若tan(θ+π/4)=2,则tan(2θ+π/4)等于?求详细过程!
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若tan(θ+π/4)=2,则tan(2θ+π/4)等于?
求详细过程!
tan(θ+π/4)
=[tanθ+tan(π/4)]/[1-tanθtan(π/4)]
=(tanθ+1)/(1-tanθ)
=2
即tanθ+1=2(1-tanθ)
解得:tanθ=1/3
tan(2θ+π/4)
=tan[θ+(θ+π/4)]
=[tanθ+tan(θ+π/4)]/[1-tanθtan(θ+π/4)]
=(1/3+2)/[1-(1/3)*2]
=(7/3)/(1/3)
=7
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tanπ/4=1
所以tan(θ+π/4)=2
则(tanθ+1)/(1-tanθ)=2
解得tanθ=1/3
所以tan2θ=2tanθ/(1-tan²θ)=3/4
所以原式=(tan2θ+1)/(1-tan2θ)=7
结果是7 过程打太费劲了。。。0 0
tan(θ+π/4)=(tanθ+tanπ/4)/(1-tanθtanπ/4)=(tanθ+1)/(1-tanθ)=2
解得:tanθ=1/3
tan2θ=3/4
tan(2θ+π/4)=7
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