已知90小於b小於a小於135,cos(a-b)=12/13,sin(a b)=-3/5,求sin2a 網友廻答:
- 90所以0又因爲sin²(a-b) cos²(a-b)=1
所以sin(a-b)=5/13;
同理cos(a b)=-4/5
sin(2a)=sin(a-b a b)
=sin(a-b)cos(a b) cos(a-b)sin(a b)
=5/13*(-4/5) 12/13 (-3/5)
=-56/65
網友廻答:
- Sin(a-b)=1-(12/13)^2=5/13=Sinacosa-sinbcosb 1式 sin(a b)=sinacosa sinbcosb=-3/5, 2式 1式 2式得 2sinacosa=-14/65
網友廻答:
- 9090180所以a b在第三象限
cos(a b)sin²(a b) cos²(a b)=1
cos(a b)=-4/5
-13590-45a>b,a-b>0
所以0所以sin(a-b)>0
sin²(a-b) cos²(a-b)=1
sin(a-b)=5/13
sin2a=sin[(a b) (a-b)]
=sin(a b)cos(a-b) cos(a b)sin(a-b)
=-56/65
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