F (x) = - 1 / 2 + sin (PAI / 6-2x) + cos (2x Pai / 3) + cos ^ 2x Max Simplify f (x) = - 1,2 + sin (π - 6-2x) + cos (2x - π - 3) + (cosx) square 1 to find the minimum positive period of F (x) and the maximum value of F (x) in the interval [π - 8,5 π - 8], The first question is: F (x) = - 1 / 2 + 1 / 2cos2x - √ 3 / 2sin2x + 1 / 2cos2x + √ 3 / 2sin2x + cos ^ 2x = - 1 / 2 + cos2x + cos ^ 2x = - 1 / 2 + cos ^ 2x + cos ^ 2x = - 3 / 2 + 3cos ^ 2x = - 3 / 2 + 3cos ^ 2x = - 3 / 2 + 3 (1 + cos2x) / 2 = - 3 / 2 + 3 / 2cos2x = 3 / 2cos2x t = 2pi / 2 = Pi

F (x) = - 1 / 2 + sin (PAI / 6-2x) + cos (2x Pai / 3) + cos ^ 2x Max Simplify f (x) = - 1,2 + sin (π - 6-2x) + cos (2x - π - 3) + (cosx) square 1 to find the minimum positive period of F (x) and the maximum value of F (x) in the interval [π - 8,5 π - 8], The first question is: F (x) = - 1 / 2 + 1 / 2cos2x - √ 3 / 2sin2x + 1 / 2cos2x + √ 3 / 2sin2x + cos ^ 2x = - 1 / 2 + cos2x + cos ^ 2x = - 1 / 2 + cos ^ 2x + cos ^ 2x = - 3 / 2 + 3cos ^ 2x = - 3 / 2 + 3cos ^ 2x = - 3 / 2 + 3 (1 + cos2x) / 2 = - 3 / 2 + 3 / 2cos2x = 3 / 2cos2x t = 2pi / 2 = Pi

1) F (x) = (3 / 2) cos2x, t = 2 π / 2 = π 2) x ∈ [π / 8,5 π / 8],  2x ∈ [π / 4,5 π / 4] y = cosx on [π / 4,5 π / 4] y = (3 / 2) cos2x on [π / 8,5 π / 8] x = π / 8, then f (x) = (3 / 2) × √ 2 / 2 = 3 √ 2 / 4
Remember to adopt it

f(x)=sin(2x-pai/6)
G (x) = f (x + FAI). If the function g (x) is an even function, find the value of the smallest integer Fai satisfying the condition

There is no minimum integer Fai that satisfies the condition. If it is the minimum positive number Fai that satisfies the condition, then there exists
From F (x) = sin (2x - π / 6), f (x + FAI) = sin (2x + 2fai - π / 6)
Then G (x) = f (x + FAI) = sin (2x + 2fai - π / 6)
Because the function g (x) is even, G (x) = g (- x), that is sin (2x + 2fai - π / 6) = sin (- 2x + 2fai - π / 6)
And because sin (2x + 2fai - π / 6) = sin [(2k + 1) π - (2x + 2fai - π / 6)] (where k is an integer)
So (2k + 1) π - (2x + 2fai - π / 6) = - 2x + 2fai - π / 6, Fai = (3K + 2) π / 6 (where k is an integer)
The minimum positive number Fai = π / 3 satisfying the condition when k = 0

If (COS 2a) / (1 + sin2a) = 1 / 5, then Tan a =?

'2 means square
cos2a/(1 sin2a)=1/5
→5(cosa'2-sina'2)=sina'2 cosa'2 2sina*cosa
→4cosa'2-6sina'2=2cosa*sina
→2-3tana'2=tana
Let t denote Tana and solve the equation 3T'2 T-2 = 0
T = 2 / 3 or - 1

Tan a + 1 / Tan a = 5, cos 2A + sin2a-1 / 1-tan a
(Tan a + 1) / (Tan a) = 5, find (COS 2A + sin2a-1) / (1-tan a)

