How to simplify √ 3cos ^ x + sinxcosx - √ 3 / 2 to sin (2x + π 3)

How to simplify √ 3cos ^ x + sinxcosx - √ 3 / 2 to sin (2x + π 3)

√3cos^2x+sinxcosx-√3/2
=√3/2(1+cos2x)+1/2*2sinxcosx-√3/2
=√3/2cos2x+1/2sin2x
=sinπ/ 3cos2x+cosπ/ 3sin2x
=sin(2x+ π /3 )

Let f (x) = 5sinxcosx-5 root 3cos ^ 2x + 5 root 3 / 2, (x ∈ R), find the value range of F (x) Let f (x) = 5sinxcosx-5 root 3cos ^ 2x + 5 root 3 / 2, (x ∈ R), find the value range of F (x)

So f (x) = 5 / 2sin2x-5 √ 3 (1 + cos2x) / 2 + 5 √ 3 = (5 / 2) sin2x - (5 √ 3 / 2) cos2x = 5 (1 / 2sin2x - √ 3 / 2cos2x) = 5 (sin2xcos π / 6-cos2xsin π / 6) = 5sin (2X - π / 6) 1

Given that the function f (x) = 5sinxcosx-5 √ 3cos ^ 2x + 5 / 2 √ 3 (where x ∈ R), we find that 1. The minimum positive period 2. Monotone interval 3. Symmetry axis and center of F (x) image

F (x) = 5 / 2 * sin2x - 5 √ 3 / 2 * (1 + cos2x) + 5 √ 3 / 2 = 5 / 2 * sin2x - 5 √ 3 / 2 * cos2x = 5 * [1 / 2 * sin2x - √ 3 / 2 * cos2x) = 5sin (2x - π / 3) 1. T = 2 π / ψ = π 2

Let f (x) = 5sinxcosx-5 root 3cos ^ 2x + 5 root 3 / 2, (x ∈ R), find the monotone interval of F (x)

f(x)=5/2*sin2x-5√3*(1+cos2x)/2+5√3/2
=5/2*sin2x-5√3/2*cos2x
=5(sin2x*1/2-cos2x*√3/2)
=5(sin2xcosπ/3-cos2xsinπ/3)
=5sin(2x-π/3)
If SiNx increases, 2K π - π / 2 decreases, 2K π + π / 2, so monotonically increases
2kπ-π/2<2x+π/6<2kπ+π/2
2kπ-2π/3<2x<2kπ+π/3
K π - π / 3, so increasing interval (K π - π / 3, K π + π / 6)
Similarly, the minus interval (K π + π / 6, K π + 2 π / 3)

Given the function f (x) = 5sinxcosx-5 √ 3cos ^ 2x (where x ∈ R), find 1. Minimum positive period 2. Monotone decreasing interval 3. Symmetry axis of F (x) image

(x) = 5 / 2sin22x-5 √ 3 / 2cos2x-5 √ 3 / 2cos2x-5 √ 3 / 2cos2x-5 √ 3 / 2cos2x-5 √ 3 / 2 = 5sin (2x - π / 3) - 5 √ 3 / 2 / 2 (1) t = 2 π / w = 2 π / 2 = 2 = π (2 = 2 = π) 2 (π / 2 + 2K π ≤ 2x - π / 3 ≤ 3 π / 2-2k π (K ∈ z), the reduction interval is 5 π / 12 + K π π / 12 + K π π π π π / 12 + K π π π π (K ∈ Z), the interval is 5 π(K ∈ z) ≤ x ≤ 11 π / 12 + K π (K ∈ z)

Given the function f (x) = sin2x + radical 3cos (π - 2x), find the minimum positive period of the function and the value range of the function

=Sin2x root sign 3cos (π - 2x), find the minimum positive period sum of function

Let f (x) = 5sinxcosx-5 root 3cos ^ 2x + 5 root 3 / 2, (x ∈ R), find the symmetry center of F (x)

f(x)=5/2*sin2x-5√3*(1+cos2x)/2+5√3/2
=5/2*sin2x-5√3/2*cos2x
=5(sin2x*1/2-cos2x*√3/2)
=5(sin2xcosπ/3-cos2xsinπ/3)
=5sin(2x-π/3)
The center of symmetry of SiNx is the intersection with the X axis
So 5sin (2x - π / 3) = 0
2x-π/3=kπ
x=kπ/2+π/6
So the center of symmetry is (K π / 2 + π / 6,0)

Given the function f (x) = 5sinxcosx-5 √ 3cos ^ 2x + [(5 √ 3) / 2] (x ∈ R), find the center and axis of symmetry

F (x) = 5sinxcosx-5 √ 3cos ^ 2x + [(5 √ 3) / 2] = 5 / 2 (sin2x-2 √ 3cos ^ 2x + √ 3) = 5 / 2 [sin2x - √ 3 (2cos ^ 2x - √ 1)] = 5 / 2 (sin2x - √ 3cos ^ 2x] is reduced to f (x) = 5sin (a + 2x) according to the auxiliary angle formula, so the symmetry center (K π, 0) axis X = π / 2 + K π (k belongs to Z)

Find the function y = sin square x + 2Sin xcos x + 3cos square Maximum and minimum of

The square of sin + 2xs + 2xs
=1 + 2Sin xcos x + 2cos square x
=1+sin 2x+cos2X+1
=2 + √ 2Sin (2x + 45 degrees)
Because - 1

The period, amplitude and initial phase of y = - 2Sin (1 / 2x + π / 4) The answer is 4 π, 2, π / 4 (why amplitude is not - 2?) how to calculate the initial phase?

Amplitude refers to the amplitude of the vibration, how can it be negative? The amplitude is | - 2 |, and the initial phase is the value in the bracket of sin when x = 0