已知cos(π/2-a)=2*1/2 cos(3π/2 B),3*1/2 sin(3π/2-a)=-2*1/2 sin(π/2 B),求a,b的值 網友廻答:
- 先化簡方程一:
cos(π/2-A)=2*1/2 cos(3π/2 B)
cos(π/2-A)= cos[2π-(π/2-B)]
sinA= cos(π/2-B)
sinA= sinB.(1)
再化簡方程二:
3*1/2 sin(3π/2-A)=-2*1/2 sin(π/2 B)
3 sin[2π-(π/2 A)]=-2 sin(π/2 B)
-3 sin(π/2 A)=-2 sin(π/2 B)
3 sin(π/2 A)=2 sin(π/2 B)
3 sin[π-(π/2-A)]=2 sin[π-(π/2-B)]
3 sin(π/2-A)=2 sin(π/2-B)
3 cosA=2 cosB
cosA=2/3 cosB.(2)
(1)^2 (2)^2:
sin^2A cos^2A=sin^2B 4/9 cos^2B.(3)
1=5/9 sin^2B 4/9
5/9sin^2B=5/9
sinB=±1.(4)
將(4)代入(1)得:
sinA=±1
∴A=B=2kπ±π/2
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