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Given that f (x) = sin? 2x + 2sinxcosx + 3cos? X, X ∈ R, how to translate function f (x) into odd function and even function
The solution f (x) = sin ^ 2x + 2sinxcosx + 3cos ^ 2x, = sin ^ 2x + cos ^ 2x + 2sinxcosx + 2cos ^ 2x = 1 + sin2x + 2cos ^ 2x = 2 + sin2x + 2cos ^ 2x-1 = sin2x + cos2x + 2 = √ 2Sin (2x + π / 4) + 2, shift π / 8 units to the right and 2 units to the bottom
Find the value range of the function y = sin? X + 2sinxcosx + 3cos? X
y=sin²x+cos8x+2sinxcosx+2cos²x
=1+sin2x+(1+cos2x)
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
-1
Let f (x) = m? N, where M = (sin ω x + cos ω x, √ 3cos ω x), n = (COS ω x-shi) Let f (x) = m × n, where M = (sin ω x + cos ω x, √ 3cos ω x), n = (COS ω X - Shi ω x, 2shi ω x) where ω > 0, if the period of function f (x) is π. (1) Find the analytic expression of F (x). (2) In △ ABC, a, B, C are the opposite sides of angles a, B and C respectively. A = √ 3, B + C = 3, f (a) = 1. Find the area of △ ABC.
1.f(x)=（sinωx＋cosωx)（cosωx－shiωx*2shiωx
=cos2wx+√3sin2ωx
=2sin(2wx+ π/6 )
2w=2π/π=2
So w = 1
Let f (x) = vector m · vector n, where vector M = (sin ω x + cos ω x, √ 3cos ω x), vector n = (COS ω x-sin ω x, 2Sin ω x) (ω > 0). If the distance between the symmetry axes of F (x) is not less than π / 2 ① Find the value range of ω ② In △ ABC, a, B, C are opposite sides of angles a, B and C respectively, a = √ 3, B + C = 3. When ω is maximum, f (a) = 1, find the area of △ ABC
1）
F (x) = vector m · vector n
=（sinωx+cosωx）（cosωx-sinωx）+√3cosωx*2sinωx
=(cos^2ωx-sin^2ωx）+√3sin2ωx
=cos2ωx+√3sin2ωx
=2sin(2ωx+π/6)
Distance between adjacent symmetry axes = 2 π / 2 ω △ 2 = π / 2 ω
So, π / 2 ω ≥ π / 2
ω≤1
2)
When ω is maximum, ω = 1
f(x)=2sin(2x+π/6)
f(A)=2sin(2A+π/6)=1
sin(2A+π/6)=1/2
2A + π / 6 = π / 6, or, 5 π / 6
A = 0, or, π / 3
Because a > 0, a = π / 3
cosA=(b^2+c^2-a^2)/2bc
=[(b+c)^2-2bc-a^2]/2bc
=[(b+c)^2-a^2]/2bc-2
=(9-3)/2bc-1
=3/bc-1
So, 1 / 2 = 3 / BC-1
bc=2
Area of △ ABC = bcsina / 2
=2sinπ/3 /2
=sinπ/3
=√3/2
Given the function f (x), its definition domain is R and even function in the definition domain. If f (1-x) = f (1 + x), it is proved that y = f (x) is a periodic function
Let 1-x = t, then f (T) = f (2-T), because f (x) is an even function, so f (T) = f (- t), that is, f (- t) = f (2-T), and then - t = u, then f (U) = f (U + 2), so f (x) is a periodic function with period 2
The function of defined domain on R satisfies f (x + y) + F (X-Y) = 2F (x) f (y) f (0) ≠ 0 f (1 / 2) = 0. It is proved that f (x) is an even function and f (x) is a periodic function The function defined on R satisfies f (x + y) + F (X-Y) = 2F (x) f (y) f (0) ≠ 0 f (1 / 2) = 0. It is proved that f (x) is an even function and f (x) is a periodic function if the function is monotone in [0,1] =? F (1 / 6) =?
Let X and y be equal to zero
It can be concluded that f (0) = 1
When x = 0
f(y)+f(-y)=2f(y)
That is, f (- y) = f (y)
So f (x) is even
Let y = 1 / 2
Then f (x + 1 / 2) + F (x-1 / 2) = 0
That is, f (x) = - f (x + 1)
So f (x-1 / 2) + F (x-3 / 2) = 0
Subtraction of two formulas
It is found that f (x + 1 / 2) = f (x-3 / 2)
That is, f (x) = f (x + 2)
F (x) is a periodic function with period 2
Let x = 1 / 3, y = 1 / 6
f(1/2)+f(1/6)=2f(1/3)f(1/6)
That is, f (1 / 6) = 2F (1 / 3) f (1 / 6) and f (1 / 6) is not equal to 0
The solution is 2F (1 / 3) = 1 / 2
Let x = 1 / 6, y = 1 / 6
3/2=f(1/3)+f(0)=2f(1/6)f(1/6)
F (1 / 6) = sqrt3 / 2
If the even function defined as R satisfies f (x + 1) = - f (x), is it a periodic function
f(x+1)=-f(x)
So - f (x + 1) = f (x)
f(x+2)=f[(x+1)+1]=-f(x+1)=f(x)
So as long as the definition domain is r, it is a periodic function
Even function FX defined on R satisfies f (x + 1) = - f (x) period why is 2
f(x+1)=-f(x)
f(x+2)=f(x+1+1)=-f(x+1)=f(x)
So, the period is two
Let f (x) be a function defined on R with period 2 and even function. When x ∈ [2,3], f (x) = x and find x ∈ [- 2,0], the analytic expression of F (x) is obtained
When x ∈ [2,3], f (x) = x, so f (2) = f (0) = 2, f (1) = f (3) = 3
So x ∈ [0,1] f (x) = x + 2
The rest is the picture
When x belongs to 2,3, f (x) = x, X belongs to - 1,0 analytic formula
Let x ∈ [- 1,0] then - x ∈ [0,1] and f (x) is a even function; \ (- x) = f (x) = f (x) and f (x) the period of F (x) and f (x) is 2 ~ f (- x (- x) = f (- x + 2) = f (- x + 2) = f (- x + 2) and - x ∈ [0,1], - x + 2 ∈ [2,3] and X ∈ [2,3] x ∈ [2,3] has: F (x) = x \\\\\\\\\\\\\\\x
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