TAN(a+π/4)=1/2 -π/2<A<0 求(2SINA^2+SIN2A)/COS(A-π/4) 的值,
问题描述:
TAN(a+π/4)=1/2 -π/2<A<0 求(2SINA^2+SIN2A)/COS(A-π/4) 的值,
答
tan(A+π/4) =1/2(tanA+1)/(1-tanA)=1/22tanA+2 =1-tanAtanA= -1/3sinA = -1/√10cosA = 3/√10sin2A = 2sinAcosA = -3/5cos(A-π/4) = cosAcos(π/4)+sinAsin(π/4)=(1/√2)(2/√10) =1/√5[2(sinA)^2+sin2A]/ cos(...