已知sinα=2⼀3,求tanα⼀2的值

2025-04-10 23:17:36
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回答(1):

解:∵sinα/2=±√[(1-cosα)/2], cosα/2=±√[(1+cosα)/2].
tamα/2=(sinα/2)/(cosα/2).
=±√[(1-cosα)/2]/√[(1+cosα)/2].
=±√[(1-cosα)/(1+cosα)].
=√[(1-cosα)^2/(1+cosα)(1-cosα)].
=√[(1-cosα)^2/(1-cos^2α). 【1-cosα≥0, 1-cos^2α=sin^2α】
=(1-cosα)/(sinα).
∵cosα=√(1-sin^2α)=√[1-(2/3)^2]=√5/3.
tan(α/2)=(√5/3)/(2/3).
∴tan(α/2)=√5/2. 【由sinα=2/3,得α=41°即α为锐角,tanα/2>0】