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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