化简 f(α)=sin(π—α) cos(2π—α)tan(-α+3/2π)/cot(-α-π)sin(-π-α)
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若f(sinα+cosα)=sinαcosα,则f(cosπ/6)等于..麻烦讲清楚过程,辛苦各位了.
化简 f(α)=sin(π—α) cos(2π—α)tan(-α+3/2π)/cot(-α-π)sin(-π-α)
已知f(α)=(tan(π-α)*cos(2π-α)*sin(二分之π+α))/(cos(-α-π)) (1)化简f(α)
设f(α)=(2sinαcosα+cosα)/[(1+sin^2α+cos(3/2π+α)-sin^2(π/2+α)](1+sinα≠0)(1)化简f(α)(2)求f(1°)+f(2°)+f(3°)+……+f(89°)不好意思,第二问是求f(1°)*f(2°)*f(3°)*……*f(89°)
如果f(sinα+cosα)=sinαcosα那么f(cos30′)=?
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设f(sinα+cosα)=sinα·cosα,则f[sin(π/6)]=_______?要有过程
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π〈α〈2π且f(α)={√2-2 cosα}/{(1-sinα-cosα)(cosα/2+sinα/2)}请化简f(α)步骤尽量详细点