设sinβ+cosβ=1/2,求(1)sin2β,(2)cos²(π/4+β)- sin²(π/4+
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设sinβ+cosβ=1/2,求(1)sin2β,(2)cos²(π/4+β)- sin²(π/4+β)的值
sinβ+cosβ=1/2
sin²β+cos²β+2sinβcosβ=1/4
sin2β=1/4-1=-3/4
2、cos²(π/4+β)- sin²(π/4+β)
=[cos(π/4+β)+sin(π/4+β)][cos(π/4+β)- sin(π/4+β)]
=[cosπ/4cosβ-sinπ/4sinβ+sinπ/4cosβ+cosπ/4sinβ][cosπ/4cosβ-sinπ/4sinβ-sinπ/4cosβ-cosπ/4sinβ]
=√2/2×√2/2×[cosβ-sinβ+cosβ+sinβ][cosβ-sinβ-cosβ-sinβ]
=1/2×[2cosβ][-2sinβ]
=-2sinβcosβ
=-sin2β
=-(-3/4)
=3/4
sin²β+cos²β+2sinβcosβ=1/4
sin2β=1/4-1=-3/4
2、cos²(π/4+β)- sin²(π/4+β)
=[cos(π/4+β)+sin(π/4+β)][cos(π/4+β)- sin(π/4+β)]
=[cosπ/4cosβ-sinπ/4sinβ+sinπ/4cosβ+cosπ/4sinβ][cosπ/4cosβ-sinπ/4sinβ-sinπ/4cosβ-cosπ/4sinβ]
=√2/2×√2/2×[cosβ-sinβ+cosβ+sinβ][cosβ-sinβ-cosβ-sinβ]
=1/2×[2cosβ][-2sinβ]
=-2sinβcosβ
=-sin2β
=-(-3/4)
=3/4
设sinβ+cosβ=1/2,求(1)sin2β,(2)cos²(π/4+β)- sin²(π/4+
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