin&
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已知sin(3π+a)=2sin[(3π/2)+a],求sin^2a+sin2a的值
证明:sinπ/9[(sinπ/9)+sin(3π/9)]=sin2(2π/9)
已知sin(3π−α)=−2sin(π2+α),则sinαcosα=( )
已知函数fx=(1+1/tanx)sin^x-2sin(x+π/4)sin(x-π/4)
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
已知(3sinα+cosα)/(3cosα-sinα)=2,则2-3sin(α-3π)sin(1.5π-α)-[cos(
若(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)等于?