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已知tanA=112,求下列2式的值.(1)sina-3cosa/sina+cosa(2)sin的平方+sinacosa+2
人气:363 ℃ 时间:2019-12-15 10:46:25
解答
tanA=112,则 (cosA)^2 = 1/[1+(tanA)^2] = 1/12545
(1)(sina-3cosa)/(sina+cosa) 分子分母同除以cosx
=(tanx - 3)/(tanx + 1) = 109/113
(2) (sinA)^2 +sinAcosA+2 = (cosA)^2 [(tanA)^2 + tanA] + 2 = (1/12545)*[112^2 + 112]
= 12656/12545
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