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化简sin(π/2+α)cos(3π-α)tan(π+α)/cos(π/2-α)cos(-α-π)
人气:437 ℃ 时间:2020-04-11 21:28:22
解答
sin(π/2+α)cos(3π-α)tan(π+α)/cos(π/2-α)cos(-α-π)
=[cosα*(-cosα)*tanα]/[sinα*(-cosα)]
=[-sinα*cosα]/[sinα*(-cosα)]
=1.
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