1,若θ=15度,则tan(θ/2)+tan(θ/2)*tan(5θ/2)+tan(5θ/2)=?

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1,若θ=15度,则tan(θ/2)+tan(θ/2)*tan(5θ/2)+tan(5θ/2)=?1,若θ=15度,则tan(θ/2)+tan(θ/2)*tan(5θ/2)+tan(5θ/2)=?1,

1,若θ=15度,则tan(θ/2)+tan(θ/2)*tan(5θ/2)+tan(5θ/2)=?
1,若θ=15度,则tan(θ/2)+tan(θ/2)*tan(5θ/2)+tan(5θ/2)=?

1,若θ=15度,则tan(θ/2)+tan(θ/2)*tan(5θ/2)+tan(5θ/2)=?
1,tan(θ/2)+tan(θ/2)*tan(5θ/2)+tan(5θ/2)=1
tan(θ/2+5θ/2)=(tanθ/2+tan5θ/2)÷[1-tanθ/2tan5θ/2]=tan3θ=tan45º=1……①
由①式可以得出tanθ/2+tan5θ/2=1-tanθ/2tan5θ/2……②
又原式为:tanθ/2+tanθ/2tan5θ/2+tan5θ/2,
把②代入原式=1-tanθ/2tan5θ/2+tanθ/2tan5θ/2=1.