1. Find the maximum, minimum and period of the following functions: (1). Y = SiNx + cosx (2). 3sinx + 4cosx (3). Y = sin2x + cos2x (4).y=5sin6x+12cox6 2. Translate the image f with function y = SiNx left by π / 3 units along the X axis, and then translate it up by 2 units along the Y axis to obtain f ', find the functional formula of image f', and find the maximum, minimum and period of the new function

1. Find the maximum, minimum and period of the following functions: (1). Y = SiNx + cosx (2). 3sinx + 4cosx (3). Y = sin2x + cos2x (4).y=5sin6x+12cox6 2. Translate the image f with function y = SiNx left by π / 3 units along the X axis, and then translate it up by 2 units along the Y axis to obtain f ', find the functional formula of image f', and find the maximum, minimum and period of the new function

1. Is to sum the squares of the coefficients and then square them
(1) 2^0.5
(2) 5
(3) 2^0.5
(4) 13
2. Translate x plus to the left and y plus to the up
F':y=sin(x+pi/3)+2
The maximum value is 3, the minimum value is 1, and the cycle is 2pi unchanged

It is known that the function y = cos2x + (SiNx) ^ 2-cosx (i.e. y = cos2x + SiNx ^ 2 x - cosx) Find the maximum and minimum values of 1. Y 2. In [0360], the set of X when the function takes the maximum and minimum values

This kind of question focuses on transformation
y=cos2x+(sinx)^2-cosx
=(cosx)^2-(sinx)^2+(sinx)^2-cosx
=(cosx)^2-cosx
=(cosx-1/2)^2-1/4
1. When cosx = 1 / 2, y (min) = - 1 / 4
When cosx = - 1, y (max) = 2
2.y(min)=-1/4 x=k × 360 + 60 in [0360] {60300}
y(max)=2 x=k × 360 + 180 in [0360] {180}

Given the function FX = 2 (SiNx + cosx). Cosx, the minimum positive period of FX is

Because f (x) = 2 (SiNx + cosx). Cosx = 2sinxcosx + 2 (cosx) ^ 2 = sin2x + 2 (cosx) ^ 2-1 + 1
=sin2x+cos2x+1
So the minimum positive period of F (x) is π

Minimum positive period of function FX = (SiNx cosx) ^ 2 To detail the process

f(x)=(sinx-cosx) ²
=sin ² x-2sinxcosx+cos ² x
=1-2sinxcosx
=1-sin2x
So the minimum positive period T = 2 π / 2 = π
Answer: π

The minimum positive period of the function FX = SiNx - (cosx SiNx) is

f(x)=2sinx+cosx
=√5sin(x+ φ） (including COS) φ= 2/√5,sin φ= 1/√5）
The minimum positive period is t = 2 π / 1 = 2 π

Given the function FX = 2cosx (SiNx cosx) + 1, X ∈ r.. 1. Find the minimum positive period of FX 2. Find the minimum and maximum value of function FX in the interval [π / 8,3 π / 4]

(1) Questions in this form are generally transformed into a trigonometric function of 2x, so the period is π
F (x) = 2sinxcosx-2cosx ^ 2 + 1 = sin2x cos2x = root 2 / 2Sin (2x - π / 4)
(2)x∈[π/8,3π/4] （2x-π/4）∈[0,5π/4]
Max = root 2 / 2
Min = - 0.5 (when x = 3 π / 4)