Find the function y = 1 2cos2x+ Three The minimum positive period, maximum value and minimum value of 2sinxcosx + 1

Find the function y = 1 2cos2x+ Three The minimum positive period, maximum value and minimum value of 2sinxcosx + 1

Function y = 1
2cos2x+
Three
2sinxcosx+1=1+cos2x
4+
Three
4sin2x+1=1
2 sin（2x+π
6）+5
4，
Therefore, the minimum positive period of the function T = 2 π
2 = π with a maximum of 1
2+5
4=7
4, the maximum value is − 1
2+5
4=3
4．

What is the minimum period of the function y = 2cos squared 2x

y=2cos^2(2x)=(1+cos4x)
T = 2 π / 4 = π / 2 -- the minimum period of the function

If the minimum positive period of the function f (x) = 1-2 cos 2 ω x is twice the minimum positive period of the function g (x) = cos4x, then ω=

The minimum positive period of G (x) = cos4x is
T=2π/4=π/2
The minimum positive period of F (x) = 1-2 cos 2 ω x is twice that of the function g (x) = cos4x
That is, the minimum positive period of function f (x) is
T=π
And f (x) = 1-2cos? ω x = - (2cos? ω x-1) = - cos2wx
That is, t = 2 π / / 2W / = π, so w = ± 1
(/ 2W / is absolute)
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The minimum positive period of the function y = SiNx + 2cos? X / 2

Y=sinx+2cos²(x/2)
=sinx+cosx+1
=√2sin(x+π/4)+1
So the minimum positive period is t = 2 π / 1 = 2 π
If you don't understand, I wish you a happy study!

Find the maximum value of the function y = sinxcosx-2cos ^ 2x

0

0

y=(2cos²x-1)+1-2sinxcosx
=-sin2x+cos2x+1
=-(sin2x-cos2x)+1
=-√2sin(2x-π/4)+1
So t = 2 π / 2 = π

Find the period of the following function (1) y = 2sinx + 3 (2) y = 2Sin (π / 3 + X / 2) (3) y = 2cos3x (4) y = - 2cos (1 / 2x + π / 3) Find the period of the following function (1) y = 2sinx + 3 ⑵ y=2sin(π/3 + x/2) ⑶ y=2cos3x ⑷ y=-2cos(1/2x + π/3） Need to have process, thank you!

The period of y = asin (ω x + φ) + B or y = ACOS (ω x + φ) + B is t = 2 π / ω
(1)T＝2π
(2)T＝2π/(1/2)＝4π
(3)T＝2π/3
(4)T＝2π/(1/2)＝4π

Let f (x) = cos (2X-4 π / 3) + 2cos? X (1) to find the maximum value of F (x), and write the solution set of X which makes f (x) take the maximum value (2) It is known that the opposite sides of angles a, B and C in Δ ABC are a, B, C respectively. If f (B + C) = 3 / 2, B + C = 2, find the minimum value of A

（1）、
f(x)=cos2xcos4π/3+sin2xsin4π/3+cos2x+1
=-1 / 2cos2x radical 3 / 2sin2x + cos2x + 1
=1 / 2cos2x radical 3 / 2sin2x + 1
=cos（2x+π/3）+1
The maximum value of F (x) is 2
Let 2K π - π / 2 < 2x + π / 3 < 2K π + π / 2
K π - 5 π / 12 < x < K π + π / 12
The solution set {X / K π - 5 π / 12 ＜ x ＜ K π + π / 12, K ∈ r}
(2) It is known that f (B + C) = 3 / 2
Then f (B + C) = cos [2 (B + C) + π / 3] + 1 = 3 / 2
cos[2（B+C）+π/3]=1/2
Because cosine values are positive in the first and fourth quadrants, 2 (B + C) + π / 3 = π / 3 (round off) or 2 π - π / 3
B + C = 2 π / 3, a = π / 3
From cosine theorem
a²=b²+c²-2bccosA=（b+c）²-2bc-2bccosπ/3
Because B + C = 2, then (B + C) 2 = 4
B + C ≥ 2 root sign BC, i.e. 0 < BC ≤ 1
a²=（b+c）²-2bc-2bccosπ/3
≥4-2-2×1/2=1
The minimum value of a is 1

Let the function FX = cos (2X-4 / 3) + 2cos ^ 2x (1) to find the maximum value of FX, the set of X (2 If f (B + C) = 3 / 2. B + C = 2, find the minimum value of A

When cos (2x-2 π / 3) = 1,1,2x-2-2 π / 3 = 2K π + 1 = 2cos (2x-2 π / 3) Cos2 π / 3 + 1 = 2cos (2x-2 π / 3) cos (2x-2 π / 3) 1.when cos (2x-2 π / 3) = 1,1,2x-2 π / 3 = 2K π + π, x = k π + 5 π / 6F (x) max = 1 + √ 32.b + C = π - AF (AF) is (1 + √ 32), B + C = π - AF (C = π - AF) is (1) + 32.b + C = π - AF (C = π - AF (AF) is a (x) B + C) = 1 - √ 3cos (2 (π - a) - 2 π / 3) = 3 / 2

The minimum positive period of the function y = 2sinxcosx-2sin2x + 1 is () A. π Four B. π Two C. π D. 2π

The function f (x) = 2sinxcosx-2sin2x + 1 = sin2x + cos2x=
2sin（2x+π
4）．
So the minimum positive period of the function: T = 2 π
2=π；
Therefore, C