Question

已知tanα=2，求（sin(π-α）cos(2π-α)sin(-α+3π/2))/tan(-α-π)sin(-π-α)的值

Answer 1

原式=[sinα*cosα*(-cosα)] / tan[-(π+α)*sin[-(π+α)]
=-(sinα*cosα*cosα) / [-tan(π+α)]*[-sin(π+α)]
=-(sinα*cosα*cosα) / (-tanα)*sinα
=(cosα)^2 / 2
=1/ [2*(secα)^2]
=1/ {2*[1 + (tanα)^2]}
=1/ (2*5)
=1/10
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Answer 2

原式=[sinα*cosα*(-cosα)] / tan[-(π+α)*sin[-(π+α)]
=-(sinα*cosα*cosα) / [-tan(π+α)]*[-sin(π+α)]
=-(sinα*cosα*cosα) / (-tanα)*sinα
=(cosα)^2 / 2
=1/ [2*(secα)^2]
=1/ {2*[1 + (tanα)^2]}
=1/ (2*5)
=1/10

Answer 3

原式=[sinacosa(－cosa)]/[(－tana)sina]=[1/tana][cos²a]=[1/tana][(cos²a)/(cos²a＋sin²a)]=[1/tana][(1)/(1＋tan²a)]=1/10。

Answer 4

1/9

Answer 5

1/5