Radical 3sinx cosx =? - 1 / 2sinx cosx =? - SiNx root 3cosx =? - SiNx?

Radical 3sinx cosx =? - 1 / 2sinx cosx =? - SiNx root 3cosx =? - SiNx?

analysis,
√3sinx-cosx
=2sin(x-π/6)
-1/2*sinx-cosx
=-√5/2sin(x+a),【cosa=√5/5,sina=2/√5】
-sinx-√3cosx
=-2sin(x+π/3)

Known a＝(1−cosx，2sinx 2)， b＝(1+cosx，2cosx 2) (1) If f (x) = 2 + SiNx − 1 4| A- B|2, find the expression of F (x) (2) If the images of function f (x) and function g (x) are symmetric about the origin, find the analytic formula of G (x) (3) If h (x) = g (x) - λ f (x) + 1 at [− π 2，π 2] The value range of real number λ is calculated

Solution (1): F (x) = 2 + SiNx − 1
4[4cos2x+4(sinx
2−cosx
2)2]，
=2+sinx-cos2x-1+sinx=sin2x+2sinx
(2) Let any point m (x0, Y0) of the graph of function y = f (x)
On the symmetric point of n
Then x0 = - x, Y0 = - y,
∵ the point m is on the graph of the function y = f (x)
ν - y = sin2 (- x) + 2Sin (- x), that is, y = - sin2x + 2sinx
The analytic formula of the function g (x) is g (x) = - sin2x + 2sinx
（3）∵h（x）=-（1+λ）sin2x+2（1-λ）sinx+1，
Let SiNx = t,
∵x∈[−π
2，π
2]
∴-1≤t≤1，
Then H (T) = - (1 + λ) T2 + 2 (1 - λ) t + 1 (- 1 ≤ t ≤ 1)
① When λ = - 1, H (T) = 4T + 1 is an increasing function on [- 1, 1],  λ = - 1,
② When λ ≠ - 1, the equation of symmetry axis is a straight line t = 1 − λ
1+λ
I) when λ is less than - 1, 1 − λ
1 + λ ≤− 1, the solution is λ < - 1
When λ is greater than - 1, 1 − λ
1 + λ ≥ 1, the solution is - 1 ＜ λ ≤ 0. In summary, λ ≤ 0

Given the vector a = (2sinx, cosx) B = (√ 3cosx, 2cosx), define f (x) = vector a * B-1 to find the symmetry axis

f(x) = a.b-1
=（2sinx,cosx）.（√3cosx,2cosx） -1
= 2sinxcosx + 2(cosx)^2 -1
= sin2x+cos2x
= sin(π/2-2x)+ cos(π/2-2x)
= sin2(π/4-x) + cos2(π/4-x)
f(x) = f(π/4-x)
f(π/8-x) = f(π/4-(π/8-x)
=f(π/8+x)
Axis of symmetry: y = π / 8

Let a = (cosx, 2sinx), B = (2cosx, 3cosx), f (x) = vector A. vector B, find the minimum positive period and monotone increasing interval

F (x) = 2cos square x plus 2 times root three sinxcosx = cos2x plus root three sin2x plus one = 2Sin (2x + card / 6) plus one. So the positive period is the card, and the increasing interval is [K card - 60 degrees, K card + 30 degrees],

The minimum positive period of the function f (x) = sin (1 / 2x + π / 3) + cos (1 / 2x - π / 6) is It's better to attach the formula in detail. I don't know anything about it

F (x) = sin (1 / 2x + π / 3) + cos (1 / 2x - π / 6) = sin (1 / 2x + π / 3) + sin (1 / 2x - π / 6 + π / 2) (induction formula) = 2Sin (1 / 2x + π / 3)
Then y = asin (Wx + FAI), the minimum positive period T = 2 π / | w |, so t = 2 π / (1 / 2) = 4 π

Given the function f (x) = cos (2 π - x) cos (π / 2-x) - Sin ^ 2x (1), find the minimum positive period of function f (x) (2) When x ∈ [- π / 8,3 / 8 π], find the range of the function f (x)

