化简(1)cos(α+π)sin(-α)⼀cos(-3π-α)sin(-α-4π)

2024-12-19 09:57:15
推荐回答(2个)
回答1:

第一个= cosasina/-cos(3π+a)sin(a+4π)=cosasina/-cosa*(-sina)=1
第二个=[cos(π-a)/sin(π/2+a)]*sina*cosa=[-cosa/cosa]*sinacosa=-sinacosa=-sin2a/2

回答2:

(1)cos(α+π)sin(-α)/cos(-3π-α)sin(-α-4π)
=cos(α+π)*(-sinα)/cos(3π+α)[-sin(α+4π)]
=cos(α+π)*(-sinα)/cos(π+α)(-sinα)
=1
(2)[cos(α-π)/sin(5π/2+α)]*sin(α-2π)*cos(2π-α)
=[(-cosα)/sin(π/2+α)]*sinα*cos(-α)
=[(-cosα)/cos(-α)]*sinα*cosα
=-sinα*cosα