We know that the opposite sides of the inner angles a, B and C in the acute angle △ ABC are a, B, C, vector M = (2sinb, root 3), vector n = (2cos ^ 2B / 2-1, cos2b), and m ⊥ n, (1) find the size of B, (2) if B = 2, find the maximum area of △ ABC

We know that the opposite sides of the inner angles a, B and C in the acute angle △ ABC are a, B, C, vector M = (2sinb, root 3), vector n = (2cos ^ 2B / 2-1, cos2b), and m ⊥ n, (1) find the size of B, (2) if B = 2, find the maximum area of △ ABC

∵ vector M = (2sinb, root 3), vector n = (2cos ^ 2B / 2-1, cos2b),
And m ⊥ n
∴m●n=0
2, i. e. 0-2co2 B + 0
∵2cos²B/2-1=cosB
∴2sinBcosB+√3cos2B=0
∴sin2B+√3cos2B=0
∴sin2B=-√3cos2B
∴tan2B=-√3
∵ acute angle △ ABC
∴2B=120º,B=60º
(2)
∵B=60º,b=2
According to the cosine theorem
b²=a²+c²-2accosB
∴4=a²+c²-ac
∵a²+c²≥2ac
∴4=a²+c²-ac≥ac
∴ac≤4
∴SΔABC=1/2acsinB=√3/4*ac≤√3
The maximum area of △ ABC is √ 3

In △ ABC, a, B, C are the opposite sides of angles a, B and C, respectively. The vector M = (2sinb, 2-cos2b), n = (2Sin 2 (π / 4 + B / 2), - 1), and M is perpendicular to n (1) Find the size of angle B, (2) find the value range of sina + sinc

1) M * n = 0 = 2sinb * 2Sin 2 (π / 4 + B / 2) - 2 + cos2b = 2sinb * (1-cos (B + π / 2)) - 2 + cos2b = 2sinb * (1 + SINB) - 2 + 2cosb * cosb-1 = 2sinb-1sinb = 1 / 2 b = 30 or 1502) when B = 30, Sina + sinc = Sina + sin (150-a) = Sina + sin150cosa-sinacos150 = 1 / 2cosa +

Given SiNx + cosx / 2sinx-3cosx = 3, find the value of 1, Tan (π - x), 2,5 / 2Sin? X-3cos? X = 3

First of all, the numerator and denominator of TaNx are divided by cosxtanx + 1 / 2tanx-3 = 3, and then the value of TaNx is 21

Given (2sinx + cosx) / (sinx-3cosx) = - 5, find (1) (SiNx + cosx) / (SiNx cosx) (2) 3sin2x + 4cos2x

1: (2sinx + cosx) / (sinx-3cosx) = - 5. TaNx = 2 can be obtained from cosx
The denominator of (SiNx + cosx) / (SiNx cosx) is also divided by cosx to obtain (1 + TaNx) divided by (tanx-1) = 3
2: Universal displacement formula:
Sin2x = 2tanx divided by (1 + tanx2) = 4 / 5
Cos2x = (1-tanx) divided by (1 + TaNx) = - 3 / 5
So the answer is zero

If Tan (π / 4-x) = - 1 / 2, find (2sinx cosx) / (SiNx + 3cosx)

tan(π/4-x)
=(tanπ/4-tanx)/(1+tanπ/4tanx)
=(1-tanx)/(1+tanx)
=-1/2.
TaNx = do it yourself
(2sinx cosx) / (SiNx + 3cosx) is divided by cosx to obtain (2tanx-1) / (TaNx + 3)=
Take Tan in

It is known that the minimum positive period of the function f (x) = √ 3sin ω xcos ω x-cos? ω x + 3 / 2 (ω = R, X ∈ R) is π and the image is symmetric about x = π / 6; (1) find the analytic formula of F (x); (2) if the equation 1-f (x) = a has only one real root on [0, π / 2], find the range of real number a

