半小时内回答加三十分】高一数学题已知tan(四分之派+α)=2.1.求tanα的值 2.求(2sin²α+sin2α)/(1+tanα)的值
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半小时内回答加三十分】高一数学题已知tan(四分之派+α)=2.1.求tanα的值 2.求(2sin²α+sin2α)/(1+tanα)的值
半小时内回答加三十分】高一数学题
已知tan(四分之派+α)=2.1.求tanα的值 2.求(2sin²α+sin2α)/(1+tanα)的值
半小时内回答加三十分】高一数学题已知tan(四分之派+α)=2.1.求tanα的值 2.求(2sin²α+sin2α)/(1+tanα)的值
tan(A+B) = (tanA+tanB)/(1-tanAtanB) 且 tan(四分之派) = 1
tan(四分之派+a)=2 =>
(tan(四分之派)+ tan(a))/(1-tan(四分之派)tan(a)) = 2 =>
(1+tan(a))/(1-tan(a)) = 2 =>
1+tan(a)=2-2tan(a) =>
3tan(a)=1 =>
tan(a) = 1/3
第一问和答案是 tan(a)的值是1/3
第二问:(2sin²α+sin2α)/(1+tanα)
=(2sinαsinα+2sinαcosα)/(1+tanα)
= 2sinαcosα(sinα/cosα+1)/(1+tanα)
= sin2α(tanα+1)/(1+tanα)
= sin2α
依据万能公式 sinα=[2tan(α/2)]/{1+[tan(α/2)]^2} 可得
sin2α=2tanα/(1+tanαtanα)=2*(1/3)/(1+(1/3)*(1/3))= 2/3/(10/9)=6/10=3/5
所以第二问的得值为3/5
原式=1+tana/(1-tana)=2
得tana=1/3
2.(2sina*sina+2sina*cosa)/(1+tana)
=[(2+2/tana)/(1+1/tana)]/(1+tana)
=3/2