sin(π/2 +2a)/cos(π/4 +a)是否=2sin(π/4 +a)

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sin(π/2+2a)/cos(π/4+a)是否=2sin(π/4+a)sin(π/2+2a)/cos(π/4+a)是否=2sin(π/4+a)sin(π/2+2a)/cos(π/4+a)是否=2si

sin(π/2 +2a)/cos(π/4 +a)是否=2sin(π/4 +a)
sin(π/2 +2a)/cos(π/4 +a)是否=2sin(π/4 +a)

sin(π/2 +2a)/cos(π/4 +a)是否=2sin(π/4 +a)
∵sin[π/2 +2a)]=sin[2(π/4 +a)]=2sin(π/4 +a)cos(π/4 +a)
两边同除cos(π/4 +a)得:
sin(π/2 +2a)/cos(π/4 +a)=2sin(π/4 +a)

sin(π/2 +2a)/cos(π/4 +a)
=2sin(π/4 +a) cos(π/4+a)/cos(π/4 +a)
= 2sin(π/4 +a)