cos(π/4+x)=3/5,则sin(π/4-x)=?
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cos(π/4+x)=3/5,则sin(π/4-x)=?
cos(π/4+x)=3/5,则sin(π/4-x)=?
cos(π/4+x)=3/5,则sin(π/4-x)=?
sin(π/4-x)=cos【π/2-(π/4-x)]=cos(π/4+x)=3/5
这里用了公式互余的角正弦=余弦
sin(π/4-x) =cos[π/2-(π/4-x)] =cos(π/4+x) =3/5
sin(3π/4+x)sin(3π/4-x)
=﹙sin3π/4cosx+cos3π/4sinx﹚﹙sin3π/4cosx-cos3π/4sinx﹚
=﹙√2/2cosx-√2/2sinx﹚﹙√2/2cosx+√2/2sinx﹚
=﹙√2/2﹚²﹙cosx-sinx﹚﹙cosx+sinx﹚
=½﹙cos²x-sin²...
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sin(3π/4+x)sin(3π/4-x)
=﹙sin3π/4cosx+cos3π/4sinx﹚﹙sin3π/4cosx-cos3π/4sinx﹚
=﹙√2/2cosx-√2/2sinx﹚﹙√2/2cosx+√2/2sinx﹚
=﹙√2/2﹚²﹙cosx-sinx﹚﹙cosx+sinx﹚
=½﹙cos²x-sin²x﹚
=½cos2x=3/10
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