已知sin(π/4-x)=5/13,x属于(0,π/4),求(cos2x)/cos[(π/4)+x]谢啦,急用
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![已知sin(π/4-x)=5/13,x属于(0,π/4),求(cos2x)/cos[(π/4)+x]谢啦,急用](/uploads/image/z/1618657-25-7.jpg?t=%E5%B7%B2%E7%9F%A5sin%28%CF%80%2F4-x%29%3D5%2F13%2Cx%E5%B1%9E%E4%BA%8E%280%2C%CF%80%2F4%29%2C%E6%B1%82%28cos2x%29%2Fcos%5B%28%CF%80%2F4%29%2Bx%5D%E8%B0%A2%E5%95%A6%2C%E6%80%A5%E7%94%A8)
已知sin(π/4-x)=5/13,x属于(0,π/4),求(cos2x)/cos[(π/4)+x]谢啦,急用
已知sin(π/4-x)=5/13,x属于(0,π/4),求(cos2x)/cos[(π/4)+x]
谢啦,急用
已知sin(π/4-x)=5/13,x属于(0,π/4),求(cos2x)/cos[(π/4)+x]谢啦,急用
x属于(0,π/4)
0<π/4-x<π/4
所以cos(π/4-x)>0
[sin(π/4-x)]^2+[cos(π/4-x)]^2=1
所以cos(π/4-x)=12/13
sin(π/2-2x)=2sin(π/4-x)cos(π/4-x)=120/169
所以cos2x=sin(π/2-2x)=120/169
cos(π/4+x)=sin[π/2-(π/4+x)]=sin(π/4-x)=5/13
所以
(cos2x)/cos(π/4+x)
=(120/169)/(5/13)
=24/13
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