If f (x) = sin (π + x) cos (2 π - x) / cos (π / 2 + x), find the value of F (π / 3)

If f (x) = sin (π + x) cos (2 π - x) / cos (π / 2 + x), find the value of F (π / 3)

The results are as follows
F (x) = cosx is 1 / 2 of π / 3

We know that f (x) = 2 √ 3 sinxcosx + (COS ^ 2) x - (sin ^ 2) x is coming soon!
It is known that f (x) = 2 √ 3 sinxcosx + (COS ^ 2) x - (sin ^ 2) X
1. Find the minimum positive period of function

f(x)=2√3 sinxcosx+(cos^2)x-(sin^2)x
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
So the minimum positive period T = 2 π / 2 = π

The known function f (x) = cos ^ 2 (x / 2) - sin (x / 2) * cos (x / 2) - 1 / 2
If f = (α) = 3 * √ 2 / 2, find sin2 α
Cos ^ 2 (x / 2) = (1 + cosx) / 2, isn't it

F (x) = cos & # 178; (x / 2) - sin (x / 2) * cos (x / 2) - 1 / 2 = 1 / 2 * [2cos & # 178; (x / 2) - 1] - 1 / 2 * SiNx = 1 / 2 * cosx-1 / 2 * SiNx = √ 2 / 2 * (√ 2 / 2 * cosx - √ 2 / 2 * SiNx) = √ 2 / 2 * cos (x + π / 4) the minimum positive period is t = 2 π / 1 = 2 π∵ - 1 ≤ cos (x + π / 4) ≤ 1 ∵ - √ 2 / 2 ≤ f (x)

Let f (x) = cos & # 178; x-2sinxcos Sin & # 178; X (1) find the minimum positive period of F (x) (2) find the maximum and minimum value of F (x)

Solution
f(x)=cos²x-2sinxcosx-sin²x
=(cos²x-sin²x)-2sinxcosx
=cos2x-sin2x
=√2(√2/2cos2x-√2/2sin2x)
=√2(cos2xcosπ/4-sinπ/4sin2x)
=√2cos(2x+π/4)
The minimum positive period is:
T=2π/2=π
∵cos(2x+π/4)∈[-1,1]
∴f(x)∈[-√2,√2]
The maximum value of F (x) is: √ 2
The minimum value is: - √ 2

Why is the derivative of F (x) = sin (x) f (x) = cos (x)?

That's right

Given the function f (x) = sin (π / 3 + ω x) + cos (ω X - π / 6) (ω > 0), the minimum positive period of F (x) is π. (1) find the analytic expression of F (x). (2) find the monotone decreasing interval of F (x)

1. F (x) = 1 / 2 * sin2 ω x + √ 3 * (1 + Cos2 ω x) / 2 + a = 1 / 2 * sin2 ω x + √ 3 / 2 * Cos2 ω x + √ 3 / 2 + a = sin2 ω xcos π / 3 + Cos2 ω xsin π / 3 + √ 3 / 2 + a = sin (2 ω x + π / 3) + √ 3 / 2 + a the distance from the origin to the first highest point is t / 4, so t / 4 = π / 6T = 2 π / 2 ω = 2 π / 3 ω = 3 / 22, f

Trigonometric function problem, known Tan (β) = f (x), find cos ^ 2 (β) and sin ^ 2 (β)
Cos ^ 2 (β) and sin ^ 2 (β) are required to be functions of X only

A:
f(x)=tanβ=sinβ/cosβ
sinβ=f(x)cosβ
Substituting sin ｛ 178; β + cos ｛ 178; β = 1, we have the following formula:
f²(x)cos²β+cos²β=1
So:
cos²β=1/[1+f²(x)]
sin²β=1-1/[1+f²(x)]

Given the function f (x) = cos (- X / 2) + sin (π - X / 2), X ∈ R, if f (a) = (2 √ 10) / 5, a ∈ (0, π / 2), find Tan (2a + π / 4)

F (x) = cos (- X / 2) + sin (π - X / 2) = cos (x / 2) + sin (x / 2) f (a) = cos (x / 2) + sin (x / 2) = (2 √ 10) / 5 cos (x / 2) + sin (x / 2) = (2 √ 10) / 5 square 1 + Sina = 8 / 5 Sina = 3 / 5 cosa = 4 / 5 tan2a = 2tana / (1-tan ^ 2a) = 24 / 7 Tan (2a + π / 4) = (1 +...)

Y = 1-sin (2x) / sin (x) - cos (x) for T. maximum

y = { 1-sin(2x) } / { sin(x)-cos(x) }
= { (sinx)^2+(cosx)^2-2sinxcosx } / { sin(x)-cos(x) }
= (sinx - cosx)^2 / { sin(x)-cos(x) }
= sinx - cosx
=Radical 2 (sinxcos π / 4 - cosxsin π / 4)
=Radical 2Sin (x - π / 4)
T=2π
Minimum root 2
Maximum root 2

(1+cos x-sin x)/(1-sin x-cos x)+(1-cos x-sin x)/(1-sin x+cos x)=?

The original formula = ((1 + cosx SiNx) ^ 2 + (1-cosx-sinx) ^ 2) / ((1-sinx-cosx) (1-sinx + cosx)) = (1 + sin ^ 2 (x) + cos ^ 2 (x) - 2sinx + 2cosx-2sinxcosx + 1 + sin ^ 2 (x) + cos ^ 2 (x) - 2sinx-2cosx + 2sinxcosx) / ((1-sinx) ^ 2-cos ^ 2 (x)) = (4-4sinx) / (2sinx (SiNx -...)