已知sina+cosa=(3根号下5)⼀5,a∈(0,π⼀4),sin(b-π⼀4)=3⼀5,b∈(π⼀4,π⼀2)

2025-03-26 01:57:58
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回答1:

sina+cosa=(3根号下5)/5 两边平方展开sin2a=4/5cos2a=3/5tan2a=4/3 sin(b-π/4)=3/5 b-π/4∈(0,π/4)cos2a=3/5 2a∈(0,π/2) 2a=b-π/4b=2a-π/4sina+cosa=(3根号下5)/5sinacosa=2/5 a∈(0,π/4),由上面三式得sina=(根号下5)/5cosa= (2根号下5)/5sin4a=2sin2acos2a=2*4/5*3/5= 24/25cos4a=2(cos2a)2-1= -7/25 cos(a+2b)=cos(5a-π/2)=sin5a=sin(a+4a)=sinacos4a+cosasin4a=(根号下5)/5 * (-7/25)+(2根号下5)/5 * 24/25=41根号下5/125