Simplification (sin α + cos α) / (1 + cos α + Cos2 α)=

Simplification (sin α + cos α) / (1 + cos α + Cos2 α)=

The formula is: Cos2 α = 2cos α - 1; sin α / cos α = Tan α
=(sin α + cos α) / (1 + cos α + 2cos α - 1)
= (sinα+cosα)/cosα(2cosα+1)
= (tanα+1)/(2cosα+1)

Given sin θ + cos θ = 2Sin α, sin θ cos θ = (sin β) ^ 2, it is proved that 4 (Cos2 α) ^ 2 = (Cos2 β) ^ 2 High school mathematics, to step, thank you!

sinθ+cosθ=2sinα
(sinθ+cosθ)^2=1+2sinθcosθ=4(sina)^2
1+2(sinβ)^2=4(sina)^2
2-cos2β=2-2cos2a
2cos2a=cos2β
4(cos2α)^2=(cos2β)^2

If f (A / 2) = 3 √ 2 / 5 and a belongs to (π / 2, π), try to find Sina

f(x)
=a.b
=(1,sin2x).(cos2x,1)
=cos2x+sin2x
f(a/2) = cosa + sina = 3√2/5
=> (cosa)^2 = (3√2/5 - sina)^2
2(sina)^2 - (6√2/5) sina -7/25 =0
50(sina)^2 - 30√2sina -7 =0
sina = (30√2 + 40√2)/100
= 7√2/10

(1+ sin2x)/(1+ cos2x+ sin2x)=(1/2)tanx+ 1/2

(1+ sin2x)/(1+ cos2x+ sin2x)
=（sin²x+cos²x+2sinxcosx)/(1+2cos²x-1+2sinxcosx)
=(sinx+cosx)²/[2cosx(sinx+cosx)]
=(sinx+cosx)/(2cosx)
=(1/2)tanx+ 1/2

Given TaNx = 2, find sin2x + cos2x of cos2x-sin2x

Cos2x-sin2x = (cosx) ^ 2 - (SiNx) ^ 2-2sinx * cosx denominator sin2x + cos2x = 2sinx * cosx + (cosx) ^ 2 - (SiNx) ^ 2 the molecular denominator of (cosx) ^ 2 is removed: 1 - (TaNx) ^ 2-2tanx = 1-4-4 = - 7, molecule: 2 * TaNx + 1 - (TaNx) ^ 2 = 1, so the original formula = - 1 / 7

Simplification of sin2x (1 + TaNx · TaNx) 2) The result is______ .

sin2x(1+tanx•tanx
2)=sin2x（1+sinxsinx
Two
cosxcosx
2）=sin2x（1+2sinx
2sinx
Two
cosx）=sin2x（1+1−cosx
cosx）=2sinx
So the answer is: 2sinx

Simplify 1 + cos2x divided by (TaNx / 2) - (1 / TaNx / 2), A -1/2sin2x B 1/2sin2x C -2sinx D 2sin2x

1+cos2x=2(cosx)^2
tanx/2-1/(tanx/2)=(sinx/2)/(cosx/2) - (cosx/2)/(sinx/2)
=[(sinx/2)^2-(cosx/2)^2]/[(sinx/2)*(cosx/2)]
=-cosx/(1/2sinx)
=-2cosx/sinx
Original formula = 2 (cosx) ^ 2 / (- 2cosx / SiNx)
=-cosx*sinx=-1/2sin(2x)

Let f (x) = AB, where vector a = (m, cos2x), B = (1 + sin2x, 1), X ∈ R, and the image of function y = f (x) passes through point (π / 4,2) (1) Find the value of real number M; (2) Find the minimum value of function f (x) and the set of x value at this time, and write its increasing interval

(1)f(x)=ab=m(1+sin2x)+cos2x
Substituting (π / 4,2) into f (x)
M (1 + 1) + 0 = 2 is obtained
M = 1
(2) Therefore, f (x) = sin2x + cos2x + 1 = √ 2Sin (2x + π / 4) + 1
Let 2x + π / 4 = 2K π - π / 2 (k is an integer) to obtain x = k π - 3 π / 8 (k is an integer)
The minimum value of F (x) is 1 - √ 2
Let 2K π - π / 2

The image passing through point (π / 4,2) with vector a = (m, cos2x), B = (1 + sin2x, 1), X ∈ R and function y = f (x) is adopted If the function f (x + φ) - 1 is an even function, find φ

If (x) = AB = AB = m (1 + sin2x) + cos2x image over point (π / 4,2), that is 2 = m (1 + sin π / 2) + cos π π / 22m = 22m = 2m = 1F (x) = 1 + sin2x + cos2x + cos2x = √ 2Sin (2x + π / 4) + 1F (x + φ) - 1 = = √ 2Sin [2 (x + φ) + π / 4] + 1-1 = √ 2Sin (2x + 2 φ + π / 4) makes g (x) = f (x + X + φ) - 1 + π / 4) Let G (x) = f (x + X + X + φ) - 1 + π / 4) make g (x) = f (x) = f (x + X + X + X is the even function 2 φ + π / 4 = π / 2 +

Let f (x) = A.B, where vector a = (m, cos2x), B = (1 + sin2x, 1), X ∈ R, and y = f (x) pass through a point (π, 2)

Let the function f (x) = A.B, where vector a = (m, cos2x), B = (1 + sin2x, 1), X ∈ R, and y = f (x) pass through point (quarter π, 2). (1) find the value of function m; (2) find the minimum value of function f (x) and the set of values of X at this time