已知tanα=2,求cos(α-7π/2)+2sin(π-3α)/csc(π+α)+sec(π/2+α)的值
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已知tanα=2,求cos(α-7π/2)+2sin(π-3α)/csc(π+α)+sec(π/2+α)的值
我先化简了 可还做不出
我先化简了 可还做不出
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cos(α-7π/2)+2sin(π-3α)/csc(π+α)+sec(π/2+α)
=cosα cos(7π/2) + sinα sin(7π/2) + 2[sinπ cos(3α) - cosπsin(3α)]sin(π+α)+ 1/cos(π/2+α)
=cosα cos(π/2) + sinα sin(π/2) + 2[sinπ cos(3α) - cosπsin(3α)](sinπcosα +cosπsinα)
+ 1/[cos(π/2)cosα - sin(π/2)sinα ]
=0+sinα+2sin(3α) (-sinα) + 1/[0 - sinα ]
= sinα-2sinαsin(3α) - 1/sinα
=sinα-2sinαsin(α+2α) - 1/sinα
=sinα-2sinα(sinαcos2α+cosαsin2α) - 1/sinα
=sinα-2sinα[sinα(1-2sin²α)+2cos²αsinα] - 1/sinα
=sinα-2sinα[sinα-2sin³α+2cos²αsinα] - 1/sinα
tanα=2 sinα/cosα = 2 sin²α/cos²α =4
sin²α+cos²α =1
sin²α = 4/5
cos²α =1/5
sinα= 2√5/5 ,cosα =√5/5
或sinα= -2√5/5 ,cosα = -√5/5
原式=自己算
cos(α-7π/2)+2sin(π-3α)/csc(π+α)+sec(π/2+α)
=cosα cos(7π/2) + sinα sin(7π/2) + 2[sinπ cos(3α) - cosπsin(3α)]sin(π+α)+ 1/cos(π/2+α)
=cosα cos(π/2) + sinα sin(π/2) + 2[sinπ cos(3α) - cosπsin(3α)](sinπcosα +cosπsinα)
+ 1/[cos(π/2)cosα - sin(π/2)sinα ]
=0+sinα+2sin(3α) (-sinα) + 1/[0 - sinα ]
= sinα-2sinαsin(3α) - 1/sinα
=sinα-2sinαsin(α+2α) - 1/sinα
=sinα-2sinα(sinαcos2α+cosαsin2α) - 1/sinα
=sinα-2sinα[sinα(1-2sin²α)+2cos²αsinα] - 1/sinα
=sinα-2sinα[sinα-2sin³α+2cos²αsinα] - 1/sinα
tanα=2 sinα/cosα = 2 sin²α/cos²α =4
sin²α+cos²α =1
sin²α = 4/5
cos²α =1/5
sinα= 2√5/5 ,cosα =√5/5
或sinα= -2√5/5 ,cosα = -√5/5
原式=自己算
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