Known vector a = (1 + sin2x, SiNx cosx), vector b = (1, SiNx + cosx), f (x) = vector a * vector B Find the value range of F (x)

Known vector a = (1 + sin2x, SiNx cosx), vector b = (1, SiNx + cosx), f (x) = vector a * vector B Find the value range of F (x)

F (x) = a · B = (1 + sin (2x), SiNx cosx) · (1, SiNx + cosx) = 1 + sin (2x) + (SiNx cosx) (SiNx + cosx) = 1 + sin (2x) + SiNx ^ 2-cosx ^ 2 = 1 + sin (2x) - cos (2x) = 1 + √ 2Sin (2x - π / 4) sin (2x - π / 4) ∈ [- 1,1] √ 2Sin (2x - π / 4) ∈ [- √ 2, √ 2] so: F (x) ∈ [1 -

Given vector a = (SiNx, 1), B = (cosx, - 1 / 2), if a ⊥ B, then sin2x= As above, please write down the process

Because a ⊥ B
So a · B = SiNx · cosx + 1 * (- 1 / 2) = 0
SiNx · cosx = 1 / 2
Because sin2x = 2sinx · cosx
So 1 / 2sin2x = 1 / 2
sin2x=1

Known vectors a = (cosx, SiNx), B = (sin2x, 1-cos2x), C = (0,1), X ∈ (0, π) (1) Are vectors a and B collinear? Please explain the reasons (2) Find the maximum value of function f (x) = B - (a + b) · C

(1) Because cosx (1-cos2x) - sinxsin2x = cosx (1 - (1-2sin ^ 2x)) - SiNx * 2sinxcosx = cosx * 2Sin ^ 2x-2sinxsinxcosx = 0
So vectors a and B are collinear

The value range of function f (x) = cosx / (COS (x / 2) - sin (x / 2)) is?

Molecular cos (2 * 1 / 2x)
It becomes the square of COS (1 / 2x) - the square of sin (1 / 2x)
You can use the square difference formula to resolve it, and the original formula can be approximately
Cos (x / 2) + sin (2 / x)
Then extract the root 2 and sort out √ 2 [sin（ д/ 4-2 / x)]
So the value range is [- √ 2, √ 2]

What is the derivative of Cotx

Cot ` x = - CSC ^ 2 X. (formula)
So cot ` x = - 1 / sin ^ 2 x
Similar are:
tan` X = sec^2 X .
sec` X = tan X sec X .
csc` X = - cot X csc X .

Proving the derivative of secx by definition

f(x)=secx=1/cosxf(x+ Δ x)=sec(x+ Δ x)=1/cos(x+ Δ x)f(x+ Δ x)-f(x)=1/cos(x+ Δ x)-1/cosx=[cosx-cos(x+ Δ x)]/[cosx*cos(x+ Δ x)]=[2sin(x+ Δ x/2)*sin( Δ x/2)]/[cosx*cos(x+ Δ x)][f(x+ Δ x)-f(x)]/ Δ x=[2sin(x+ Δ x/2)*sin...