In the triangle ABC, if sin (2 π - a) = - radical 2 (π - b) and root 3cosa = - radical 2cos (π - b), find the three inner angles of triangle ABC

In the triangle ABC, if sin (2 π - a) = - radical 2 (π - b) and root 3cosa = - radical 2cos (π - b), find the three inner angles of triangle ABC

-sinA=-√2sinB,sinA=√2sinB
√3cosA=√2cosB,cosA=√(2/3)cosB
(sinA)^2+(cosA)^2=1
SO 2 (SINB) ^ 2 + (2 / 3) (COSA) ^ 2 = 1
(4/3)(sinB)^2+(2/3)[(sinB)^2+(cosA)^2]=1
(4/3)(sinB)^2+(2/3)=1
(sinB)^2=1/4
The interior angle of the triangle is between 0 and 180 degrees
So SINB > 0
So SINB = 1 / 2, B = 30 or 150 degrees
Sina = √ 2sinb = √ 2 / 2, a = 45 degrees or 135 degrees
If B = 150 degrees, then a + C = 30 degrees, which contradicts a = 45 degrees or 135 degrees
So B = 30 degrees
So CoSb = √ 3 / 2
cosA=√(2/3)cosB=√2/2
So a = 45 degrees
All in all
A = 45 degrees
B = 30 degrees
C = 105 degrees

In the acute triangle ABC, we know that the inner angles a, B and C are a, B, C. The vector M = (2Sin (a + C), root 3), n = (cos2b, 2cos square B / 2-1) And the vectors m, n are collinear (1) (2) if B = 1, find the largest area of triangle ABC

(1)
m. N collinear
>
2sin(A+C)/√3 = cos2B/(2(cos(B/2))^2-1)
2sin(A+C) .cosB = √3cos2B
2sinBcosB = √3cos2B
sin2B=√3cos2B
tan2B = √3
B = π/6
(2)
B=1
by sine-rule
a/sinA =b/sinB
a = (b/sinB)sinA
=2sinA
And
c= 2sinC = 2sin(5π/6-A)
A1=area of ABC
=(1/2)acsinB
= (1/2)(2sinA)(2sin(5π/6-A))(1/2)
= sinAsin(5π/6-A)
=(1/2)(cos(2A-5π/6) - cos5π/6)
=(1/2)(cos(2A-5π/6) +√3/2)
max A1 at cos(2A-5π/6)=1
max A1 = (1/2)(1+√3/2)

TaNx = 3 find the value of 2 / 3 sin square and 1 / 4 cos square X

tanx=sinx/cosx=3
sinx=3cosx
sin^2x+cos^2x=1
9cos^2x+cos^2x=1
cos^2x=1/10
2/3(sin^2x)+(cos^2x)/4
=2/3*9cos^2x+(cos^2x)/4
=25/4 cos^2x
=25/4*1/10
=5/8

Given cos (π / 4 + x) = 4 / 5 x ∈ (- π / 2, - π / 4), find the value of (sin2x-2sin square x) / (1 + TaNx)

At the same time, we can get the following results: (1 / 4 + x) = 4 / 5 x ∈ (- π / 2, - π / 4), sin (π / 4 + x) = - 3 / 5 x ∈ (- π / 2, - π / 4), SiNx = sin (π / 4 + x) cos π / 4-sin (π / 4) cos (π / 4 + x) = (- 7 √ 2) / 10; cosx = √ 2 / 10; sin2x-2sin square x) / (1 + TaNx) = (2sinx cosx-2 sinxsinx) / ((1 + TaNx) = (2sinx cosx-2 sinxsinx) / ((1 + TaNx) / ((2) sinxsinx) / ((1 + TaNx) / ((2) sincosx + s

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

sinx+cosx/2sinx-3cosx=3 sinx+cosx=3(2sinx-3cosx)=6sinx-9cosx5sinx=10cosxsinx=2cosxsin²x=4cos²x1-cos²x=4cos²x5cos²x=1cos²x=1/52sin²x-3sin（3π+x）cos（π-x）-3cos&sup...

It is known that the negative dichotomy is less than X and less than zero, SiNx + cosx = 1 / 5 (1) find the value of SiNx cosx (2) find the square of 3sin x-2sin It is known that the negative dichotomy is less than X and less than zero, and SiNx + cosx = one fifth (1) Find the value of SiNx cosx (2) Find the value of the square of 3sin x-2sin bisection xcos bisection x + cos bisection X

1. Square 1 + 2sinxcosx = 1 / 25, 2sinxcosx = - 24 / 25 (SiNx cosx) 2 = 1-2sinxcosx = 49 / 25
Then SiNx cosx = - 7 / 5
2.sinx=-3/5 tanx=-3/4
The original formula = (3tan? X-2tanx + 1) / (tan? X + 1) = 67 / 25

Let f (x) = 2Sin ^ 2 (π / 4 + x) -√ 3cos2x 1. Find the minimum positive period and monotone decreasing interval of F (x)

2Sin ^ 2 (π / 4 + x) can be converted to 1-cos (π / 2 + 2x)
Therefore, the original formula can be changed into: F (x) = 1-cos (π / 2 + 2x) -√ 3cos2x, that is: F (x) = 1 - [cos (π / 2 + 2x) + √ 3cos2x]
=1-(SIN2X+√3cos2x)
=1-2*SIN(2X+π/3)
Therefore: 1. The minimum positive period of F (x) is: 2 π / 2 absolute value = π
2. The monotone decreasing interval of F (x) and the monotone increasing interval of sin (2x + π / 3) are required, that is, 2K π - π / 2

It is known that the function f (x) = 2Sin (x + π / 4) ^ 2 - √ 3cos2x-1. X belongs to R. (1) find the maximum value and the minimum positive period of F (x); (2) if the image H (x) = f (x + T) is symmetric about the point (- π / 6,0), and t belongs to (0, π), find the value of T (3) if x belongs to [π / 4, π / 2], there is always if (x) - MI

1) (x) = 2Sin (x + π / 4) ^ 2 - √ 3cos2x-1 = - √ 3cos2x - [1-2sin (x + π / 4) ^ 2] = - √ 3cos2x-sin2x = -2sin (2x + π / 3) has a minimum value of - 2, a maximum value of 2, a minimum value of 2, a minimum positive cycle of T = π 2) H (x) H (x) = f (x + T + T) = - 2Sin (2x + 2T + π / 3) about the point (- π / 6,0) symmetry, so 2 × (- π π / 6,0) symmetry, so 2 × (- ππ π π π, 6,0) symmetry, so 2 × (- ππ π π/ 6) + 2T + π / 3

Given the function f (x) = 4cos square x + 4 times root sign 3sin xcos x-3 (x included in R), find the minimum period of function f (x), and write the symmetry axis of the function

f(x)=4cos²x+4√3sinxcosx-3
=2(2cos²x-1)+2√3(2sinxcosx)-1
=2cos2x+2√3sin2x-1
=4sin(2x+π/3)-1
The minimum positive period is 2 π / 2 = π
The axis of symmetry of a function
That is, the value of X when the function gets the maximum or minimum value
2x-π/3=kπ+π/2
2x=kπ+5π/6
x=kπ/2+5π/12 k∈z
So the axis of symmetry is
x=kπ/2+5π/12 k∈z

The symmetric axis equation of the function f (x) = radical 3sin? X is

f(x)=√3*(1-cos2x)/2
The axis of symmetry of COS is where the maximum value is taken
That is, cos2x = ± 1
2x=kπ+π/2
So it's x = k π / 2 + π / 4