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化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)]
人气:177 ℃ 时间:2020-04-10 21:09:52
解答
[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)]=[(-sinα)*(-cosα)*(-tgα)]/[(-sinα)*ctgα*(tgα)]=[-sinα*cosα*tgα]/[-sinα*ctgα*tgα]=cosα/ctgα=cosαtgα诱导公式:s...
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