1樓:愛吃魚與魚
f(b)=1 b=60, a=30 c=90
已知函式f(x)=2√3sinxcosx-2cos²x+1 問函式的最小正週期和單調遞增區間?
2樓:墨書香閣
f(x)=2√3sinxcosx+1-2(sinx)^2=√3sin2x+cos2x=2sin(2x+π/6)所以最小正週期=2π/2=π-π/2+2kπ<=2x+π/6<=π/2+2kπ時遞增,所以遞增區間為[-π/3+kπ,π/6+kπ]
∵f(x)=sinxcosx+sin²x=0.5sin2x+0.5(1-cos²x)=0.5√2sin(2x-π/4)+0.5
∴最小正週期t=2π/2=π
∴當2x-π/4∈[-π/2+2kπ,π/2+2kπ](k∈z)時f(x)單調遞增
解得x∈[-π/8+kπ,3π/8+kπ],(k∈z)∴f(x)單調遞增區間[-π/8+kπ,3π/8+kπ],(k∈z)
已知函式f(x)=2√3sinx×cosx+2cos²x-1(x∈r) 10
3樓:韓增民鬆
已知函式f(x)=2√3sinx×cosx+2cos²x-1(x∈r)
<1>求函式f(x)的最小正週期及在區間[0,π/2]上的最值
<2>若f(x0)=6/5,x0∈[π/4,π/2],求cos2x0的值
(1)解析:∵函式f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6)
∴f(x)的最小正週期為π
f(0)=2sin(π/6)=1,f(π/2)=2sin(π+π/6)=-1,f(π/6)=2sin(π/3+π/6)=2
在區間[0,π/2]上的最大值為2,最小值為-1
(2)解析:∵f(x0)=2sin(2x0+π/6)=6/5==>sin(2x0+π/6)=3/5
∴sin2x0*√3/2+cos2x0*1/2=3/5
√3sin2x0+cos2x0=6/5
sin2x0=2√3/5-√3/3cos2x0==> (sin2x0)^2=(2√3/5-√3/3cos2x0)^2=12/25-4/5(cos2x0)+1/3(cos2x0)^2
代入(sin2x0)^2+(cos2x0)^2=1
4/3(cos2x0)^2-4/5(cos2x0)^2-13/25=0
解得cos2x0=3/10-2√3/5或cos2x0=3/10+2√3/5
4樓:
簡化為f=√3sin2x+cos2x=2sin(2x+π/6)所以f(x)的最小正週期=2π/2=π
2x+π/6 ∈(π/6,7π/6) 所以 最大值為2 最小值為-1
f(x0)=6/5 =2sin(2x0+π/6),sin(2x0+π/6)=3/5,觀察 2x0+π/6 為鈍角 觀察下圖
設x 現有餘弦定理求直角邊為之長度 然後用 勾股定理 算出x 最後cos2x0=-40/x
已知函式f(x)=sin²x+2根號3sinxcosx-cos²x
5樓:小百合
f(x)=sin²x+2√3sinxcosx-cos²x=2√3sinxcosx-(cos²x-sin²x)=√3sin(2x)-cos(2x)
=2sin(2x-π/6)
1)2kπ-π/2≤2x-π/6<2kπ+π/2kπ-π/6≤x 2kπ+π/2≤2x-π/6<2kπ+3π/2kπ+π/3≤x 單增區間:[kπ-π/6,kπ+π/3] 單減區間:[kπ+π/3,kπ+5π/6]2)f(x)最大值為2 sin(2x-π/6)=1 2x-π/6=2kπ+π/2 x=kπ+π/3 已知函式f(x)=2√3sinx·cosx+cos^2x-sin^2x-1(x∈r) 6樓:匿名使用者 答:f(x)=2√3sinxcosx+cos²x-sin²x-1=√3sin2x+cos2x-1 =2*[sinx(√3/2)+cos2x(1/2)-1=2sin(x+π/6)-1 (1)f(x)=2sin(x+π/6)-1的單調增區間滿足:-π/2+2kπ<=x+π/6<=π/2+2kπ 所以:函式的單調增區間為[2kπ-2π/3,2kπ+π/3],k∈z(2)若-5π/12<=x<=π/3 則:-π/4<=x+π/6<=π/2 所以:-√2/2<=sin(x+π/6)<=1所以:-√2<=2sin(x+π/6)<=2所以:-1-√2<=f(x)<=1 所以:f(x)的值域為[-1-√2,1] f x 3sin2x cos2x 2sin 2x 6 sin 2x0 6 3 5 xo屬於 4,2 2x0 6 2 3,7 6 cos 2x0 6 4 5 cos2xo cos 2x0 6 6 4 3 10 3 10 3 4 3 10 解 f x 2 3sinxcosx 2cos 2x 1 3sin... a 1 4,f x 2x 3 x 2 3x 令 f x 2x x 3 0,得駐點 x 1,x2 3 2 兩點均處於指定區間內部 f 2 16 3 4 2 3 2 4 3 f 1 2 3 1 2 3 1 11 6 f 2 16 3 4 2 3 2 8 3 f 3 2 2 27 8 3 9 4 2 3 ... f x 2 sinxsin45 cosxsina45 f x 2sin x 45 當x 2 135 時,f x 最大 所以f x0 2,f 2x0 1,f 3x0 0所以f x0 f 2x0 f 3x0 2 1 f x 2sin x 4 當x 2k 3 4時,f x 最大 所以x0 2k 3 4,f...已知函式f x 2 3sinxcosx 2cos 2x
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