If f (x) = 2Sin (π / 2x + π / 3), then the value of F (1) + F (2) +... + F (2009) is given

If f (x) = 2Sin (π / 2x + π / 3), then the value of F (1) + F (2) +... + F (2009) is given

Notice that the sum of four cycles is one cycle

Given the function f (x) = 2Sin (2x + pi / 3), find the period and the maximum value of F (PI / 8)

From the formula of the period
T=2π/2=π
The maximum values of sine function are - 1 and 1, respectively
So - 1

Find the monotone interval of y = log ^ 2 [2Sin (PI / 3-2x)]

 with 2 as the base, log2 is the increasing function ᙽ only the monotone interval of 2Sin (π / 3-2x) is required, but y increases when it increases, y decreases when it decreases

The maximum value of F (x) = cos ^ 2x / [2Sin (π / 2 + x)] + (A / 2) SiNx is 2. Try to determine the value of A The maximum value of F (x) = (cosx) ^ 2 / [2Sin (π / 2 + x)] + (A / 2) SiNx is 2. Try to determine the value of A

By using the induction formula, the maximum value of sin (π / 2 + x) = cosxf (x) = cos ^ 2x / 2cosx + (A / 2) SiNx = 1 / 2cosx + (A / 2) SiNx = root (1 + A ^ 2) / 2Sin (x + ψ) ∵ the maximum value of F (x) is 1  the maximum value of sin (x + ψ) is 1  the maximum value of sin (x + ψ) is 1  the maximum value of sin (x + ψ) is 1  root sign (1 + A ^ 2) / 2 = 2  root sign (1 + A ^ 2) = 4 ﹥ 1 + A ^ 2 = 16

It is known that f (x) = 2Sin (π + x) cos (π - x) - cos (π + x) divided by 1 + SiNx ^ 2 + sin (π - x) - cos (π + x) ^ 2 (dividing by the front is the numerator, dividing by the back is the denominator) 1. Simplification 2. When x = - 35 π / 6, find the value of F (x)

f(x)=(2sinxcosx+cosx)/(1+sin²x+sinx-cos²x)
=cosx(2sinx+1)/(2sin²x+sinx)
=cosx/sinx
=cotx
f(-35π/6)
=cot(π/6-6π)
=cot(π/6)
=√3

If f (x) = (√ 3 + √ 3cos2x) / (2Sin (π / 2-x)) - 2A (SiNx / 2) cos (π - X / 2) (a > 0), the maximum value is 2 1. Try to determine the value of constant a 2. If f (α - π / 3) - 4cos α = 0, find (COS α ^ 2 + 0.5sin2 α) / (sin α ^ 2-cos ^ 2)

(1)
f(x)=(√3+√3cos2x)/(2sin(π/2-x))-2a(sinx/2)cos(π-x/2)
=√3（1+cos2x)/(2cosx)-2asinx/2(-cosx/2)
=√3*2cos²x/(2cosx)+asinx
=√3cosx+asinx
=√(3+a²)sin(x+φ)
Where cos φ = A / √ (3 + a?), sin φ = √ 3 / √ (a 2 + 3)
∵ the maximum value of F (x) is 2
∴√(3+a²)=2 ∴a²=1
∵a>0
∴a=1
(2)
From (1)
f(x)=2sin(x+π/3)
∵f(α-π/3)-4cosα=0
∴2sinα-4cosα=0
∴sinα=2cosα,tanα=2
∴（cos²α+0.5sin2α)/(sin²α-cos²α)
=(cos²α+sinαcosα)/(sin²α-cos²α)
=(1 + Tan α) / (Tan 2 α - 1) (the numerator and denominator divide at the same time cos 2 α, chemical cleavage)
=（1+2)/(4-1)
=1

Given the function f (x) = f '(π / 4) cos + SiNx, then f (π / 4) =?

F '(x) = - f' (π / 4) SiNx + cosx. F '(π / 4) = - f' (π / 4) (1 / 2) ^ 2 + (1 / 2) ^ 2. The solution f '(π / 4) = radical 2-1

(1 / 2) in the triangle ABC, the known angles ABC are ABC, vector M = (2sinb, - radical 3), n = (cos2b, B-1 of 2cos Square), and m

M / / N, then: 2sinb / cos2b = (- √ 3) / [2cos 2 (B / 2) - 1], that is, 2sinb / cos2b = (- √ 3) / CoSb, √ 3cos2b + 2sinbcosb = 0, √ 3cos2b + sin2b = 0, sin (2B + π / 3) = 0, then: B = π / 3

(1 / 2) the opposite sides of the inner angles a, B and C in the triangle ABC are a, B, C, respectively. The vector M = (2sinb, - radical 3), the vector n = (cos2b, 2cos ^ 2B / 2-1) and

If M / / N, then 2sinb: cos2b = (- √ 3): (2cos? B / 2-1), - √ 3cos2b = 2sinbcosb, - √ 3cos2b = sin2b, tan2b = - √ 3, 2b = 2 π / 3, then B = π / 3

In triangle ABC, the opposite sides of angle ABC are ABC. Vector M = (2sinb, - radical 3), n = (cos2b, 2 squared - 1) And the vector m is parallel to n. (1). Find the size of the acute angle B; (2) if B = 2, find the maximum area of the triangle ABC

(1) M = (2sinb, - radical 3),
n=(cos2B,cosB)
M / / N, then 2sinb / cos2b = - (radical 3) / CoSb
That is, 2sinbcosb + (root 3) cos2b = 0
2 sin (2B + 60) = 0
So 2B = 120, B = 60 degrees
(2) Triangle area
S = (1 / 2) acsinb = (radical 3 / 4) AC
=(root 3 / 8) (a ^ 2 + C ^ 2);
The necessary and sufficient condition of the equal sign is a = C;
On the other hand, B = 2: by cosine theorem 4 = a ^ 2 + C ^ 2-ac
So when a = C, there is a = C = 2;
S