已知cos(a+π/4)=3/5,π/2

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已知cos(a+π/4)=3/5,π/2已知cos(a+π/4)=3/5,π/2已知cos(a+π/4)=3/5,π/2因为cos(a+π/4)=3/5>0,a∈[π/2,3π/2]a+π/4∈[3π

已知cos(a+π/4)=3/5,π/2
已知cos(a+π/4)=3/5,π/2

已知cos(a+π/4)=3/5,π/2
因为cos(a+π/4)=3/5>0,a∈ [π/2,3π/2]
a+π/4∈ [3π/4,7π/4],
所以a∈ [5π/4,3π/2]
cos(2a+π/2)=2[cos(a+π/4)]^2-1=-7/25
a∈ [5π/4,3π/2]
2a+π/2∈ [3π,7π/2]
sin(2a+π/2)=-24/25
cos(2a+π/4)=cos(2a+π/2-π/4)
=cos(2a+π/2)cosπ/4+sin(2a+π/2)sinπ/4
=(-7/25-24/25)(根号2)2
=-31倍(根号2)/50

π/2<=a<3π/2
所以3π/4<=a+π/4<7π/4
cos(a+π/4)>0
3π/2即SIN(a+π/4)=-4/5 小于0 由cos(a+π/4)=3/5和SIN(a+ π/4)=-4/5可以算COS(2a+π/2)=-7/25即SIN(2a)=7/25
因为π/2<=a<3π/2
所以π<2a<3π即...

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π/2<=a<3π/2
所以3π/4<=a+π/4<7π/4
cos(a+π/4)>0
3π/2即SIN(a+π/4)=-4/5 小于0 由cos(a+π/4)=3/5和SIN(a+ π/4)=-4/5可以算COS(2a+π/2)=-7/25即SIN(2a)=7/25
因为π/2<=a<3π/2
所以π<2a<3π即-π<2a<π cos2a=-24/25或24/25
代入cos(2a+π/4)得原式=(17根号2)/50 或 (-31根号2)/50

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