Let f (x) = 2 + sin2x + cos2x, X ∈ R (1) Find the maximum value of function f (x) and the set of independent variable x which obtains the maximum value; (2) Find the monotone increasing interval of function f (x)

Let f (x) = 2 + sin2x + cos2x, X ∈ R (1) Find the maximum value of function f (x) and the set of independent variable x which obtains the maximum value; (2) Find the monotone increasing interval of function f (x)

（1）f（x）=2+sin2x+cos2x=2+
2sin(2x+π
4) (4 points)
ν when + π
4＝2kπ+π
2, that is, x = k π + π
When 8 (K ∈ z), f (x) has a maximum value of 2+
2．
Therefore, the set of independent variables X with the maximum value of F (x) is {x | x = k π + π
8, K ∈ Z}; (8 points)
（2）f(x)＝2+
2sin(2x+π
4)，
2K π - π
2≤2x+π
4≤2kπ+π
2(k∈Z)，
That is, K π - 3
8π≤x≤kπ+π
8(k∈Z)．
Therefore, the monotone increasing interval of F (x) is [K π - 3 π
8，kπ+π
8](k∈Z)．　… (12 points)

Given sin2x + cos2x > 0, what is the value range of X? Write down the steps. X is a real number······

sin2x+cos2x=√2sin(2x-π/4)>0
2kπ<2x-π/4<2kπ+π
kπ+π/8
Homework help users 2017-10-17
report

(SiNx) ^ 2 on the formula of cosx

(sinx)^2+(cosx)^2=1
(sinx)^2=1-(cosx)^2

Find all SiNx, cosx, TaNx mutual conversion formula since high school The square of (SiNx) and the like, all of them! Better give me a link

The square relation of trigonometric function is: sin ^ 2 (α) + cos ^ 2 (α) = 1 cos ^ 2 (a) = 1-sin ^ 2 (a) Tan ^ 2 (α) + 1 = 1 / cos ^ 2 (α) 2Sin ^ 2 (a) = 1-cos2 (a): sin α = Tan α × cos α cos α = cot α × sin α Tan α = sin α × sec α

SiNx cosx formula transformation

Interpretation of sin -- cos 1. Inverted delta as sin --- cos (SiNx) ^ 2 + (COX) ^ 2 = 1
- - - - - -
tan --- 1 --- cot 1
- - - -
---- 2. In this hexagon, the sign in the corresponding corner is reciprocal
3. Among the three connected angles, the middle corner is the product of the two sides of the angle, namely sin --- COS in -- tansinx = cosx times TaNx

If f (cosx) = cos2x, then the expression of F (SiNx) How can f (COS (π / 2-x)) be equal to - cos2x?

f(cosx)=cos2x
sinx=cos(π/2-x)
f(sinx)=f[cos(π/2-x)]=cos(π-2x)=-cos2x

How to turn the sine double angle formula into the same angle with different angles to prove 2Sin (π / 4 + a Inverse use of sine double angle formula How do different angles turn into the same angle It is proved that 2Sin (π / 4 + a) cos (π / 4 + a) = cos2a It is proved that 2Sin (π / 4 + a) cos (π / 4-A) = cos2a The second bracket is a minus sign

It is proved that: 2Sin (π / 4 + a) cos (π / 4-A) = sin [(π / 4 + a) + (π / 4-A)] + sin [(π / 4 + a) - (π / 4-A)] = sin (π / 2) + sin2a = 1 + sin2a should not be your result

1 + sin 2x + cos 2x,

Solution 1 + sin2x + cos2x
=1+sin2x+2cos^2x-1
=sin2x+2cos^2x
=2sinxcosx+2cos^2x
=2cosx（sinx+cosx）
=2√2cosxsin（x+π/4).

In order to get the image of the function y = sin2x, we only need to translate the image of the function y = cos (2x-3): where to translate and how many units? Detailed process The main formula to be deformed

y=cos(2x-π/3)=cos[-π/2+(2x+π/6)]=sin(2x+π/6)=sin[2(x+π/12)]
The image of the function y = sin2x is shifted to the left π / 12 units to obtain the image of y = sin [2 (x + π / 12)]
The y = cos (2x - π / 3) image is shifted to the right by π / 12 units to get y = sin2x image

Cos ^ 4x sin ^ 4x = cos2x why

cos^4x-sin^4x
=(cos^2x+sin^x)(cos^2x-sin^2x)
=1*(cos^2x-sin^2x)
=cos^2x-sin^2x
=cos2x