Question

急求解——已知tan（派/4+a）=1/3,求tanα的值,和sin2α-cos^2α/1+cos2α的值

已知tana/2=-3,求(sina-3cosa)/(sina+cosa)的值
已知函数f(x)=2sinx(sinx+cosx),求f(x)的最小周期和f(x)的单调递增区间

答案满意我会追加悬赏的，急——

Answer 1

1. tan（派/4+a）=(tanπ/4+tana)/(1-tanπ/4*tana)=1/3
即(1+tana)/(1-tana)=1/3
解得 tana=-1/2
(sin2a-cos²a)/(1+cos2a)
=cosa(2sina-cosa)/(2cos²a)
=tana-1/2
=-1/2-1/2
=-1
2. tan(a/2)=-3
tana=[2tan(a/2)]/[1-tan²(a/2)]=2(-3)/(1-9)=3/4
代入(sina-3cosa)/(sina+cosa)
=[(sina/cosa)-3]/[(sina/cosa)+1]
=(tana-3)/(tana+1)
=(3/4-3)/(3/4+1)
=-9/7
3. f(x)=2sin²x+2sinxcosx
=sin2x+1-cos2x
=√2[(√2/2)sin2x-(√2/2)cos2x]+1
=√2sin(2x-π/4)+1
所以最小正周期T=2π/2=π
单调递增区间2x-π/4∈[2kπ-π/2, 2kπ+π/2]
x∈[kπ-π/8, kπ+3π/8] k∈Z
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

已知tana/2=-3,求(sina-3cosa)/(sina+cosa)的值
tana=2tana/2/(1-tan^2 a/2)=(-6)/(-8)=3/4
(sina-3cosa)/(sina+cosa)
=（tana-3)/(tana+1)
=(-9/4)/(7/4)
=-9/7

已知函数f(x)=2sinx(sinx+cosx),求f(x)的最小周期和f(x)的单调递增区间
f(x)=2sinx(sinx+cosx),
=2sin^2 x+2sinx cosx
=1-cos^2 x+sin^2 x+sin2x
=1-cos2x+sin2x
=1-√2sin(45度-2x)
周期是π
x∈[kπ-π/8, kπ+3π/8] 是增区间

Answer 3

tan(π/4+a)=(tanπ/4+tana)/ (1-tanπ/4tana)=1/3
tana=-1/2

tana/2=-3 tana=2tana/2/(1-tan²a/2)=3/4
=(tana-3)/(tana+1)
=-9/7

f(x)=2sin²x+2sinxcosx=1-cos2x+sin2x=√2sin(2x+π/4)+1
T=2π/2=π
-π/2+2kπ<2x+π/4<π/2+2kπ