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Let f (x) = sin2x + root sign 3cos2x (x R). 1. Find the minimum positive period of function f (x) 2. Find the minimum value of function f (x), and point out the value of X at this time
f(x)=√[1^2+(√3)^2]sin(2x+z)=2sin(2x+z)
Where Tanz = √ 3 / 1 = √ 3
z=π/3
So f (x) = 2Sin (2x + π / 3)
So t = 2 π / 2 = π
Obviously, sin (2x + π / 3) = - 1 is the smallest
So 2x + π / 3 = 2K π - π / 2,2x = 2K π - 5 π / 6
So when x = k π - 5 π / 12, f (x) is minimum = - 2
The known function f (x) = sin2x radical 3cos2x 1： Find the value of F (PAI / 3). 2. Find the monotone increasing interval of function f (x). 3. Explain how the image of function f (x) can be obtained from the image of y = SiNx. Please help me -Pai / 2 + 2kpai ＜ 2x Pai / 3 ＜ Pai / 2 + 2kpai ＇ it's not like this, is it?
1. Transform the original formula into 2 (sin2x / 2-radical 3cos2x / 2) = 2Sin (2x Pai / 3)
So f (PAI / 3) = 3 under the radical
2, let - Pai / 2 + 2kpai < 2x Pai / 3 < Pai / 2 + 2kpai
The single increasing interval is - Pai / 12 + KPAI < x < 5pai / 12 + KPAI
3. First move Pai / 3 to the right, then double the abscissa, and finally double the ordinate
I'm sorry, I just made a mistake. I changed it!
Given the function f (x) = Sina + radical 3sinxcosx + 2cosx, X ∈ R (1), how to find the minimum period and monotone interval (2) from sin2x, X ∈ r
The original formula = 1 / 2 - (cos2x) / 2 + (Radix 3sin2x) / 2 + 1 + cos2x = (Radix 3sin2x) / 2 + (cos2x) / 2 + 3 / 2 = sin (2x + π / 6) + 3 / 2, so the minimum positive period T = 2 π / 2 = π
Given the function f (x) = 2sin2xcos2x + (cos2x) ^ 2 - (sin2x) ^ 2, find the minimum positive period of function f (x)
Using the formula: from sin2x = 2sinxcosx, we can get 2sin2xcos2x = sin4x, from cos2x = (cosx) ^ 2 - (SiNx) ^ 2 = cos4xf (x) = sin4x + cos4x = √ 2 (√ 2 / 2sin4x + √ 2 / 2cos4x) = √ 2Sin (4x + π / 4) t = 2 π / w = 2 π / 4 = π / 2
The minimum positive period and maximum value of the function f (x) = (sin2x-cos2x) ^ 2 are A π,1 B π,2 C π/2,2 D π/2,3
f(x)=(sin2x-cos2x)^2
=sin²2x+cos²2x-2sin2xcos2x
=1-sin4x
So the minimum positive period T = 2 π / 4 = π / 2,
When sin4x = - 1, the original function has a maximum value of 2
So choose C
What are the minimum positive periods and maximum values of the square of the function f (x) = (sin2x cos2x)
F(x)=(sin2x-cos2x)²
=(sin2x)²-2sin2xcos2x+(cos2x)²
=1-2sin2xcos2x
=1-sin(4x)
The minimum positive period is 2 π / 4 = π / 2
The maximum value is 1 - (- 1) = 2
When π / 2
When π / 2
Given the function f (x) = 3cos2x + 2cosxsinx + sin2x, find the maximum value and monotone increasing interval of F (x)
f（x）= 3cos2x + 2cosxsinx + sin2x
= 3cos2x + sin2x + sin2x
= 3cos2x + 2sin2x
= √13 sin【2x + arc tan（2/3）】
The maximum value is root 13
Increasing interval: 2K π - π / 2 ≤ 2x + arc Tan (2 / 3) ≤ 2K π + π / 2
kπ - π/4 - （1/2）arc tan（2/3）≤ x ≤ kπ + π/4 +（1/2）arc tan（2/3）
Subtraction interval: 2K π + π / 2 ≤ 2x + arc Tan (2 / 3) ≤ 2K π + 3 π / 2
kπ + π/4 - （1/2）arc tan（2/3）≤ x ≤ kπ + 3π/4 +（1/2）arc tan（2/3）
Given the function = sin2x + 2sinxcosx + 3cos2x, ∈ r the set of independent variables X when finding the maximum value of (1) f (x), and (2) the monotone increasing interval of function f (x)
Y = 2sin2x + 3cos2x = √ 13 * sin (2x + a), where cosa = 2 √ 13 / 13, a = arccos2 √ 13 / 13
When the maximum value is 2x + a = (2k + 1 / 2) TT, K is an integer,
x=(k+1/4)TT+a/2,
When 2x + a ∈ [2ktt, (2k + 1) TT], the function increases monotonically
x∈[kTT+a/2,(k+1/2)TT+a/2]
The monotone increasing interval of the function f (x) = √ 3cos2x-sin2x is
The solution is obtained by F (x) = √ 3cos2x-sin2x
=2（√3/2cos2x-1/2sin2x)
=2（cosπ/6cos2x-sinπ/6sin2x)
=2cos（2x+π/6）
If π + π + 2 is π + 2, then π + 2 is π + 2
When k π + 5 / 12 π ≤ x ≤ K π + 11 / 12 π, K belongs to Z, y = f (x) is an increasing function
In other words, the monotone increasing interval of the function f (x) = √ 3cos2x-sin2x is [K π + 5 / 12 π, K π + 11 / 12 π], K belongs to Z
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