已知α是第三象限角,
且f(a)=[sin(5π-α)cos(α+3π/2)cos(π+α)]/[sin(α-3π/2)cos(α+π/2)tan(α-3π)]1,化简f(a)2,已知cos(...
且f(a)=[sin(5π-α)cos(α+3π/2)cos(π+α)]/[sin(α-3π/2)cos(α+π/2)tan(α-3π)]
1,化简f(a)
2,已知cos(π-α)=1/5,求f(a)的值
求详细步骤 谢谢大家 展开
1,化简f(a)
2,已知cos(π-α)=1/5,求f(a)的值
求详细步骤 谢谢大家 展开
1个回答
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1)f(α)= [sin(5π-α)cos(α+3π/2)cos(π+α)]/[sin(α-3π/2)cos(α+π/2)tan(α-3π)]
= [cosαsinα(-cosα)]/[cosα(-sinα)tanα]
= cotαcosα
2)cos(π-α) = -cosα = 1/5,cosα = -1/5
又α是第三象限角
所以sinα = -2√6/5, cotα = cosα/sinα = √6/12
f(α) = √6/12·(-1/5) = -√6/60
= [cosαsinα(-cosα)]/[cosα(-sinα)tanα]
= cotαcosα
2)cos(π-α) = -cosα = 1/5,cosα = -1/5
又α是第三象限角
所以sinα = -2√6/5, cotα = cosα/sinα = √6/12
f(α) = √6/12·(-1/5) = -√6/60
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