已知cos(π/4+x)=5/13,0<x<π/4,求sin(π/4-x)/cos2x
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已知cos(π/4+x)=5/13,0<x<π/4,求sin(π/4-x)/cos2x
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![已知cos(π/4+x)=5/13,0<x<π/4,求sin(π/4-x)/cos2x](/uploads/image/z/1765156-4-6.jpg?t=%E5%B7%B2%E7%9F%A5cos%28%CF%80%2F4%2Bx%29%3D5%2F13%2C0%EF%BC%9Cx%EF%BC%9C%CF%80%2F4%2C%E6%B1%82sin%28%CF%80%2F4-x%29%2Fcos2x)
因为π/4+x和π/4-x是互余的(和是π/2),
所以cos(π/4+x)=sin(π/4-x)=5/13
因为0<x<π/4
所以sin(π/4+x)=12/13 (因为π/4+x为第一象限角)
cos2x=sin(π/2+2x)
=2*sin(π/4+x)*cos(π/4+x)
=2×(5/13)×(12/13)
=120/169
所以sin(π/4-x)/cos2x
=(5/13)/(120/169)
=13/24
所以cos(π/4+x)=sin(π/4-x)=5/13
因为0<x<π/4
所以sin(π/4+x)=12/13 (因为π/4+x为第一象限角)
cos2x=sin(π/2+2x)
=2*sin(π/4+x)*cos(π/4+x)
=2×(5/13)×(12/13)
=120/169
所以sin(π/4-x)/cos2x
=(5/13)/(120/169)
=13/24
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