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Given that the partial image of the function FX = asin (Wx + φ) (a > 0, w > 0, and φ less than π / 2) is shown in the figure, then the function analytic formula of FX is Given that the partial image of the function FX = asin (Wx + φ) (a > 0, w > 0, and φ less than π / 2) is shown in the figure, then the function analytic formula of FX is
A = 3
T=4(π/2-(-π/2))=4π
And T = 2 π / W
SO 2 π / w = 4 π
So w = 1 / 2
So f (x) = 3sin (1 / 2x + φ)
Its image crossing point (- π / 2,3)
We know that 3sin (1 / 2x (- π / 2) + φ) = 3
That is sin (- π / 4 + φ) = 1
That is - π / 4 + φ = 2K π + π / 2, K belongs to Z
That is, φ = 2K π + 3 π / 4, K belongs to Z
It is also found that the ratio of φ is less than π / 2
When k = - 1,
φ=-5π/4
Therefore, f (x) = 3sin (1 / 2x-5 π / 4)
The function FX = asin (Wx + ω (a > 0, ω > 0, - π / 2
The period of F (x) is 1 / 4, that is, t / 4 = 2 π / 3 - π / 6 = π / 2
ν t = 2 π, then 2 π / w = 2 π, w = 1
And the maximum value is 1, so a = 1
∴f(x)=sin(x+φ)
By substituting (π / 6,1) into,
π/6+φ=2kπ+π/2,k∈Z
∴φ=2kπ+π/3,k∈Z
∵-π/2
It is known that the function f (x) = asin (Wx + pi / 4) (where x belongs to R, a > 0, w > 0) has a maximum value of 2 and a minimum positive period of 8. (1) find the analytic formula of the function f (x) (2) if the abscissa of two points P and Q on the image of function f (x) are 2,4 and O is the origin of coordinates, the area of triangle poq can be obtained
One
A> 0, w > 0
fx=Asin(wx+pai/4) （A>0,w>0)
The maximum value is 2, a = 2,
The minimum positive period is 8,
From 2 π / w = 8, w = π / 4
∴f(x)=2sin(π/4*x+π/4)
Two
When x = 2, f (2) = 2Sin (π / 2 + π / 4) = √ 2
When x = 4, f (4) = 2Sin (π + π / 4) = - √ 2
∴P(2,√2),Q(4,-√2)
Midpoint m (3,0) of line PQ
The area of the triangle poq
S=SΔPOM+SΔQOM
=3×√2×1/2+3×√2×1/2
=3√2
】Let f (x) = asin (Wx + φ), (a > 0,0)
(1) The minimum positive period is 2 π w = 1 f (π / 3) = 2Sin (π / 3 + φ) = 1 sin (π / 3 + φ) = 1 sin (π / 3 + φ) = 1 / 2sin5 π / 6 = 1 / 2 φ = π / 2 F (x) = 2Sin (x + π / 2) = 2cosx (2) 2cos θ - sin θ = 0 2cos θ = sin θ Tan θ = 2 cos θ = 1 / root sign (2? + 1 & SU)
Under root sign (sin "" x + 4cos "x) - under radical sign (COS" "x + 4sin" x) =? "To the fourth power
√[(sinx)^4+4(cosx)^2 ] - √[(cosx)^4+4(sinx)^2 ]
=√[ ((sinx)^2-2)^2] - √[ ((cosx)^2-2)^2]
= (sinx)^2-2 - [(cosx)^2 -2]
= (sinx)^2 - (cosx)^2
= -cos2x
Reduction of radical 2 / 4sin (π / 4-x) + radical 6 / 4cos (π / 4-x) The root 2 / 4 is a quarter root two
√2/4 sin（π/4-x）+√6/4 cos（π/4-x）
=√2/2 [1/2 sin（π/4-x）+√3/2 cos（π/4-x）]
=√2/2 [cosπ/3 sin（π/4-x）+sinπ/3 cos（π/4-x）]
=√2/2 [ sin（π/4-x+π/3）]
=√2/2 [ sin（7π/12-x）]
=√2/2 [ sin（π/2+π/12-x）]
=√2/2 [ cos（π/12-x）]
It is known that the minimum positive period of the function f (x) = root 3sinwxcoswx + sin ^ 2wx (W > 0) is u, (1) Find the value of W (2) find the maximum value of function and the corresponding x value (3) if the image of function f (x) is shifted to the left by 3 unit length, the image of function g (x) is obtained. The monotone decreasing interval of function g (x) is obtained
F (x) = Radix 3sinwxcoswx + sin ^ 2wx = [(radical 3) / 2] sin2wx + (1 - cos2wx) / 2 = sin (2wx - Wu / 6) - 1 / 2 (1) t = 2 Wu / (2 W) = Wu w = 1 f (x) = sin (2x - Wu / 6) - 1 / 2 (2) if sin (2x - Wu / 6) = 1, then 2x - Wu / 6 = 2K
It is known that the distance between two adjacent maxima of the radical 3sinwxcoswx cos ^ 2wx + 3 / 2 is X Find the analytic formula of F (x) (W, X ∈ R)
f(x)=（√3）sinwxcoswx-（coswx）^2+3/2
=[(√3）/2]sin2wx-(1/2)cos2wx+1
=sin(2wx-π/3)+1,
The distance between the two adjacent maximum points is 2 π / | 2W | = π, w = soil 1
ν f (x) = sin (soil 2x - π / 3) + 1
The minimum positive period of the function f (x) = cos ^ Wx plus root sign 3sinwxcoswx (W > 0) is known to be PI Find the value of F (2 / 3)
f(x)=cos^wx+√3sinwxcoswx=[1+cos(2wx)]/2+√3(2sinwxcoswx)/2=cos(2wx)/2+√3sin(2wx)/2+1/2=sin(2wx+π/6)+1/2T=2π/2w=π/w=πw=1f(x)=sin(2x+π/6)+1/2f(2π/3)=sin(3π/2)+1/2= -1/2
It is known that the minimum positive period of the radical sign 3sinwxcoswx cos ^ 2wx + 3 / 2, (x ∈ R, w ∈ R) is π, and when x = π / 6, the function has a minimum value To find the analytic formula of F (x), thank you
f（x）
=Root 3 times sinwxcoswx cos 2 Wx + 3 / 2
=Radical 3 / 2 * sin2wx-1 / 2 (1 + cos2wx) + 3 / 2
=sin（2wx-∏/6）+1
Because t = Π, w = 1
When x = π / 6, the function has a minimum value!
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