已知cotb=√5,sina/sinb=sin(a+b),求cot(a+b)=?
已知cotb=√5,sina/ainb=sin(a+b),求cot(a+b)=?
人气:113 ℃ 时间:2020-10-01 19:16:22
解答
∵sina/sinb=sin(a+b) ==> sina/sinb=sina*cosb+cosa*sinb
==> 1/sinb=cosb+cota*sinb
==> 1/sin²b=cotb+cota
==> csc²b=cotb+cota
==> 1+cot²b=cotb+cota
==> cota=1+cot²b-cotb
又cotb=√5
∴cota=1+5-√5=6-√5
故 cot(a+b)=(cota*cotb-1)/(cota+cotb)
=[(6-√5)*√5-1]/(6-√5+√5)
=6(√5-1)/6
=√5-1.
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