1. F (x) = cos (2 π - x) cos (π / 2-x) - Sin ^ 2x = - cosxsinx sin ^ 2x = - half sin2x - (1-cos2x) / 2 = - 1 / 2sin2x + 1 / 2cos2x-1 / 2 = - √ 2 / 2Sin (2x - π / 4) - 1 / 2, so the minimum positive period of the function f (x) is π 2. When x ∈ [- π / 8,3 π / 8], - π / 2 ≤ 2x - π / 4 ≤ π

Let f (x) = 2sinxcosx − cos (2x − π) 6)． (1) Find the minimum positive period of function f (x); (2) Find the monotone increasing interval of function f (x); (3) When x ∈ [0, 2 π 3] The maximum value of function f (x) and the value of X when the maximum value is obtained

f（x）=2sinxcosx-cos（2x-π
6）=sin2x-（cos2xcosπ
6+sin2xsinπ
6）=-cos（2x+π
6）
（1）T=2π
2=π
(2) The monotone increasing interval of function f (x) is 2x + π
6∈[2kπ，π+2kπ]k∈Z
∴x∈[−π
12+kπ，5π
12+kπ]k∈Z
In other words, the monotone increasing interval of function f (x) is x ∈ [− π)
12+kπ，5π
12+kπ]k∈Z
(3) When x ∈ [0, 2 π
3] When, 2x + π
6∈[π
6，3π
2]
ν when + π
When 6 = π, f (x) takes the maximum value, that is, x = 5 π
At 12, f (x) max = 1

Let f (x) = 2sinxcosx − cos (2x − π) 6)． (1) Find the minimum positive period of function f (x); (2) Find the monotone increasing interval of function f (x); (3) When x ∈ [0, 2 π 3] The maximum value of function f (x) and the value of X when the maximum value is obtained

f（x）=2sinxcosx-cos（2x-π
6）=sin2x-（cos2xcosπ
6+sin2xsinπ
6）=-cos（2x+π
6）
（1）T=2π
2=π
(2) The monotone increasing interval of function f (x) is 2x + π
6∈[2kπ，π+2kπ]k∈Z
∴x∈[−π
12+kπ，5π
12+kπ]k∈Z
In other words, the monotone increasing interval of function f (x) is x ∈ [− π)
12+kπ，5π
12+kπ]k∈Z
(3) When x ∈ [0, 2 π
3] When, 2x + π
6∈[π
6，3π
2]
ν when + π
When 6 = π, f (x) takes the maximum value, that is, x = 5 π
At 12, f (x) max = 1

It is known that f (x) = cos ^ 2x sin ^ x + 2sinxcosx. ① find the minimum positive period of function; ② when x ∈ [0, π / 2], find the maximum and minimum value of function f (x)

(x) = cos ^ 2x-sin ^ x + 2sinxcosx = cos2x + sin2x = 2x + sin2x = 2 (sinπ / 4cos2x + cos π / 4sin2x / 4sin2x) = √ 2Sin (2x + π / 4) function minimum positive period = 2 π / 2 = π / 2 = π x ∈ [0, π / 2] 2x ∈ [0, π / 2] 2x ∈ [0, π] 2x + π / 4 ∈ [π / 4,5 π / 4] 2x + π / 4 ∈ [π / 4, π / 4] 2x + π / 4 ∈ [π / 4, π / 2] when the monotonmonotonmonotonmonotonmonotonmonotonmonotonmonotonmonotonmonotonmonotonmonotonmonotonmonotone 2x +

Given the function f (x) = 2sinxcosx cos (2x + π / 6) 1, find the minimum positive period of function f (x) 2 Given the function f (x) = 2sinxcosx cos (2x + π / 6) 1, find the minimum positive period of function f (x), find the monotone increasing interval of function f (x)

f(x)=sin2x-(√3/2)cos2x+(1/2)sin2x=√3sin(2x-π/6)
So the minimum positive period = π;
When: 2K π - π / 2 ≤ 2x - π / 6 ≤ 2K π + π / 2
kπ-π/6≤x≤kπ+π/3
So the increasing interval is: [K π - π / 6, K π + π / 3]