F (x) = - root 2 sin 2 x + root 2 SiN x cosx

F (x) = - root 2 sin 2 x + root 2 SiN x cosx

By using the formula of double angle and auxiliary angle, f (x) = - root 2Sin? X + radical 2sinxcosx = - (√ 2 / 2) (1-cos2x) + (√ 2 / 2) sin2x = (√ 2 / 2) sin2x + (√ 2 / 2) cos2x - √ 2 / 2 = sin2x * cos (π / 4) + cos2x * sin (π / 4) - √ 2 / 2 = sin (2x + π / 4) -√ 2 / 2

F (x) = [2Sin (x + π / 3) + SiNx] cosx radical 3sin2x, simplify

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin2x
=[2sinx+√3cosx]cosx-√3sin2x
=sin2x+(√3/2）（1+cos2x)-√3sin2x
=(1-√3）sin2x+(√3/2）cos2x+√3/2
=Numerous!
Please check the title

It is proved that: (1-sin2x) / cos2x = Tan (π / 4-x)

By (1-sin2x) / cos2x
=（1-cos（π/2-2x）/sin（π/2-2x）
=[1-cos2（π/4-x）]/sin2（π/4-x）
=[1-1+2sin²（π/4-x)]/2sin(π/4-x)cos(π/4-x)
=sin(π/4-x)/cos(π/4-x)
=tan(π/4-x)

Let f (x) = TaNx (1 + sin2x + cos2x), (1) Tan (α + π / 4) = 2, find f (α) Let f (x) = TaNx (1 + sin2x + cos2x), (1) Tan (α + π / 4) = 2, find f (α) (2) If f (β) = 2, β∈ [0,2 π], find β

f(x)=tanx(1+sin2x+cos2x)
=tanx(1+2sinxcosx+2cos²x-1)
=tanx(2sinxcosx+2cos²x)
=2sin²x+2sinxcosx
（1） tan(α+π/4)=2
tan(α+π/4)
=(tanα+1)/(1-tanα)=2
∴tanα=1/3
f(α)=2sin²α+2sinαcosα
=（2sin²α+2sinαcosα）/1
=（2sin²α+2sinαcosα）/(sin²α+cos²α)
The denominator of the numerator is the same as the division of COS 2 α
=(2tan²α+2tanα)/(tan²α+1)
=4/5
（2）
f(β)=2,β∈[0,2π],
2sin²β+2sinβcosβ=2
sin²β+sinβcosβ=1
∵sin²β+cos²β=1
∴sinβcosβ=cos²β
∴cosβ(sinβ-cosβ)=0
β=π/2,3π/2
Or β = π / 4,5 π / 4

Simplify (1 + sin2x-cos2x) / (1 + sin2x + cos2x)

（1+sin2x-cos2x)/(1+sin2x+cos2x)
=(2sin^2x +2sinxcosx)/(2cos^2x+2sinxcosx)
=2sinx(sinx+cosx)/2cosx(cosx+sinx)
=tanx

Verification: (sin2x / 1 + cos2x) × (cosx / 1 + cosx) = Tan (x / 2) How to translate this tan (x / 2)?

tan(x/2)=sin(x/2)/cos(x/2)=[2sin(x/2)cos(x/2)]/[2cos^2(x/2)]
=sinx/(1+cosx)

Given sin2x = m.cos2x = n. find the value of Tan (x + π / 4)

tan(x+π/4)
=(tanx+1)/(1-tanx)
Cosx
=(sinx+cosx)/(cosx-sinx)
sin2x=2sinxcosx
1+2sinxcosx
=sin²x+cos²x+2sinxocsx
=(sinx+cosx)²
=1+m
cos2x
=cos²x-sin²x
=(cosx-sinx)(cosx+sinx)
=n
∴(sinx+cosx)²/[(cosx-sinx)(cosx+sinx)]
=(sinx+cosx)/(cosx-sinx)
=(1+m)/n
therefore
tan(x+π/4)
=(1+m)/n

Proof (sin4x) / (1 + cos4x) * (cos2x) / (1 + cos2x) * (cosx) / (1 + cosx) = Tan (x / 2)

For this problem, raise the power. Cos2x = 2 (cosx) ^ 2-1, use 3 times
(sin4x)/(1+cos4x)=2sin2xcos2x/[2(cos2x)^2-1+1]=sin2x/(cos2x)
The last two are the same as the first fraction, and then half - angle
[sin2x/(cos2x)]*(cos2x)/(1+cos2x)
=sinx/cosx
[sinx/cosx])*(cosx)/(1+cosx)
=[sin(x/2)]/[cos(x/2)]
=tan(x/2)

It is known that SiN x / 2 -- 2cos X / 2 = 0. (1) find the value of Tan X. (2) find the value of cos2x / (√ 2cos (π / 4 + x) * SiNx)

(1) Because SiN x / 2 = 2cos X / 2, TaNx / 2 = (SiN x / 2) / (COS X / 2) = 2, so TaNx = [2tan (x / 2)] / [1-tan ^ 2 (x / 2)] = - 4 / 3 (2) primitive = [cos ^ 2x-sin ^ 2x] / [(cosx SiNx) * SiNx] = (cosx SiNx) (cosx + SiNx) / (cosx SiNx) * SiNx = (cosx + SiNx) / SiNx = Cotx

Tan (x + π / 4) = - 1 / 2, calculate the value of cos2x

tan(x+π/4)=2tanx/(1-tanx)=-1/2
∴ tanx=-1.
∴tan²x=1
cos2x=cos²x-sin²x=1-tan²x=0
Or, TaNx = - 1
X = k π - π / 4
∴ 2x=2kπ-π/2
So cos2x = 0
Very detailed students, give the best