Reduction of F (x) = cos (2x - π / 3) - cos2x-1

Reduction of F (x) = cos (2x - π / 3) - cos2x-1

f(x)=cos(2x-π/3)-cos2x-1
= cos2xcosπ/3+sin2xsinπ/3-cos2x-1
= 1/2cos2x+√3/2sin2x-cos2x-1
= -1/2cos2x+√3/2sin2x-1
= -cos2xcosπ/3+sin2xsinπ/3-1
= -cos(2x+π/3)-1

Simplify f (x) = cos (2x - π / 3) - cos2x

Sum difference product formula
cos(a)-cos(b) = -2sin[(a+b)/2]sin[(a-b)/2]
f(x)=cos(2x-π/3)-cos2x=-2sin(2x-π/6)sin-π/6
=sin(2x-π/6)

How to simplify the function f (x) = cos ^ 2 (2x) - 2 / 3 π cos2x

(1) (x) = (cos2x-1 / 3 π) ^ 2 + 1 / 9 π ^ 2cos2x ∈ [- 1,1] f (x) max = (- 1-1 / 1 / 3 π ^ 2 + 1 + 1 / 9 π ^ 2 = 1 + 2 / 3 π + 2 / 9 π ^ 2F (x) min = (1-1 / 3 π) ^ 2 + 1 / 1 / 3 π ^ 2 = 1-2 / 3 π + 2 / 9 π ^ 2F (x) the value domain [1-2 / 3 π + 2 / 9 π ^ 2,1 + 2 + 2 + 2 + 2 + 2 + 2 / 3 π ^ 3 π ^ 2,1 + 2 + 2 / 3 / 3 π ^ 2,1 + 2 + 2 + 2,1 + 2 / 3 π+ 2 / 9 π ^ 2] (2) f (x) ′ = - 4cos2x

Simplification: radical 3 / 2sin2x - (1 + cos2x) / 2-1 / 2

(√3/2)sin2x－(1＋cos2x)/2－(1/2)
=(√3/2)sin2x－(1/2)cos2x－1
=sin2xcos(π/6)－cos2xsin(π/6)－1
=sin(2x－π/6)－1

Given that cos2x = 4 / 5, and X belongs to the (7 school / 4,2 school), the evaluation: (1) (sin ^ 4) x + (COS ^ 4) x; (2) Tan (x / 2)

(1) Using the same angle trigonometric function relation is sin ^ 2x + cos ^ 2x = 1 and cos2x = cos ^ 2x-sin ^ 2x = 4 / 5, the solution is: sin ^ 2x = 1 / 10, cos ^ 2x = 9 / 10; so (sin ^ 4) x + (COS ^ 4) x = (1 / 10) ^ 2 + (9 / 10) ^ 2 = 41 / 50 (2) because x belongs to (7 π / 4,2 π) and (1) has SiNx = - radical 10 /

(1 / 2) given that x, y satisfy sinxcosx 1-cos2x = 1, Tan (X-Y) = - 2 / 3. (1) find the value of TaNx (2) find Tan (2)

Y is transformed into the expression of X

Given cos2x = 4 / 5 and X ∈ (7 π / 4,2 π), evaluate 1, sin ^ 4 + cos ^ 4.2, Tan (x / 2)

If the known cos2x = 4 / 5, and X ∈ (7 π / 4,2 π), then 2x ∈ (7 π / 2,4 π), in the fourth quadrant, we can get sin2x = - 3 / 5cos ^ 2; X = (1 + cos2x) / 2 = 9 / 10, get cosx = 3 / 10 √ 10, SiNx = - 1 / 10 √ 101, sin ^ 4x + cos ^ 4x = (sin ^ 2x + cos ^ 2x) ^ 2-2sin ^ 2xcos ^ 2x = 1-1 / 2Sin ^ 2; 2; 1-1 / 2Sin ^ 2; 2; 2; 2; 1 / 2, 2; 2; 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2; 2; 2x = 1-1

How much is 1-cos2x = how much? Find all the formulas about the transformation between sin cos Tan arctan and so on. I forgot all the previous studies

1-cos2x=2sin^2x
sin^2x+cos^2x=1
sinx/cosx=tanx

Simplify f (x) = (SiNx + cosx) ^ 2 / 2 + 2sin2x - (cos2x) ^ 2 And find the definition domain and value range of F (x)

F (x) = (SiNx + cosx) ^ 2 / 2 + 2sin2x - (cos2x) ^ 2 = (SiNx) ^ 2 + (cosx) ^ 2 + 2sinxcosx / 2 + 2sin2x - (cos2x) ^ 2 = 1 / 2 + 1 / 2 * sin2x + 2sin2x - (cos2x) ^ 2 = (sin2x) ^ 2 + 5 / 2 * sin2x-1 / 2 the domain is x ∈ R. the value domain is (sin2x + 5 / 4) ^ 2-33 / 4 when sin2x = 1 gets the maximum value

The result of simplifying [1 + sin2x-cos2x] / [1 + sin2x + cos2x]

The original formula = (- 2cos2x / 1 + sin2x + cos2x) + 1 = (- 2cos ^ 2x + 2Sin ^ 2x) / (1 + 2sinxcosx + cos ^ 2x - Sin ^ 2x) + 1 = [2 (SiNx + cosx) (SiNx cosx)] / [(cosx + SiNx) (cosx + SiNx) + 1 = 2 (SiNx cosx) / 2cosx + 1 = tanx-1 + 1 = TaNx