已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于() A.-2009 B.-1/2009 C.1/2009 D.2009
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已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于() A.-2009 B.-1/2009 C.1/2009 D.2009
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于() A.-2009 B.-1/2009 C.1/2009 D.2009
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于() A.-2009 B.-1/2009 C.1/2009 D.2009
【参考答案】D
[sin(π/2-x)+sin(π-x)]/[cos(-x)+sin(2π-x)]=2009
根据诱导公式,化简
(cosx+sinx)/(cosx-sinx)=2009
左边分子分母同时除以cosx:
(1+tanx)/(1-tanx)=2009
∵tan5π/4=tan(π+π/4)=1
∴(tan5π/4+tanx)/(1-tan5π/4tanx)=2009
根据两角和正切公式
tan(x+5π/4)=(tan5π/4+tanx)/(1-tan5π/4tanx)=2009
∴tan(x+5π/4)=2009
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