If Tan α = 2, then the square of sin α + sin α * cos α=

If Tan α = 2, then the square of sin α + sin α * cos α=

(Sina) 2 + sinacosa = sin? 2A + sinacosa = (sin? 2A + sinacosa) / (sin? 2A + cos? A) = (tan? 2A + Tana) / (tan? 2A + 1) = (2? 2) / (2? 2 + 1) = 6 / 5 = 6 / 5

The results were as follows: sin α (1 + Tan α) + cos α (1 + 1) tanα）=1 sinα+1 cosα．

It is proved that the left side = sin α + sin2 α
cosα+cos α+cos2α
sinα
=sin2α+cos2α
sinα+sin2α+cos2α
cosα
=1
sinα+1
Cos α = right
That is, the original equation holds

It was proved that Tan half α = sin α / 1 + cos α = 1-cos α / sin α

tan(α/2)=sin(α/2)/cos(α/2)
(1)=[2sin(α/2)cos(α/2)] / [2cos²(α/2)]
=sinα/[2cos²(α/2)-1 +1]
=sinα/(cosα+1)
(2)=[2sin²(α/2)] / [2sin(α/2)cos(α/2)]
=[1+(2sin²(α/2)-1)] /sinα
=(1-cosα)/sinα
(they are the inverse of the formula of sine and cosine double angle)

It was proved that sin ^ 2 / (sin COS) - (sin + COS) / (Tan ^ 2 - 1) = sin + cos

sin^2/(sin-cos) - (sin+cos)/(tan^2 -1)
=sin^2/(sin-cos) -(sin+cos)/[(sin^2/cos^2)-1]
=sin^2/(sin-cos) -(sin+cos)cos^2/(sin^2-cos^2)
=sin^2/(sin-cos) -cos^2/(sin-cos)
(sin^2-cos^2)/(sin-cos)
=sin-cos

It was proved that Tan (α / 2) = (sin α) / (1 + cos α) Please write down the detailed process

On the right = sin α / (1 + cos α) = 2Sin (α / 2) cos (α / 2) / 2cos? 2 = sin (α / 2) / cos (α / 2) = Tan (α / 2) = left

1. Verification: sin θ (1 + Tan θ) + cos θ (1 + 1 / Tan θ) = 1 / Tan θ + 1 / cos θ 2. Given Tana = - 1 / 3, find (4sina-2cosa) / (5cosa + 3sina) 3. Simplification: radical (1-2sin10 ° cos10 °) / (sin10 ° - radical (1-sin 2 10 °)) =1 / Tan θ + 1 / cos θ = 1 / sin θ + 1 / cos θ

sinθ(1+tanθ)+cosθ(1+1/tanθ)
=sinθ(1+sinθ/cosθ)+cosθ(1+cosθ/sinθ)
=sinθ(cosθ+sinθ)/cosθ+cosθ(sinθ+cosθ)/sinθ
=(sinθ+cosθ)(sinθ/cosθ+cosθ/sinθ)
=(sinθ+cosθ)(sin²θ+cos²θ)/(cosθsinθ)
=(sinθ+cosθ)/(cosθsinθ)
=1/sinθ+1/cosθ
2.∵tana=-1/3 ∴sina/cosa=-1/3
∴cosa=-3sina
∴（4sina-2cosa)/(5cosa+3sina)
=(4sina+6sina)/(-10sina+3sina)
=-10/7
3. Simplification
√（1-2sin10°cos10°）/（sin10°-√(1-sin²10°））
=√(sin²10º+cos²10º-2sin10°cos10°)/sin10°-√cos²10°）
=√(cos10º-sin10º)/(sin10º-cos10º)
=|cos10º-sin10º|/(sin10º-cos10º)
=(cos10º-sin10º)/(sin10º-cos10º) (cos10º>sin10º)
=-1
∴tana=sina/cosa=-12/5

If this is the root of the equation 2 x sin a, then it is a set

sin a

It is known that @ is an acute angle, and sin^ 2@-sin@cos@-2cos ^2 = 0, find the value of Tan @ and sin (@ - 3.14 / 3)

sin²a-sinacos-2cos²a=0
Divide both sides of the equation by cos? A at the same time
tan²a-tana-2=0
(tana-2)(tana+1)=0
Because Tana + 1 > 0
So tana-2 = 0, Tana = 2
Sina = 2 / (root 5), cosa = 1 / (root 5)
sin(a-pi/3)
=sinacos(pi/3)-cosasin(pi/3)
=1 / (root 5) - (root 3 / 2) (1 / (root 5))
=(1 - (Radix 3) / 2) / (Radix 5)

Given that α is an acute angle and Tan α = 2, calculate the value of (sin α - 2) \ (2cos α + sin α)

The numerator and denominator of the original formula are divided by cos α
The original formula = [Tan α - (2 / cos α)] / (2 + Tan α)
And ∵ 1 / cos ∵ α = 1 + Tan ∵ α
∴cos²α=1/5
And ∵ α is an acute angle
∴cosα=1/√5
The original formula = (1 - √ 5) / 2
This type of topic is often divided by the top and the bottom, so it is important to master the method,

If Tan α = 2, then sin square α + sin α cos α - 2cos square α=

All divided by the square of (ConA), the result is Tana, and then bring it in and get the result directly