cos(a+π/4)=3/5,a属于[π/2,3π/2),求cos(2a+π/4)值?
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cos(a+π/4)=3/5,a属于[π/2,3π/2),求cos(2a+π/4)值?
cos(a+π/4)=3/5,a属于[π/2,3π/2),求cos(2a+π/4)值?
cos(a+π/4)=3/5,a属于[π/2,3π/2),求cos(2a+π/4)值?
因为cos(a+π/4)=3/5>0,a∈ [π/2,3π/2]
a+π/4∈ [3π/2,7π/4],
所以a∈ [5π/4,3π/2]
cos(2a+π/2)=2[cos(a+π/4)]^2-1=-7/25
a∈ [5π/4,3π/2]
2a+π/2∈ [3π,7π/2]
sin(2a+π/2)=-24/25
cos(2a+π/4)=cos(2a+π/2-π/4)
=cos(2a+π/2)cosπ/4+sin(2a+π/2)sinπ/4
=(-7/25-24/25)(根号2)2
=-31倍(根号2)/50
因为cos(a+π/4)=3/5>0, a属于[π/2,3π/2)
所以3π/2所以sin(a+π/4)<0,sin(a+π/4)=-根号(1-9/25)=-4/5
因此sin(2a+π/2)=2sin(a+π/4)cos(a+π/4)=-24/25,
cos(2a+π/2)=2cos(a+π/4)^2-1=2*9/25-1=-7/25
cos(2a+π/4)=cos[(2a+π/2)-π/4]=(-7/25)(√2/2)+(-24/25)(√2/2)=-31√2/50
两个正弦 、余弦定理的题目1.cos a =1/17,cos (a+b)= -47/51 ,a、b属于(0,π/2),求cos b.2.cos (a+b)=4/5,cos (a-b)= -4/5,a+b属于(7π/4,2π),a-b属于(3π/4,π),求cos 2a.谢八辈祖宗!
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