帮我解一道简单的“三角函数”题

cos(2a)=cos(a+b+a-b)=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)

又因为ab的取值范围， 所以sin(a+b)=4/5 sin(a-b)=5/13
所以cos(2a)=3/5*12/13-4/5*5/13=16/65

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cos(a+b)=3/5 ，sin(a+b)=4/5
cos(a-b)=12/13
sin(a-b)=5/13
cos2a=cos(a+b+a-b)
=cos(a+b)*cos(a-b)-sin(a+b)*sin(a-b)
=16/65

cos2a=cos(a+b+a-b)=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=3/5 *12/13- 4/5 *5/13
=16/65

∵cos(a+b)=3/5 则a+b=43°或-43°
∵0<b<a<π/2
∴a+b=43° sin(a+b)=4/5
cos(a-b)=12/13 ∴sin(a-b)=+/-5/13
而a>b ∴sin(a-b)=5/13
∴cos(2a)=cos[(a+b)+(a-b)]=3/5*12/13-4/5*5/13=16/65

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