tanα=2,则(sinα的三次方+cosα)÷(sinα的三次方+sinα)

问题描述:

tanα=2,则(sinα的三次方+cosα)÷(sinα的三次方+sinα)

=[sin³a+cosa(sin²a+cos²a)]/[sin³a+sina(sin²a+cos²a)]=(sin³a+sin²acosa+cos³a)/(2sin³a+sinacos²a)=(tan³a+tan²a+1)/(2tan³a+tana)=(8+4...