搜狗做题网求证:tan(30°+α/2)tan(30°-α/2)=(2cosα-1)/(2cosα+1)
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求证:tan(30°+α/2)tan(30°-α/2)=(2cosα-1)/(2cosα+1)
RT
RT
左边=[ (tan30º+ tanα/2) / (1-tan30º tanα/2)][(tan30º- tanα/2)/(1+tan30º tanα/2)]
=[(tan30º)^2-(tanα/2)^2] / [1-(tan30º tanα/2)^2]
=[1/3-(tanα/2)^2] / [1-(tanα/2)^2/3]
=[1-3(tanα/2)^2] / [3-(tanα/2)^2]
=[(cosα/2)^2-3(sinα/2)^2] / [3(cosα/2)^2-(sinα/2)^2]
=[(1/2)[(1+cosα)-3(1-cosα)] / {[(1/2)[3(1+cosα)-(1-cosα)]}
=[(1+cosα)-3(1-cosα)] / [[3(1+cosα)-(1-cosα)]
=[4cosα-2] / [4cosα+2]
=(2cosα-1) / (2cosα+1)
=右边
=[(tan30º)^2-(tanα/2)^2] / [1-(tan30º tanα/2)^2]
=[1/3-(tanα/2)^2] / [1-(tanα/2)^2/3]
=[1-3(tanα/2)^2] / [3-(tanα/2)^2]
=[(cosα/2)^2-3(sinα/2)^2] / [3(cosα/2)^2-(sinα/2)^2]
=[(1/2)[(1+cosα)-3(1-cosα)] / {[(1/2)[3(1+cosα)-(1-cosα)]}
=[(1+cosα)-3(1-cosα)] / [[3(1+cosα)-(1-cosα)]
=[4cosα-2] / [4cosα+2]
=(2cosα-1) / (2cosα+1)
=右边
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