(1) tanα
cos(α+β)+cos(α-β)=4/5,sin(α+β)+sin(α-β)=3/5
cos(α+β)+cos(α-β)
=cosαcosβ-sinαsinβ+cosαcosβ+sinαsinβ
=2cosαcosβ=4/5
cosαcosβ=2/5
sin(α+β)+sin(α-β)
=sinαcosβ+cosαsinβ+sinαcosβ-cosαsinβ
=2sinαcosβ=3/5
sinαcosβ=3/10
sinαcosβ/cosαcosβ=3/4
sinα/cosα=3/4
tanα=3/4
,(2)2cos²(α/2)
cosα=±3/√(3²+4²)=±4/5
2cos²(α/2)=1+cosα=1±4/5
(2)[2cos²(α/2)-3sinα-1]/{√2sin[α+(π/4)]}
[2cos²(α/2)-3sinα-1]/{√2sin[α+(π/4)]}
=(1+cosα-3sinα-1)/[√2sinαcos(π/4)+√2cosαsin(π/4)]
=(cosα-3sinα)/(sinα+cosα)
=(cosα-3sinα)(sinα-cosα)/[(sinα+cosα)(sinα-cosα)]
=(sinαcosα+3sin²α-cos²α+3sinαcosα)/(sin²α-cos²α)
=(3sin²α-cos²α+4sinαcosα)/(sin²α-cos²α)
=(2sin²α-1+1+sin²α-cos²α+4sinαcosα)/(sin²α-cos²α)
=(-cos2α+1-cos2α+2sin2α)/(-cos2α)
=(2cos2α-2sin2α-1)/cos2α
=2-2tan2α-1/cos2α
=2-2*2tan2α/(1-tan²α)-1/(2cos²α-1)
=2-2*2*(3/4)/(1-9/16)-1/(2*16/25-1)
=-8+3/7
