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The function f (x) is defined by π 2 is a periodic even function, and f (π) 3) If = 1, then f (− 17 π) 6)=______ .
∵ the function f (x) is defined by π
2 is a periodic even function,
∴f（-x）=f（x），f（x）=f（x+π
2），
Then f (− 17 π)
6)=f（17π
6）=f（5×π
2+π
3）=f（π
3），
∵f(π
3)＝1，∴f(−17π
6)=1
So the answer is: 1
When the function y = asin (ω x + φ) (a > 0, ω > 0), what is the value of φ, y is odd function, even function The odd function is k π, because when the value of φ is 0, y is an odd function. How can I calculate the even function independent of the value of φ? y=Asin(ωx+φ)=y=Asin(-ωx+φ) ωx+φ=-ωx+φ ωx=-ωx What's wrong with this?
f(x)=y=Asin(ωx+φ)(A>0,ω>0)
1. F (x) is an odd function
f(0)=0,(A>0,ω>0)
f(0)=Asin(φ)=0
φ=kπ
2. F (x) is even function
F (0) = a or - A
f(x)=Asin(φ)=±A
sin(φ)=±1
φ=±π/2+2kπ
A necessary and sufficient condition for the function y = asin (AX + b) (a ≠ 0, a ≠ 0) to be even?
Ax + B = k π + π / 2 K is an integer
Then we get x = (K π + π / 2-B) / A
The reason is that K π + π / 2 is the symmetry axis of this function
If it is an odd function, the same is true, ax + B = k π (the derivation process is reversible)
Given that the function y = asin (Wx + φ) (a > 0, w > 0) is even, then φ = --_ If the even function y = f (x) defined on R is a periodic function with the minimum positive period of π, and f (x) = SiNx when x ∈ [0, π / 2], then the value of F (5 π / 3) is obtained
Y = asin (Wx + φ) (a > 0, w > 0) is an even function, f (- x) = asin (- Wx + φ) = asin (Wx + φ) = f (x) sin (α + β) = sin α cos β + cos α sin β sin (- Wx + φ) = sin (- Wx) * cos φ + cos (- Wx) * sin φ = sin (Wx + φ) = sin (Wx) * cos φ + cos (Wx) * sin φ - sin (Wx) * C
Given that the function y = asin (Wx + φ) (a > 0, w > 0) is even, then φ=————
φ=±π/2 +kπ
Given the function y = asin [ψ x + φ], where ψ > 0, φ∈ [0, π] is an even function on R, and its image is symmetric about M [3 π / 4,0], and monotone on [0, π / 2], find the value of ψ, φ
The transverse problem has nothing to do with a, so let a = 1
Even functions are symmetric about the y-axis, so when x = 0, y takes the maximum or minimum value, so f (0) = sin, φ = 1 or - 1. Because of the range of φ, we get φ = π / 2
This function is symmetric about a point, then the sin function is substituted through this point
F (4 π / 3) = sin (4 ψ π / 3 + π / 2) = 0, so ψ = 3K / 4-3 / 8 K is a positive integer
T > π, so K
Given that f (x) = SiNx + asin (x + b), f (x) is even function, and the maximum value is 2, find a, B Detailed process
A = radical 3
B = - 30 degrees
The process is indescribable
The main reason is that it can't be typed
Given the function f (x) = 2x ^ 2-3x + 1, G (x) = asin (x - π / 6) (a > 0), when o < = x < = π / 2, find the maximum value of y = f (SiNx) If for any x 1 ∈ [0,3], there is always x 2 ∈ [0, π], so that f (x1) = g (x2) holds, we can find the value range of real number a If the equation f (SiNx) = a-SiNx has two solutions on [0,2 π), find the value range of real number a
The known function f (x) = 2x ^ 2-3x + 1, G (x) = asin (x - π / 6) (a > 0), (1) when ox3 = - π / 2, X4 = π / 2F '"(x) = 4cos2x + 3sinx = 4-8 (SiNx) ^ 2 + 3sinx = = > F"' (x1) = f "'(x1) = f"' (x2) = 4-9 / 2 + 2 + 9 / 4 = 7 / 4 > 4 > 0 (f (x) take the minimum value at x1, X2, f ＇ (x3) = 4-8-8-3 3 3 (x3) = 4-8-8-3-3-8-3-3-8-3-8-3-3-8-3-3-8-3-3-8-3-3= - 7
The known function f (x) = asin (ω x + F) [a > 0, w > 0,0
（1）a=2,w=2
F (x) is even function, so f (0) = 2 or - 2
So sinf = 1 or - 1
So f = π / 2 + K π (k is an integer)
Zero
If the function f (x) = asin (x + π / 4) + 3sin (x - π / 4) is even, find the maximum value of F (x)
Sin (a-b) = sinacosb cosasinbcos (a-b) = cosacosb + sinasinb can be easily known from the above process, sin φ = a, cos φ = - 3, so tan φ = - A / 3. In addition, we can also do the following: F (x) = asin (x + π / 4) + 3sin (x - π / 4) = ACOS (x - π / 4) + 3sin (x - π / 4) = √ (a ^ 2 + 9) sin (x - π / 4) = √ (a ^ 2 + 9) sin (x - π / 4) = √ (a ^ 2 + 9) sin (x - π / 4) = √ (a ^ 2 + 9) sin (x - π / 4) = √ (x - π / 4
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