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已知tanA=2 求(1)4sinA-2cosA/5cosA+3sinA (2)sinAcosA
人气:268 ℃ 时间:2020-06-07 12:11:40
解答
1.4sinA-2cosA/5cosA+3sinA
=4-2tanA/5+3tanA (上下两边同除以cosA)
=4-2*2/5+3*2
=0
2.sinAcosA=sinAcosA/sinA²+cosA²
=tanA/tanA²+1(上下两边同除以 cosA²)
=2/2²+1
=2/5
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