It is proved that the 2nd power of Tan α - sin α = the 2nd power of sin α * the 2nd power α of COS

It is proved that the 2nd power of Tan α - sin α = the 2nd power of sin α * the 2nd power α of COS

(Tan @) square - (sin @) square = (Tan @) square (sin @) square
Let x = (sin @) square
Then (COS @) square = 1-x,
(Tan @) square = x / (1-x)
.
So left = right

1. It is proved that Tan α / 2 = sin α / (1 + cos α) = (1-cos α) / sin α 2. Find the minimum positive period, increasing interval and maximum value of the following functions: (1) y = sin2xcos2x; (2) y = 2cos (square) x / 2 + 1; 3) y = radical 3 cos4x + sin4x

1. It is proved that Tan (α / 2) = sin α / (1 + cos α) = (1-cos α) / sin α
tan(α/2)
=sin(α/2) / cos(α/2)
=sin²(α/2) / sin(α/2)cos(α/2)
=2sin²(α/2) / 2sin(α/2)cos(α/2)
=2Sin 2 (α / 2) / sin α
=[2-2cos²(α/2)] / sinα
=[1+1-2cos²(α/2)] / sinα
=(1-cos α) / sin α [cosine double angle formula]
=(1-cosα)(1+cosα) / sinα(1+cosα)
=sin²α / sinα(1+cosα)
=sinα / (1+cosα)
2. Find the minimum positive period, increasing interval and maximum value of the following functions:
(1)y=sin2xcos2x
=(1/2)*2*sin2xcos2x
=(1/2)sin4x
Minimum positive period: 2 π / 4 = π / 2
Increasing interval: [- π / 8 + K π / 2, π / 8 + K π / 2], K ∈ Z
Maximum: 1 / 2
(2)y=2cos²x/2+1
=2cos²x/2-1+2
=cosx+2
Minimum positive period: 2 π / 1 = 2 π
Increasing interval: [- π + 2K π, 2K π], K ∈ Z
Maximum: 3
(3)y=√3 cos4x+sin4x
=2 [(√3/2)cos4x+(1/2)sin4x]
=2 sin(4x+π/3)
Minimum positive period: 2 π / 4 = π / 2
Increasing interval: [- 5 π / 24 + K π / 2, π / 24 + K π / 2], K ∈ Z
Maximum: 2
Hope to adopt~~~~
What is not clear can be asked~~~

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

Given Tana = 2, why is Tana / 2 equal to (- 1 + Radix 5) / 2?

tanA=2
tan[2×(A/2)]=2
2tan(A/2)/[1-tan²(A/2)]=2
tan(A/2)=1-tan²(A/2)
tan²(A/2)+tan(A/2)-1=0
tan²(A/2)+2×(1/2)×tan(A/2)+(1/2)²-(1/2)²-1=0
[tan(A/2)+1/2]²=5/4
tan(A/2)+1/2=±(√5)/2
tan(A/2)=-1/2±(√5)/2
tan(A/2)=(-1±√5)/2
It can be seen that the landlord's problem is not very accurate
It should be: Tan (A / 2) = - 1 ± √ 5) / 2

Under the radical sign (1 + sin α / 1-sin α) - under the radical sign (1-sin α / 1 + sin α) = - 2, Tana is the set of angles that determine the equality

Firstly, according to the definition domain: 1-sin α ≠ 0; 1 + sin α ≠ 0
∴sinα ≠ ±1
Second, according to: the root sign [(1 + sin α) / (1-sin α)] - the radical sign [(1-sin α) / (1 + sin α)] = - 2 root sign Tan ^ 2A
Radical [(1 + sin α) ^ 2 / (1-sin ^ 2 α)] - radical [(1-sin α) ^ 2 / (1-sin ^ 2 α)] = - 2|tana|
(1+sinα) / |cosα| - (1-sinα) / |cosα| = -2|tana|
(1+sinα-1+sinα) / |cosα| = -2|tana|
2sinα / |cosα| = -2|tana|
sinα=-|tanacosα| = -|sinα|
That is, sin α ≤ 0
In addition, sin α ≠± 1
∴ -1 ＜ sinα ≤ 0
ν α∈ [(2k π - π, 2K π - π / 2), (2k π - π / 2,2k π], where k ∈ Z

Sin (2x-3 π / 4) = 1 / radical 2, X is acute angle

Because x is an acute angle
Sin (2x-3 π / 4) = 1 / radical 2
=Radical 2 / 2
So 2x-3 π / 4 = π / 4
2x=1
x=1/2
Wish you all the best!

The angle α is known to be acute and sin2 α - sin α cos α - 2cos2 α = 0 (I) find the value of Tan α; (II) find sin (α - π) 3)．

(1) From sin2 α - sin α cos α - 2cos2 α = 0, (sin α - 2cos α) (sin α + cos α) = 0
∵ angle α is an acute angle, ᙽ sin α > 0, cos α > 0, sin α - 2cos α = 0, so tan α = 2
(2) From (I), sin α = 2
Five
5，cosα＝
Five
Five
sin(α−π
3)＝sinαcosπ
3−cosαsinπ
Three
=2
Five
5×1
2−
Five
5 x
Three
2＝2
5−
Fifteen
10．

Given that α is an acute angle, Tan α = 3, find sin α - cos α The value of sin α + 2cos α

∵ α is an acute angle,
∴cosα≠0，
∴sinα−cosα
sinα+2cosα＝sinα
cosα−1
sinα
cosα+2＝tanα−1
tanα+2＝3−1
3+2＝2
5．

It is known that α is an acute angle and sin ^ 2 α + sin α cos α - 2cos ^ 2 = 0. (1) find the value of Tan α. (2) find sin [α - (π / 3)]

(1)
sin^2α＋sinαcosα-2cos^2=0
(sinα+2cosα)(sinα-cosα)=0
Because alpha is an acute angle
therefore
sinα=cosα
tanα=1
therefore
α=π/4
(2)
sin(α-π/3)
=sinαcosπ/3-cosαsinπ/3
=√2/2*1/2-√2/2*√3/2
=(√2-√6)/4

It is known that sin α is an acute angle and sin α ^ 2-sin α cos α - 2cos α ^ 2 = 0. (1) find the value of Tan α. (2) find sin (α - π / 3)

By the title
tanα^2-tanα-2=0
Tan α = 2 or
tanα=-1
From the question Tan α = 2
Sin α = root 5, cos α = two fifths root sign five
Sin (α - π / 3) = root 5 * half root 3-0.5 * two fifths root 5
=Five tenths fifteen minus two roots five