The vector is known to be vector (2x, SX), which belongs to vector (2x, SX), vector (2x, SX) = vector 1) Find the minimum positive period, maximum value and minimum value of function f (x) (2) Monotone increasing interval of function f (x)

（1）f(x)=sin2x-cos2x=2^0.5*sin(2x-π/4),T=π,fmax=2^0.5,fmin=-2^0.5
(2)(-1/8+k）π,（3/8+k)π）

Known vector m=（2sinx 2，1）， n=（cosx 2,1), Let f (x)= m• n-1． (1) The value field of (f) is obtained; (2) It is known that △ ABC is an acute triangle and a is the inner angle of △ ABC if f (a) = 3 5, find f (2A - π) 3) Value of

(1) By F (x)=
m•
N-1, f (x) = 2sinx
2cosx
2+1-1=sinx，
So the range of y = f (x) is [- 1,1];
(2) It is known that a is an acute angle, f (a) = Sina = 3
5，
Then cosa=
1-(3
5)2=4
5, sin2a = 2sinacosa = 2 × 3
5×4
5=24
25，
cos2A=1-2sin2A=1-2×(3
5)2=7
25，
So f (2A - π)
3）=sin（2A-π
3）=sin2Acosπ
3-cos2Asinπ
3=24
25×1
2-7
25×
Three
2=24-7
Three
50．

The image f of function y = 3sin (x-a) is translated by vector (PAI / 3,3) to get the image f ', If one of the symmetry axes of F 'is a straight line x = Pai / 4, then one possible value of is A.5pai/12 B.-5pai/12 C.11pai/12 D.-11pai/12

The symmetry axis of y = 3sin (x-a) is 3pai / 2 (I replaced Pai / 2 first, and a = 17pai / 12 is not in the option)
3pai/2+a+pai/3=pai/4 a=5pai/12

If the graph of a function is pressed a=（-π 3, - 2) the image of the function y = cosx is obtained by translation, then the analytic expression of the original image is () A. y=cos（x+π 3）+2 B. y=cos（x-π 3）-2 C. y=cos（x+π 3）-2 D. y=cos（x-π 3）+2

A kind of
a＝(−π
3，−2)∴-
a＝(π
3，2)，
Let the function y = cosx be vector-
a＝(π
3,2) to obtain y = cos (x − π)
3) + 2 is the analytic expression of the function y = f (x)
Therefore, D

Find: is f (x) = sin (x + л / 4) cos (x + л / 4) even or odd?

f(x)=1/2*2sin(x+л/4）cos(x+л/4）
=1/2sin[2(x+л/4)]
=1/2sin(2x+л/2)
=1/2cos2x
So it's even

Let a be greater than 0 and less than Pai. If the function f (x) = sin (x + a) + cos (x-2a) is even, then the value of a is?

The function f (x) = sin (x + a) + cos (x + 2a) function f (x) = sin (x-a) + cos (x + 2a) function f (x) = sin (x + a) + cos (x-2a) is the even function, f (- x) = f (f) = f (x) f (x) cos (x + a) + cos (x + 2a) - sin (x-a) = sin (x + a) + cos (x-2a) cos (x + 2a) - cos (x-2a) - cos (x-2a) = sin (x + X + a) + sin (x-a) + sin (x-a) - 2sinxsin2a = 2sinx Cosa 2sin2sinxsin2a = 2sinx Cosa 2sinxsin2a = 2sinx Cosa 2sin2sinxsin2a = 2sinxsin2sinx Cosa 2sinxsxcosa + 4sin

In the following four functions, the even function with period Pai is a.y = sinxcosx b.y = cos ^ 2x sin^ Among the following four functions, the even function with period Pai is A.y=sinxcosx B.y=cos^2x-sin^2x C.y=tanx D.y=arccosx

Why sin is odd function and COS is even function

Look at the definition, in order to study the properties of some functions, people define odd function and even function, and SiNx and cosx meet the conditions respectively, so it is oh!
The image method is intuitive, but it needs to be proved by definition
Odd function: F (x) = - f (- x)
Even function: F (x) = f (- x)

The minimum positive period T of the function y = sin (π / 3) x * cos (π / 3) x=_____

y=sin(π/3)x * cos(π/3)x
=1/2sin(2π/3)x
So the minimum positive period of the function y = sin (π / 3) x * cos (π / 3) x = 2 π / 2 π / 3 = 3

What is the minimum positive period of the function y = sin π X / 3 * cos π X / 3? To process! Thank you! Use which formula also say, thank you!

Using the formula of double angle: y = (1 / 2) sin (2 π X / 3)
The minimum positive period T = 2 π / (2 π / 3) = 3

The vector is known to be vector (2x, SX), which belongs to vector (2x, SX), vector (2x, SX) = vector 1) Find the minimum positive period, maximum value and minimum value of function f (x) (2) Monotone increasing interval of function f (x)