(1) F (x) = √ 3sin ω xcos ω x – Cos2 ω x + 3 / 2 = (√ 3 / 2) sin2 ω x – (1 / 2) (1 + Cos2 ω x) + 3 / 2 = (√ 3 / 2) sin2 ω x – (1 / 2) Cos2 ω x + 1 = sin (2 ω x – π / 6) + 1, the minimum positive period of function f (x) is π, so 2 π / | 2 ω| = π / | ω =

Known vector a = (2cosx / 2, Tan (x / 2 + π / 4)), B = (√ 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)) seek F (x) = vector a * vector B, find the maximum value of F (x), the minimum positive period, and find the monotone interval of F (x) on [0, π]!

(x) = 2cosx / 2 × ((x / 2 + π / 4) + Tan (x / 2 + π / 4) + Tan (x / 2 + π / 4) × Tan (x / 2 - π / 4)) = √ 2 [sin (x + π / 4) + sin (π / 4)) + [1 + Tan (x / 2)] / [1-tan (x / 2)] / [1-tan (x / 2)] / [Tan (x / 2)] / [1 + Tan (x / 2)] = √ 2Sin (x + π / 4) maximum value = √ 2, minimum positive cycle = 2 π, 2 π = 2 π = 2 π 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π the increasing region of SiNx

Given the vector a = (2cosx / 2,1), B = (√ 2Sin (x / 2 + Π / 4), - 1), Let f (x) = a * B 1. Simplify f (x) 2. Find the value of F (Π / 12)

1.f(x) = a•b = (2cos(x/2),1)•(√2sin(x/2 + π/4),-1)
= 2√2cos(x/2)sin(x/2 + π/4) - 1
=(2) [{sin (2) + (2) + sin (2) * (2) + sin (2) * (2) + sin (2) * (2)] (x) + sin (2) / (2)] (x) + sin (2) / (2) + (2) / (2)] sin (2) * (2) + (2) + (2) + (2) + (2) * (2) + sin (2) * (2)
= √2[sin(x + π/4) - sin(-π/4)] - 1
= √2 sin(x + π/4)
2.f(π/12) = √2 sin(π/12 + π/4)
= √2 sin(π/3)
= √6/2

The vector a = (sin ω x, 3sin ω x under the radical), B = (sin ω x, cos ω x), ω > 0, f (x) = vector a · vector B, and the minimum positive period of F (x) is π (1) Finding monotone decreasing interval of F (x) (2) Find the value range of F (x) in the interval [0,3 / 2 π]

f(x)=a*b=sin²ωx＋√3sinωxcosωx
=(√3/2)sin2ωx－(1/2)[1－cos2ωx]
=(√3/2)sin2ωx＋(1/2)cos2ωx－(1/2)
=sin(2ωx＋π/6)－(1/2)
1. If the minimum positive period is 2 π / | 2 ω| = π, then: ω = 1;
2. F (x) = sin (2x + π / 6) - (1 / 2)
2K π + π / 2 ≤ 2x + π / 6 ≤ 2K π + 3 π / 2
If K π + π / 6 ≤ x ≤ K π + 2 π / 3, then the decreasing interval is: [K π + π / 6, K π + 2 π / 3], where k ∈ Z
3、x∈[0,2π/3]
Then: 2x + π / 6 ∈ [π / 6,3 π / 2]
The results show that sin (2x + π / 6) ∈ [- 1,1]
Then: F (x) ∈ [- 3 / 2,1 / 2]

Find the symmetry axis and center of y = 1-cos (π / 3 - 3x)

Axis of symmetry:
Let π / 3 - 3x = k π (K ∈ z)
x=π/9-kπ/3(k∈Z)
Center of symmetry:
Let π / 3 - 3x = k π + π / 2 (K ∈ z)
x=-π/18-kπ/3(k∈Z)
The symmetry center is: each (- π / 18-k π / 3,1) (K ∈ z)