已知sin(π/4+2a)*sin(π/4-2a)=1/4,a∈(π/4,π/2) 求2sin^a+tan-1/tan-1
问题描述:
已知sin(π/4+2a)*sin(π/4-2a)=1/4,a∈(π/4,π/2) 求2sin^a+tan-1/tan-1
sin(π/4+2a)sin(π/4-2a) =1/4
2sin(π/4+2a)cos(π/4+2a)=1/2
sin[2(π/4+2a)]=1/2
sin(π/8+4a)]=1/2
sin4a=1/2
接下来怎么化简。
答
a∈(π/4,π/2)2a∈(π/2,π)cos2a04a∈(π,2π)π/2+4a∈(3π/2,5π/2)sin(π/4+2a)sin(π/4-2a) =1/42sin(π/4+2a)(-cos(π/4+2a))=1/2-sin[2(π/4+2a)]=1/2-sin(π/2+4a)]=1/2cos4a=1/2(你的问题补充到此步有误)...