If sin θ + cos θ = √ 2, then Tan (θ + π / 3)

If sin θ + cos θ = √ 2, then Tan (θ + π / 3)

θ= π/4
The final answer is (√ 3 1) / (√ 3-1)

Let θ be an acute angle, (1-tan θ) / (1 + Tan θ) = 3 + 2 √ 2, then sin θ cos θ =? I've already crawled, but they can't understand the formula and formula they use, So I hope we can have a detailed formula,

(cos - sin)/(cos+sin) = 3+2√2
Square of both sides (1-2sin θ cos θ) / (1 + 2Sin θ cos θ) = 17 + 12 √ 2
（1-2sinθcosθ) =(1+2sinθcosθ)(17 +12√2)
2sinθcosθ(18 +12√2)= -16-12√2
sinθcosθ = -(4+3√2)/(9+6√2) = -√2/3

Sin α + sin β = 2 / 3, cos α + cos β = 3 / 4, find Tan (α - β) / 2

Write it roughly according to the formula of double angle:
tan(α-β)=2tan(α/2-β/2)/1-tan^2(α/2-β/2）
So only Tan (α - β) is required
Tan (α - β) = sin (α - β) / cos (α - β) is equal to
=(sinαcosβ-cosαsinβ)/(cosαcosβ+sinαsinβ)

Half angle proof of Tan How to prove Tan (α / 2) = sin α / (1 + cos α) = (1-cos α) / sin α

tan(a/2)=sin(a/2)/cos(a/2)
Multiply by 2cos (A / 2),
＝2sin(a/2)cos(a/2)/2cos(a/2)^2
=sina/(1+cosa).
The following equation can be proved directly

What is Tan ^ 2A / 2 equal to Why?

=(1+cosa)/(1-cosa)
[Tan (A / 2)] ^ 2 where
Tan (A / 2) = sin (A / 2) / cos (A / 2) multiply cos (A / 2)
Cos (A / 2) sin (A / 2) / [cos (A / 2)] ^ 2 = Sina / 2 [cos (A / 2)] ^ 2 = Sina / (cosa-1)
Then [Tan (A / 2)] ^ 2 = (Sina) ^ 2 / (cosa-1) ^ 2 = (1-cosa) (1 + COSA) / (1-cosa) ^ 2
=(1+cosa)/(1-cosa)

What is Tan (2a) equal to? How can you prove it?

tan（2a)=2tan(a)/(1-tan(a)*tan(a))
Formula is the content of high school, proving that high school can learn naturally
If you don't learn, it doesn't matter if you don't know

What is the reduction of (1 + Tan ^ 2a) * cos ^ 2A?

（1+tan^2a)*cos^2a
=cos^2a+(sin^2a/cos^2a)*cos^2a
=cos^2a+sin^2a
=1

Tan half x is equal to two. Find TaNx Tan half x = 2, find Tan (x + quarter π)

TaNx = (2tan half x) / (1-tan half x squared), the answer is minus four thirds
From the above formula, and TaNx = - 4 / 3, Tan (quartile) = 1, so the answer is minus one seventh

About TaNx = 2, Tan (Y-X) = - 1 / 7, Tan (y-2x) =? I am anxious

Let Y-X = t
tan(y-x)=-1/7
=>tant=-1/7
tan(y-2x)=tan(t+x)
=(tant+tanx)/(1-tant*tanx)
=(2-1/7)/(1+2/7)
=13/9

If TaNx = 1 / 2, Tan (X-Y) = = - 2 / 5, then Tan (2x-y)

tan(2x-y)
=[tanx+tan(x-y)]/[1-tanxtan(x-y)]
=(1/10)/(1+1/5）
=1/12