α + β sin (α + β) - Sin 2?

α + β sin (α + β) - Sin 2?

sin(2α+β)-2sinαcos(α+β)=sin2αcosβ+sinβ2cos2α-2sinα(cosαcosβ-sinαsinβ)=sin2αcosβ+sinβ2cos2α-2sinαcosαcosβ+2sinαsinαsinβ=sin2αcosβ+sinβ2cos2α-sin2αcosβ+2sinαsinαsinβ=sin...

It is proved that (2-2sin (α + 3 / 4 π) cos (α + π / 4)) / cos ^ 4 α - Sin ^ 4 α = 1 = Tan α / 1-tana

Note: the last equal sign of your question should be a plus sign. In the proof process, because you can't type out the square, you can use ★ instead, 2. Proof: sin (α + 3 π / 4) * cos (α + π / 4) = sin [π / 2 + (α + π / 4)] * cos (α + π / 4)] * cos (α + π / 4) = cos (α + π / 4) = cos (α + π / 4) * cos (α + π / 4) = cos (α + π / 4) = cos (α + π / 4) = cos (α + π / 4) = cos (α + π / 4) = cos (α + 2 / 2 * (COS α - sin α)] = 1 / 2 (COS α - sin α), 2-2sin (α + 3 π / 4) * cos (α + π π π (α + π π / 4)] cos (α + π π π π π / 4/ 4) = 2 - (COS α - sin α)? = 2 - (COS ★ α + sin ★ α - 2cos α sin α) = 1 + 2cos α sin α = cos ★ α + sin ★ α + 2cos α sin α = (COS α + sin α) ★; cos^4α-sin^4α=(cos★α)★-(sin★α)★=(cos★α-sin★α)(cos★α+sin★α)=(cosα-sinα)(cosα+sinα)； [2-2sin(α+3π/4)*cos(α+π/4)]/(cos^4α-sin^4α)=(cosα+sinα)★/(cosα-sinα)(cosα+sinα)=(cosα+sinα)/(cosα-sinα)=(1+tanα)/(1-tanα)

It is proved that the following identities 1, cos ^ 2 α + 2Sin ^ 2 α + sin ^ 2 α Tan ^ 2 α = 1 / cos ^ 2 α Prove the following identities: 1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α 2、cos^2α(2+tanα)(1+2tanα)=2+5sinαcosα 3、(1+tan^2A)/(1+cot^2A)=[(1-tanA)/(1-cotA)]^2 4、(tanA-tanB)/(cotB-cotA)=tanB/cotA

1.1+tan^α=(cos^α+sin^α)/cos^α=1/cos^α,
ν left = cos ^ α + sin ^ α + sin ^ α (1 + Tan ^ α)
=1+sin^α/cos^α
=1+tan^α
=1 / cos ^ α = right
2. Left = (2cos α + sin α) (COS α + 2Sin α)
=2cos^α+5sinαcosα+2sin^α
=2 + 5sin α cos α = right
3. Left = (1 / cos ^ a) / (1 / sin ^ a) = (Sina / COSA) ^ = Tan ^ a,
Right = {[(COSA Sina) / cosa] / [(Sina COSA) / Sina]}^
=(-sinA/cosA)^
=Tan ^ a = left
4. Left = (Tana tanb) / [1 / tana-1 / tanb]
=-tanAtanB,
Right = tanatanb ≠ left
Please check question 4

It is proved that 1 + sin α - cos α 1+sinα+cosα＝tanα 2．

It is proved that the left side of the original formula = (1 − cos α) + sin α
(1+cosα)+sinα=2sin2α
2+2sinα
2cosα
Two
2cos2α
2+2sinα
2cosα
2=sinα
2(sinα
2+cosα
2)
cosα
2(sinα
2+cosα
2)=sinα
Two
cosα
2=tanα
2 = right
So the original is established

It is proved that 1 + 2 sin α cos α / cos square α - Sin square α = Tan (π / 4 - α) The school is about to start, and the old homework has not been finished

I guess the landlord won't ask questions. No one knows what a numerator is. You should learn to use brackets to enclose the relevant numerator and denominator
If so
(1 + 2sinacosa) / (COS ^ 2A - Sin ^ 2a) then
It = (sin ^ 2A + cos ^ 2A - 2sinacosa) / (COS ^ 2A - Sin ^ 2a)
= (sina-cosa)^2/(cosa-sina)(cosa+sina)
= (cosa-sina)/(cosa+sina)
= (1-tana)/(1+tana)
According to the formula Tan (PI / 4-A) = (tanpi / 4 - Tana) / (1 + tanpi / 4 Tana)
= (1-tana)/(1+tana)
So it is proved
Remember to use brackets

It is proved that 1 + sin α - cos α 1+sinα+cosα＝tanα 2．

It is proved that the left side of the original formula = (1 − cos α) + sin α
(1+cosα)+sinα=2sin2α
2+2sinα
2cosα
Two
2cos2α
2+2sinα
2cosα
2=sinα
2(sinα
2+cosα
2)
cosα
2(sinα
2+cosα
2)=sinα
Two
cosα
2=tanα
2 = right
So the original is established

Sin 2A + 2Sin square A / 1 + Tan a = 2Sin a cos a

Certificate:
sin2a=2sina*cosa
So the left side of the original
=(2sina*cosa+2sina*sina)/(1+sina/cosa)
=Cosa * (2sina * cosa + 2sina * Sina) / (Sina + COSA), that is, the numerator and denominator multiply the same cosa
=2 cosa Sina (CISA + Sina) / (Sina + COSA)
=2sinacosa
=Right
Of course, this requires that 1 + Tana is not equal to 0

If Tan α = - 1 / 2, then the value of 2Sin α cos α / sin α - cos α is the help of the great gods

The denominator of 2Sin α cos α (/ sin α - cos α) is divided by cos α = 2Sin α / cos α (/ sin α / cos α - 1) = 2tan α / (Tan α - 1) = 2 × (- 1 / 2) / [(- 1 / 2) - 1] = - 1 / (- 3 / 4) = 4 / 3 friend, if my answer helps you, please spare a second or two out of every favor,

If sin ^ 2 α - 2Sin α cos α + 2cos α ^ 2 = 1 / 2, Tan α=

Sin? A-2sinacosa + 2cos? A = (1 / 2) sin? A + (1 / 2) cos? Asin? A-4sinacosa + 3cos? A = 0 [both sides divided by cos? A] tan? A-4tana + 3 = 0 (tana-3) (tana-1) = 0, then Tana = 1 or Tana = 3

Tan α = 1 / 2, find (sin ^ 2 α + cos ^ 2 α) / 2cos ^ 2 α + 2Sin α cos α RT... evaluation

(sin^2α+cos^2α)/(2cos^2α+2sinαcosα)
=(1+tan^2α)/(2+2tan^2α)
=(1+1/4)/(2+2*1/4)
=15/8