高中数学:已知α∈(0,π),sinα+cosα=7/13,则tanα=?
已知α∈(0,π),sinα+cosα=7/13,则tanα=?已知sin(π+α)=3/5,α∈(π/-2,0),则tanα=?已知函数:f(x)=2cos²...
已知α∈(0,π),sinα+cosα=7/13,则tanα=?
已知sin(π+α)=3/5,α∈(π/-2,0),则tanα=?
已知函数:f(x)=2cos²x=2√3 sinxcosx-1.x∈R
(1)求f(π/3)的值
(2)求函数f(x)的单调递增区间 展开
已知sin(π+α)=3/5,α∈(π/-2,0),则tanα=?
已知函数:f(x)=2cos²x=2√3 sinxcosx-1.x∈R
(1)求f(π/3)的值
(2)求函数f(x)的单调递增区间 展开
3个回答
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sinα+cosα=7/13......(1)
1+2sinαcosα=49/169
2sinαcosα=-120/169
sinα-cosα=√[(sinα+cosα)^-4sinαcosα]=√[49/169+240/169]=17/13
sinα-cosα=17/13....(2)
解(1)(2)得
sinα= 12/13 , cosα=-5/13
tanα=sinα/ cosα=-12/5
2)sin(π+α)=3/5,α∈(π/-2,0),
sinα= -3/5
cosα=4/5
tanα=sinα/ cosα=-3/4
3)f(x)=2cos²x+2√3 sinxcosx-1.
=√3 sin2x+cos2x
=2sin(2x+π/6)
f(π/3)=2sin(2π/3+π/6)=2sin5π/6=2sinπ/6=1
2x+π/6单调递增区间[2kπ-π/2,2kπ+π/2]
x在[kπ-π/3,kπ+π/6]单调递增区间
1+2sinαcosα=49/169
2sinαcosα=-120/169
sinα-cosα=√[(sinα+cosα)^-4sinαcosα]=√[49/169+240/169]=17/13
sinα-cosα=17/13....(2)
解(1)(2)得
sinα= 12/13 , cosα=-5/13
tanα=sinα/ cosα=-12/5
2)sin(π+α)=3/5,α∈(π/-2,0),
sinα= -3/5
cosα=4/5
tanα=sinα/ cosα=-3/4
3)f(x)=2cos²x+2√3 sinxcosx-1.
=√3 sin2x+cos2x
=2sin(2x+π/6)
f(π/3)=2sin(2π/3+π/6)=2sin5π/6=2sinπ/6=1
2x+π/6单调递增区间[2kπ-π/2,2kπ+π/2]
x在[kπ-π/3,kπ+π/6]单调递增区间
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1 ;α∈(0,π),sinα+cosα=7/13, sinα=7/13-cosα, sin^2α=49/169-14/13cosα+cos^2α
cos^2α-7/13cosα-60/169=0, cosα=-5/13或12/13, sinα=12/13或5/13, tanα=-12/5或5/12
2; sin(π+α)=3/5,α∈(π/-2,0),sinα=-3/5, cosα=4/5, tanα=-3/4
3; f(x)=2cos²x+2√3 sinxcosx-1=cos2x+√3sin2x=2sin(2x+π/6)
(1) f(π/3)=1
(2) ,2kπ-π/2<2x+π/6<2kπ+π/2, kπ-π/3<x<kπ+π/6 ,k∈z,单调递增区间(kπ-π/3,kπ+π/6)
cos^2α-7/13cosα-60/169=0, cosα=-5/13或12/13, sinα=12/13或5/13, tanα=-12/5或5/12
2; sin(π+α)=3/5,α∈(π/-2,0),sinα=-3/5, cosα=4/5, tanα=-3/4
3; f(x)=2cos²x+2√3 sinxcosx-1=cos2x+√3sin2x=2sin(2x+π/6)
(1) f(π/3)=1
(2) ,2kπ-π/2<2x+π/6<2kπ+π/2, kπ-π/3<x<kπ+π/6 ,k∈z,单调递增区间(kπ-π/3,kπ+π/6)
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1、两边平方,可得sinacosa=-60/169<0,所以a在(π/2,π),后面自己解方程算算就可以
2、sina=-3/5,所以tana=-3/4
3、?
2、sina=-3/5,所以tana=-3/4
3、?
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