Trigonometric function SiN x is divided into three or multiplied by three. How does the period change?

Trigonometric function SiN x is divided into three or multiplied by three. How does the period change?

If the minimum positive period of the cubic root of SiN x is t, then sin (x + T) = sin X. obviously, when t = 2 π, sin (x + T) = sin X

The minimum positive period of the function y = sin (x / 2) is? () a: π B: 2 π You'd better pay for the drawing

The minimum positive period of y = │ SiNx │ is π
therefore
The minimum positive period of y = │ sin (x / 2) │ is 2 π

Find the period, monotone interval, maximum and minimum value of the function y = 2Sin (3x + 3 / π), and write the set of independent variable x when the maximum value and minimum value are obtained respectively

T = 2 π / W, so t = 2 π / 3 monotonically increasing interval: 2K π - π / 2 ≤ 3x + π / 3 ≤ 2K π + π / 2, namely: 2K π / 3 - 5 π / 18 ≤ x ≤ 2K π / 3 + π / 18 monotone decreasing interval: 2K π + π / 2 ≤ 3x + π / 3 ≤ 2K π + 3 π / 2, namely 2K π / 3 + π / 18 ≤ x ≤ 2K π 3 +

The minimum value of the function y = 2 root sign 3sinxcosx-2cos? X + 1

y=√3sin2x-cos2x
=2sin(2x-30°)
ymin=-2

Given the function y = sin ^ 2 x + root sign 3sinxcosx + 2cos ^ 2 x, X ∈ R, find the maximum and minimum value of the function

y=(1-cos2x)/2+√3/2*sin2x+(1+cos2x)
=3/2+√3/2sin2x+1/2cos2x
=3/2+sin(2x+π/6)
The maximum value is 3 / 2 + 1 = 5 / 2
The minimum value is 3 / 2-1 = 1 / 2

The value range of the trigonometric function y = sin (π - x) * cos (π + x) is

y=sin（π-x）*cos（π+x）
==-sinxcosx
=-1/2sin2x
So the range is [- 1 / 2,1 / 2]

The detailed steps of 2Sin (- 1110 °) - Sin 960 ° + √ 2cos (- 225 °) + cos (- 210 °) are obtained,

2sin(-1110°）-sin960°+√2cos(-225°）+cos(-210°）
=2sin(-1080°-30º）-sin(720°+240º)+√2cos225°+cos210°
=-2sin30º-sin(180º+60º)+√2cos(180º+45°)+cos(180º+30°)
=-1 +sin60º-√2cos45°-cos30°
=-1+√3/2-1-√3/2=-2

Known points a (1,1), B (1, - 1), C (root 2cos θ, Radix 2Sin θ), O are the coordinate origin If the absolute value (vector BC - vector BA) = radical 2, find sin2 θ

A(1,1),B(1,-1),c(√2cosθ,√2sinθ)
BC=(√2cosθ-1,√2sinθ+1)
BA=(0,2)
Vector BC vector Ba = (√ 2cos θ - 1, √ 2Sin θ - 1)
2(cosθ)^2+1-2√2cosθ+2(sinθ)^2+1-2√2sinθ=2
1-√2cosθ-√2sinθ=0
√2(sinθ+cosθ)=1
(sinθ)^2+(cosθ)^2+2sinθcosθ=1/2
1+2sinθcosθ=1/2
2sinθcosθ=sin2θ=-1/2

2 sin α - cos α = root 3 sin α, then cos α The domain of Log1 / 2 SiNx under y = radical is

∵ log1/2 sinx 》= 0
∴ sin x ∈（0,1]
∴x ∈ （2kπ,π+2kπ） k∈ R .

Prove cos (a) + Radix 3sin (a) = 2Sin (π / 6 + a)

On the right, 2Sin (π / 6 + a) = 2Sin (π / 6) cosa + 2sinacos (π / 6) = cosa + radical 3sin (a) = left formula#