sin(π/4-a)=3/5,sin(π/4+b)=12/13,其中0

问题描述:

sin(π/4-a)=3/5,sin(π/4+b)=12/13,其中0

数学人气:463 ℃时间:2020-07-23 12:18:17
优质解答
sin(π/4-a)=3/5
sin(a-π/4)=-3/5
0cos(a-π/4)=4/5
0cos(π/4+b)=-5/13
cos(a+b)=cos[(a-π/4)+(π/4+b)]
=cos(a-π/4)cos(π/4+b)-sin(a-π/4)sin(π/4+b)
=(4/5)*(-5/13)-(-3/5)*(12/13)
=16/65
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sin(π/4-a)=3/5
sin(a-π/4)=-3/5
0cos(a-π/4)=4/5
0cos(π/4+b)=-5/13
cos(a+b)=cos[(a-π/4)+(π/4+b)]
=cos(a-π/4)cos(π/4+b)-sin(a-π/4)sin(π/4+b)
=(4/5)*(-5/13)-(-3/5)*(12/13)
=16/65