Find the value range and minimum positive period of the function f (x) = 2cos (x + PIA / 4) cos (x-pia / 4) + root sign 3sin2x

Find the value range and minimum positive period of the function f (x) = 2cos (x + PIA / 4) cos (x-pia / 4) + root sign 3sin2x

F (x) = 2cos (x + PIA / 4) cos (x-pia / 4) + Radix 3sin2x
=cos(2x)+cos(pi/2)+√3sin2x
=2sin(2x+pi/6)
Range [- 2,2]
T=pi

Find the function y = 2cos (x + π 4)cos(x−π 4) + The range and minimum positive period of 3sin2x

y＝2cos(x+π
4)cos(x−π
4) +
3sin2x
=2(1
2cos2x−1
2sin2x)+
3sin2x
=cos2x+
3sin2x
=2sin(2x+π
6)
The function y = 2cos (x + π)
4)cos(x−π
4) +
The value range of 3sin2x is [- 2,2],
The minimum positive period is π;

Find the function y = 2cos (x + π 4)cos(x−π 4) + The range and minimum positive period of 3sin2x

y＝2cos(x+π
4)cos(x−π
4) +
3sin2x
=2(1
2cos2x−1
2sin2x)+
3sin2x
=cos2x+
3sin2x
=2sin(2x+π
6)
The function y = 2cos (x + π)
4)cos(x−π
4) +
The value range of 3sin2x is [- 2,2],
The minimum positive period is π;

The range and minimum positive period of F (x) = 2cos (x + π / 4) cos (x - π / 4) + √ 3sin2x Thank you. Process

There is a formula: cos (a) cos (b) = 1 / 2 [cos (a + b) + cos (a-b)]
f(x)=2cos(x+π/4)cos(x-π/4)+√3sin2x
=cos(2x)+cos(π/2)+√3sin2x
=cos(2x)+√3sin2x
=2[cos(π/3)cos(2x)+sin(π/3)sin(2x)]
=2cos(2x-π/3)
Range: [- 2,2]
Minimum positive period: π
Landlord, give me more reward~

Let f (x) = a.b-1, where vector a = (sin2x, 2cosx), B = (radical 3, cosx). In triangle ABC, the opposite sides of angles a, B and C are a, B, C, respectively, If f (A / 4) = root 3, a = 2, radical 3, B = 8, find the value of side length C

In fact, this kind of problem has two points. 1: the pure A-point multiplication B is substituted into the formula. 2: the formula obtained after that is converted according to the relationship between the triangle inner angle and 180, and the sine and cosine. Finally, the numerical value is added to get the solution, which is the case for this type of problem
I don't want you to get the so-called standard answer from the Internet, but to get a solution

It is known that f (x) = a times B-1, where vector a = (sin2x, 2cosx), B = (radical 3, cosx), (x belongs to R). In the triangle ABC, the opposite sides of angles a, B and C are a, B, C respectively (1) If the three sides a, B and C are in equal proportion sequence in turn, try to find the value range of angle B and the value range of function f (b) at this time (2) In triangle ABC, if f (a in 4) = root 3, vector AB multiplied by vector AC = 1, calculate the area of triangle ABC

If f (A / 4) = radical 3 π / 61 ≤ f (b) ≤ 22. If f (A / 4) = radical 3, then 2Sin (a

Vector a = (2cosx, 1), B = (cosx, Radix 3 sin2x), X belongs to R, and the function f (x) is equal to vector a multiplied by vector B Vector a = (2cosx, 1), B = (cosx, Radix 3sin2x), X belongs to R, and the function f (x) is equal to vector a multiplied by vector B In the triangle ABC, a, B, C are the sides opposite to the inner angles a, B and C of the triangle respectively. If f (a) = 2 and a = radical 3, find the maximum value of B + C

f(x)=2cos^2x+√3sin2x
=cos^2x＋√3sin2x＋1
＝2sin(2x+π/6)+1

Let f (x) = a * B, where vector a = (2cosx, 1), B = (cosx, radical 3 * sin2x), X belongs to R In the triangle ABC, ABC is the opposite side of angles a, B and C, f (a) = 2, a = radical 3, B + C = 3 (B is greater than C), and find the length of B and C

(x) = 2cos (x) = 2cos / x + 3 + sin2x = cos2x + 1 + √ 3sin22x = 2Sin (2x + π / 6) + 1F (a) = 2Sin (2a + π / 6) + 1 = 2 = > sin (2a + π / 6) = 1 / 2 = > 2A + π / 6 = 2K π + π / 6 or 2A + π / 6 = 2K π + 5 π / 6 (K ∈ z) = > A = k π or a = k π + π / 3 (K ∈ z) Jun 06 06 06 06 06 06, 06 06 > > A = k = k π or a = k π + π / 3 (K ∈ z) 06 06 06 06 06 06 06 06 06 06 06 06 06 06 06 06 06 06 06- 2B (3-B) = B (3-B) = > b 

Reduction function f (x) = 5 Radix 3cos? X + Radix 3sin? X-4sinxcosx Write down which formula to use in each step How did the five root three turn into three root three

Four formulas cos ^ 2x = (1 + cos2x) / 2Sin ^ 2x = (1-cos2x) / 2sin2x = 2sinxcosxasinx + bsinx = root (a ^ 2 + B ^ 2) sin (x + arctan (B / a)) f (x) = 5 root sign 3 (1 + cos2x) / 2 + radical 3 (1-cos2x) / 2-2sin2x = (5 root 3 + root 3) / 2 + (5 root 3-3 root 3) cos2x / 2 -

Find the minimum value of the function f (x) = 5 times the root 3cos ^ 2 + the root sign 3sin ^ 2x-4sinxcos (π / 4 ≤ x ≤ 7 π / 24), and find the monotone interval

f(x)=5√3(cosx)^2+√3(sinx)^2-4sinxcosx
=5√3*(1+cos2x)/2+√3*(1-cos2x)/2-2sin2x
=2√3cos2x-2sin2x+3√3
=4(√3/2*cos2x-1/2*sin2x)+3√3
=4cos(2x+π/6)+3√3 ,
Because of π / 4