ak25021047
幼苗
共回答了20个问题采纳率:90% 举报
(1) tan2α=
2tanα
1- tan 2 α =
2(
2 -1)
1- (
2 -1) 2 =1
∴ sin(2α+
π
4 )=sin2α•cos
π
4 +cos2α•sin
π
4 =
2
2 (sin2α+cos2α)
=
2
2 ×
2sinα•cosα+(co s 2 α-si n 2 α )
si n 2 α+co s 2 α (分子分母同除以cos 2 α)
=
2
2 ×
2tanα+(1-ta n 2 α)
1+ta n 2 α =1
∴f(x)=2x+1
(2)由(1)得∠A=2α=
π
4 ,而 ∠C=
π
3 ,
根据正弦定理易AB=
BC•sin
π
3
sin
π
4 =
2×
3
2
2
2 =
6 ,
sinB=sin[π-(A+C)]=sin75°=
6 +
2
4
S△ABC=
1
2 ×AB×BC×sinB=
1
2 ×
6 ×2×
6 +
2
4 =
3+
3
2
(3)∵a n+1 =2a n +1,
∴a n+1 +1=2(a n +1)
∵a 1 =1∴数列{a n +1}是以2为首项,2为公比的等比数列.
可得a n +1=2 n ,∴a n =2 n -1,
∴ S n =
2(1- 2 n )
1-2 -n= 2 n+1 -n-2
1年前
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