First, the given conditions are simplified,
[ (sina/cosa)+1 ] / (sina/cosa)
=(Sina + COSA) / Sina = 5 gives cosa = 4sina
The simultaneous solution of sin ^ 2 + cos ^ 2 = 1 shows that sina is equal to 1 / 17 of the positive and negative root sign, cosa is equal to 4 / 17 of the positive and negative root sign
Simplification (COS 2A + sin2a-1) / (1-tan a)
The original formula = (COS 2A + sin2a cos ^ 2a-sin ^ 2a) / (1-sina / COSA)
= (2sinacosa-2sin^2a)/ (1-sina/cosa)
= 2sinacosa=sin2a=8/17
Because I have left my textbook for several years, I can't remember many direct formulas clearly, so I have to use the most stupid method to solve them. I hope you can refer to it

Tan (a - π / 4) = 1 / 2 (1) find the direct of Tana (2) find the direct of the square of sin2a cosa

(1)tan(a-π/4)=1/2（tana-tanπ/4）/（1+tanatanπ/4）=1/2（tana-1）/（1+tana）=1/22tana-2=1+tanatana=3(2)sin2a-cos²a=(2sinacosa-cos²a)/(sin²a+cos²a)=(2tana-1)/(tan²a+1)=(2×3-1...

Who knows this problem? If Tan (a + 45 degree) = (?) 189;, and - 45 degree < a < 0, then 2Sin ^ 2A + sin2a / cos（
Such as the title
2Sin ^ 2A + sin2a / cos (a-45) =? This is the topic

From Tan (a + 45 °) = (1 + Tana) / (1-tana) = (?) 189, Tana = - 1 / 3 is obtained
So 1 + Tan & # 178; a = 10 / 9 = 1 / cos & # 178; a and - 45

Tan (45 degrees + a) = 3, find sin2a-2cos ^ 2A

tg(45+a)=(tg45+tga)/(1-tg45*tga)=(1+tga)/(1-tga)=3
So TGA = 1 / 2
sin2a-2cos^2a=sin2a-cos2a-1=2tga/(1+(tga)^2)-(1-(tga)^2)/(1+(tga)^2)-1=-4/5

Given Tan (45 ° + a) = 3, find the value of sin2a-2cosa ^ 2

From the solution of Tan (45 ° + α) = (tan45 ° + Tan α) / (1-tan45 ° * Tan α) = (1 + Tan α) / (1-tan α) = 3, it is obtained that if Tan α = 1 / 2, then sin2 α - 2cos α ^ 2 = (2Sin α * cos α - 2cos α ^ 2) / 1 = (2Sin α * cos α - 2cos α ^ 2) / (sin α ^ 2 + cos α ^ 2) from Tan α = sin α / cos α is meaningful

Given a B ∈ (3 π / 4, π), Tan (a - π / 4) = - 2, sin (a + b) = - 3 / 5 (1), we can find the value of sin2a
(2) Finding the value of Tan (B + π / 4)

《1》tan(a-π／4)=[tana-tan(π／4)]/[1+tana*tan(π／4)]
=(tana-1)/(1+tana)=-2
So Tana = - 1 / 3, because 1 + Tan ^ a = 1 / cos ^ A, a B ∈ (3 π / 4, π),
So cosa = (3 √ 10) / 10, Sina = √ 10 / 10
So sin2a = 2sinacosa = 2 * (3 √ 10) / 10 * √ 10 / 10 = 3 / 5
《2》 Because a B ∈ (3 π / 4, π), so (a + b) ∈ (3 π / 2,2 π),
And COS (a + b) = 4 / 5, so tan (a + b) = - 3 / 4, Tan (B + π / 4) = Tan [(a + b) - (a - π / 4)]
So tan (B + π / 4) = 1 / 2

It is known that sin (2a + x) = 2sinx. Proof: Tan (a + x) = 3tana

Because
sin(2a+x) = sin(a+x+a) = sin(a+x)cos(a)+cos(a+x)sin(a)
2sin(x) = 2sin(a+x-a) = 2sin(a+x)cos(a)-2cos(a+x)sin(a)
sin(2a+x)=2sin(x)
So it's easy to sort it out
sin(a+x)cos(a) = 3cos(a+x)sin(a)
Namely
sin(a+x)/cos(a+x) = 3 sin(a)/cos(a)
Provable
tan(a+x) = 3 tan(a)