若sin(π/6+α)=3/5 α∈(π/3,5π/6) 求tan(5π/12+α)
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若sin(π/6+α)=3/5 α∈(π/3,5π/6) 求tan(5π/12+α)
∵sin(π/6+α)=3/5,α∈(π/3,5π/6) ,∴π/6+α∈(π/2,π),∴cos(π/6+α)=√1-[sin(π/6+α)]^2=√(1-9/25)=-4/5,∴tan(π/6+α)=sin(π/6+α)/cos(π/6+α)=-3/4,
tan(5π/12+α)=tan[(π/6+α)+π/4]=[tan(π/6+α)+tanπ/4]/[1-tan(π/6+α)tanπ/4]=(1-3/4)/(1+3/4)=1/7.
tan(5π/12+α)=tan[(π/6+α)+π/4]=[tan(π/6+α)+tanπ/4]/[1-tan(π/6+α)tanπ/4]=(1-3/4)/(1+3/4)=1/7.
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