已知cot(a+b)=0求证sin(a+2b)=sina
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已知cot(a+b)=0求证sin(a+2b)=sina
证明:
∵cot(a+b)=cos(a+b)/sin(a+b)=0
又∵sin(a+b) 为分母,不能等0
∴cos(a+b)=0
∵cos(a+b)=cosacosb-sinasinb=0
所以cosacosb=sinasinb
sin(a+2b)=sinacos2b+sin2bcosa
=sina[(cosb)^2-(sinb)^2]+2sinbcosbcosa
=sina[(cosb)^2-(sinb)^2]+2sina(sinb)^2
=sina[(cosb)^2-(sinb)^2+2(sinb)^2]
=sina[(cosb)^2+(sinb)^2]
=sina
∵cot(a+b)=cos(a+b)/sin(a+b)=0
又∵sin(a+b) 为分母,不能等0
∴cos(a+b)=0
∵cos(a+b)=cosacosb-sinasinb=0
所以cosacosb=sinasinb
sin(a+2b)=sinacos2b+sin2bcosa
=sina[(cosb)^2-(sinb)^2]+2sinbcosbcosa
=sina[(cosb)^2-(sinb)^2]+2sina(sinb)^2
=sina[(cosb)^2-(sinb)^2+2(sinb)^2]
=sina[(cosb)^2+(sinb)^2]
=sina
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