Given the function f (x) = SiNx (cosx - √ 3sinx), find the minimum positive period of function f (x)

Given the function f (x) = SiNx (cosx - √ 3sinx), find the minimum positive period of function f (x)

Because f (x) = SiNx (cosx - √ 3sinx) = sinxcosx - √ 3 (SiNx) ^ 2 = 1 / 2sin2x - √ 3 / 2 (1-cos2x) = (1 / 2sin2x - √ 3 / 2cos2x) -√ 3 / 2 = (sin2xcos π / 3-cos2xsin π / 3) -√ 3 / 2 = sin (2x - π / 3) -√ 3 / 2, so the period T = 2 π / 2 = π

Given the function f (x) = cosx * cosx + 3sinx + 3, find its maximum and minimum values

f（x）=cosx*cosx+3sinx+3
=1-(sinx)^2+3sinx+3
=-(sinx-3/2)^2+25/4
When: SiNx = 1, the maximum value is: 6
When SiNx = - 1, the minimum value is: 0

The function f (x) = The minimum value of 3sinx − cosx (0 ≤ x ≤ π) is () A. -2 B. 1 C. − Three D. -1

f(x)＝
3sinx−cosx=2（
Three
2sinx-1
2cosx）=2sin（x-π
6）
∵0≤x≤π
∴-π
6x-π
6≤5π
Six
∴-1
2≤sin（x-π
6）≤1
The minimum value of the function is 2 × (- 1)
2）=-1
Therefore, D is selected

Given the function f (x) = 2acosx (Radix 3 · SiNx + cosx) + a squared (a > 0) (root 3 is together) (1) any x ∈ r of a weak team has f (x)

(1)f(x)=2acosx(√3sinx+cosx)+a^2=a(2√3sinxcosx+2cos^2a)+a^2=a(√3sin2x+2-2sin^2x)+a^2=a(√3asin2x+1+1-2sin^2x)+a^2=2a(√3/2sin2x+1/2cos2x)+a+a^2=2a(cosπ/6sin2x+sinπ/6cos2x)+a+a^2=2asin(2x+π/6)+a+a...

Find the minimum positive period of the function y = | cosx |

T＝π

The minimum positive period of the function y = | cosx | is

π. Because after folding, the image repeats every π units, so the period is π

The minimum positive period of the function y = cosx,

The minimum period of the solution is t = 2 π / / W/
so
The minimum positive period of the function y = cosx, t = 2 π / 1 = 2 π

Given the vector a = (sin (x + π / 6), cosx), B = (cosx, cos (x - π / 3)), function f (x) = vector a · B-1 / 2, ① find the minimum positive period of function f (x)

f(x)=sin(x+π/6)cosx+cosxcos(x-π/3)-1/2
cos(x-π/3)=cos(π/3-x)
=cos[π/2-(x+π/6)]
=sin(x+π/6)
Therefore, f (x) = sin (x + π / 6) cosx + cosxsin (x + π / 6) - 1 / 2
=sin(x+π/6+x)-1/2
=sin(2x+π/6)-1/2
Therefore, the minimum positive period T = 2 π / 2 = π
Wish you happy! Hope to help you, if you do not understand, please ask, I wish you progress_ ∩)O

The known function FX = sin (pai-x) - cosx (1) Find the minimum positive period of the function f (x), (2) find the maximum and minimum value of the function, (3) if f (a) = 1 / 4, a belongs to (0, Pai / 2), find the value of sina + cosa

f(x)=sinx-cosx=√2sin(x-4/π)
(1).T=2π
(2).f(x)max=√2 f(x)min=-√2
(3).sina+cosa=√2cos(a-π/4)
cos(a-π/4)=√[1-sin²(a-π/4)]
Sin 2 (a - π / 4) = [f (a) / √ 2] 2 = 1 / 32 a ∈ (0, π / 2) so (a - π / 4) ∈ (- π / 4, π / 4)
So cos (a - π / 4) = √ 31 / 32
So Sina + cosa = √ 2cos (a - π / 4) = √ 62 / 32

Given the vector a = (2sinx, cosx), B = (root 3cosx, 2cosx), function f (x) = vector a multiplied by vector B (1) Finding the minimum positive period of function f (x) (2) Monotone increasing interval of spherical function f (x)

(1) The vector vector a = (2sinx, cosx), B = (the root number 3cosx, 2cosx), f (x) = a ● B = 2 √ 3 SiNx cosx + 2cos, x = √ 3sin2x + cos2x + 1 = 2 (√ 3 / 2 * sin2x + 1 / 2 * cos2x + 1 = 1 = 2Sin (2x + π / 6) + 1 function f (x) the minimum positive cycle T = 2 π / 2 = π (2) 2K π - π / 2 ≤ 2x + π / 6 ≤ 2K π + π / 6 ≤ 2K π + π + π / 6 ≤ 2K π + π + π + π + π + π + π + π + π + π + π + π/ 2, K