Question

若f(α)=[sin(π-α）cos(2π-α）tan(-α+π）]/[-tan(-α-π）sin（-π-α）]化简

Answer 1

f(α)=[sin(π-α）cos(2π-α）tan(-α+π）]/[-tan(-α-π）sin（-π-α）]
=[sinαcosα(-tanα)]/{tanα[-sin(π+α)}
=-sinαcosαtanα/(tanαsinα)
=-cosα

追问

不是cosα吗？

追答

不是

Answer 2

利用三角函数的诱导公式的化简只需要记住一个口诀：奇变偶不变，符号看象限

Answer 3

⑴
f(α) = [sin(π-α)cos(2π-α)tan(-π-α)]/sin(-π-α)
= [sin(α)cos(-α)tan(-α)]/sin(α)
= cos(-α)tan(-α)
= cos(-α) * sin(-α) / cos(-α)
= sin(-α)
= -sin(α)
⑵
cos﹙α - 3π/2﹚=cos﹙-2π + π/2 + α ﹚= cos(π/2 + α) = -sin(α) =1/5
∴ f(α) = -sin(α) = 1/5
⑶
f(α) = -sin(α) = -sin(-1860°) = -sin(-5×360°- 60°) = -sin(-60°) = sin60°= √3/2