The known function f (x) = cos ^ 4x + cos2x sin ^ 4x (1) Finding the minimum positive period of F (x) (2) If x ∈ [0, π / 2], find the maximum and minimum of F (x)

The known function f (x) = cos ^ 4x + cos2x sin ^ 4x (1) Finding the minimum positive period of F (x) (2) If x ∈ [0, π / 2], find the maximum and minimum of F (x)

∵ f (x) = cos ^ 4x + cos2x sin ^ 4x = (COS? X + sin? X) (COS? X-sin? X) + cos2x = 2cos2x  the minimum positive period of F (x) t = 2 π / 2 = π (2) ∵ x ∈ [0, π / 2], ｛ 2x ∈ [0, π], f (x) = 2cos2x, the maximum value of F (0) = 2, the minimum value of F (π / 2) = - 2

It is proved that cos ^ 8 (x) - Sin ^ 8 (x) = cos2x [1-1 / 2Sin ^ 2 (2x)]

[SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [SiNx] = [sin

Given cos2x = √ 2 / 3, then the value of sin ^ 4x + cos ^ 4 is? (my answer is - root 2 / 3, but this option is not available) A13/18 B11/18 C7/9 D-1 sin^4X+cos^4x

sin^4X+cos^4x
=(cos^2x+sin^2x)^2-1/2(2sinxcosx)^2
=1-1/2sin^2(2x)
=1-1/2[1-cos^2(2x)]
=1-1/2*[1-(√ 2/3)^2]
=1-1/2(1-2/9)
=1-1/2*7/9
=1-7/18
=11/18
Choose B

If sin (x + 4 / π) / cos2x = - radical 2, then the value of COS (x + 4 / π) is?

(sinxcos4 / π + cosxsin4 / π) / (COS? X-sin? X) = - √ 2 √ 2 / 2 * (SiNx + cosx) / (cosx + SiNx) (cosx SiNx) = - √ 2 so cosx SiNx = - 1 / 2, so the original formula = cosxcos4 / π - sinxsin4 / π = √ 2 / 2 * (cosx SiNx) = - √ 2 / 4

How much cosx is cos2x equal to

cos2X=(cosX)^2-(sinX)^2
=2*(cosX)^2-1

If 3sinx = cosx, then cos2x + sin2x is equal to?

0

sin(-3x)=?cos(-3x)=?tan(-3x)=? Please explain the reasons in detail

Sin (- 3x) = - sin (3x) sin is an odd function
Cos (- 3x) = cos (3x) cos is an even function
Tan (- 3x) = - Tan (3x) Tan is an odd function

The primitive function of cos2x / cosx plus SiNx It's cos 2x

∫cos2x/(cosx+sinx) dx
=∫(cosx^2-sinx^2)/(cosx+sinx) dx
=∫(cosx-sinx) dx
=sinx+cosx+c
Is that what the title means

The solution equation is: (sin2x + cos2x) / (1-sin ^ 2x-2cos2x) = 2 X belongs to (π / 2, π)

sin2x+cos2x
=1-2sin^2x+2sinxcosx
2-2sin^2x-4cos2x
=-2sin^2x+2-4+8sin^2x
=-2+6sin^2x
That is, 8sin ^ 2x-2sinxcosx-3 = 0
5sin^2x-2sinxcosx-3cos^2x=0
(5sinx+3cosx)(sinx-cosx)=0
Because x belongs to (π / 2, π)
So 5sinx = - 3cosx
tanx=-3/5
x=π-arctan(3/5)

Simplification: (SiNx + sin2x) / (1 + SiNx + cos2x)

The answer is as follows: SiNx + sin2x = SiNx + 2sinxcosx = SiNx (1 + 2cosx) 1 + cosx + cos2x = 2cosx ^ 2 + cosx = cosx (1 + 2cosx)